Note On Gauss's Law

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Naveen B / Chandigarh

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ELECTROSTATICS-3 [Electric flux, Gauss's Law and its applications] ELECTRIC FLUX Electric flux is a measure of 'flow' of electric field through a surface. It is equal to the product of an area element and the perpendicular component of E integrated over a surface. (1) Flux of electric field E through any area A is defined as. = E.AcosO or (2) In case of variable electric field or curved area. = f E. dA 2 (3) It's S.I. Unit is (Volt x m) or 0 (4) For a closed body outward flux is taken to be positive while inward flux is taken to be negative. Negative flux Zero flux Positive flux Exercise:- 1. Consider a uniform electric field E = 3 x 103 i N/ C. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its plane makes a 600 angle with the x-axis? 2. What is the net flux of the uniform electric field of Exercise 1 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes? 3. A pyramid with horizontal square base, 6.00 m on each side, and a height of 4.00 m is placed in vertical electric field of 52.0 N/ C. Calculate the total electric flux through the pyramid's four slanted surfaces.[ +1.87 kN•m2/Cl 4. A cone with base radius R and height h is located on a horizontal table. A horizontal uniform field E penetrates the cone, as shown in Figure. Determine the electric flux that enters the left-hand side of the cone. [ERh] 5. A non-uniform electric field given by E = 3xi +4j pierces the Gaussian cube shown in figure. (E is in newtons per coulomb and x is in meters.) What is the electric flux through the right face, the left face, and the top face? [-12 N-m2/C, 36N- m2/C,16 N-m2/C] 6. An electric fleld given by E = 4i - 3(y2 + 2)j N/ C pierces a Gaussian cube of edge length 2.0 m and positioned as shown in figure same as for exercise 5. (The magnitude E is in newtons per coulomb and the position x is in meters.) What is the electric flux through the (a) top face, (b) bottom face, (c) left face, and (d) back face? (e) What is the net electric flux through the cube? Y 1.0m 3.0m Gaussian surface
Gauss's Law and it's Application Formal proof of Gauss's Law*** One way of deriving Gauss's law involves solid angles. Consider a spherical surface of radius r containing an area element AA . A closed surface of arbitrary shape surrounds a point charge q. The net electric flux through the surface is independent of the shape of the surface. The solid angle AQ subtended at the center of the sphere by this element is defined to be 2 From this equation, we see that AQhas no dimensions because AA and r 2 both have dimensions L2. The dimensionless unit of a solid angle is the steradian. Because the surface area of a sphere is 2 41tr 47tr2, the total solid angle subtended by the sphere is Q = = 41T steradians 2 Now consider a point charge q surrounded by a closed surface of arbitrary shape. The total electric flux through this surface can be obtained by evaluating E.AA for each small area element AA and summing over all elements. The flux through each element is AA cose = = (Ecos O)AA = kq 2 where ris the distance from the charge to the area element, 0 is the angle between the electric field E and AA for the element. In Figure, we see that the projection of the area element perpendicular to the radius vector is AA cos0. Thus, the quantity (AA cos0)/r2 is equal to the solid angle AQ that the surface element AA subtends at the charge q. We also see that AQis equal to the solid angle subtended by the area element of a spherical surface of radius r. Because the total solid angle at a point is 47t steradians, the total flux through the closed surface is dA cos0 "E = kq f 2 kq f dQ = 411 k q Thus we have derived Gauss's law. Note that this result is independent of the shape of the closed surface and independent of the position of the charge within the surface. OR (1) According to this law, the total flux linked with a closed surface called Gaussian surface. (The surface need not be a real physical surface, it can also be an hypothetical one) is (1/80) times the charge enclosed by the closed surface i.e., = f • dA = (Qenc) o (2) Electric field in f E.dA is complete electric field. It may be partly due to charge with in the surface and partly due to charge outside the surface. However if there is no charge enclosed in the Gaussian surface, then f E. dA = 0 . (3) The electric field E is resulting from all charge, both those inside and those outside the Gaussian surface.
ELECTROSTATICS-3 [Electric flux, Gauss's Law and its applications] ELECTRIC FLUX Electric flux is a measure of 'flow' of electric field through a surface. It is equal to the product of an area element and the perpendicular component of E integrated over a surface. (1) Flux of electric field E through any area A is defined as. = E.AcosO or (2) In case of variable electric field or curved area. = f E. dA 2 (3) It's S.I. Unit is (Volt x m) or 0 (4) For a closed body outward flux is taken to be positive while inward flux is taken to be negative. Negative flux Zero flux Positive flux Exercise:- 1. Consider a uniform electric field E = 3 x 103 i N/ C. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its plane makes a 600 angle with the x-axis? 2. What is the net flux of the uniform electric field of Exercise 1 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes? 3. A pyramid with horizontal square base, 6.00 m on each side, and a height of 4.00 m is placed in vertical electric field of 52.0 N/ C. Calculate the total electric flux through the pyramid's four slanted surfaces.[ +1.87 kN•m2/Cl 4. A cone with base radius R and height h is located on a horizontal table. A horizontal uniform field E penetrates the cone, as shown in Figure. Determine the electric flux that enters the left-hand side of the cone. [ERh] 5. A non-uniform electric field given by E = 3xi +4j pierces the Gaussian cube shown in figure. (E is in newtons per coulomb and x is in meters.) What is the electric flux through the right face, the left face, and the top face? [-12 N-m2/C, 36N- m2/C,16 N-m2/C] 6. An electric fleld given by E = 4i - 3(y2 + 2)j N/ C pierces a Gaussian cube of edge length 2.0 m and positioned as shown in figure same as for exercise 5. (The magnitude E is in newtons per coulomb and the position x is in meters.) What is the electric flux through the (a) top face, (b) bottom face, (c) left face, and (d) back face? (e) What is the net electric flux through the cube? Y 1.0m 3.0m Gaussian surface
(Keep in mind, the electric field due to a charge outside the Gaussian surface contributes zero net flux through the surface, because as many lines due to that charge enter the surface as leave it). Flux from surface Sl = + Flux from surface S2 — , and flux from flux from surface S4 = O Application of Gauss's law: See flux emergence in the following cases (1) If a dipole is enclosed by a surface •.•Qenc (2) The net charge Qenc is the algebraic sum of all the enclosed positive, and negative charges. If Qenc is positive the net flux is outward; if Qenc is negative, the net flux is inward. 1 O (3) If a closed body (not enclosing any charge) is placed in an electric field (either uniform or non- uniform) total flux linked with it will be zero x a a (4) If a hemispherical body surface calculated as follows "Qurved + "Circular "Qzrved = —"arcular - (E x cosi800) = +ÆR2E Sphere is placed in n z uniform (B) "tn — "out = "T = O electric field then flux linked with the curved (5) If a hemispherical body is placed in non-uniform electric field as shown below, then flux linked with the circular surface calculated as follows "Circular = —"Cürved - (E x cos00) "Ocular -2ÆR2E n
Gauss's Law and it's Application Formal proof of Gauss's Law*** One way of deriving Gauss's law involves solid angles. Consider a spherical surface of radius r containing an area element AA . A closed surface of arbitrary shape surrounds a point charge q. The net electric flux through the surface is independent of the shape of the surface. The solid angle AQ subtended at the center of the sphere by this element is defined to be 2 From this equation, we see that AQhas no dimensions because AA and r 2 both have dimensions L2. The dimensionless unit of a solid angle is the steradian. Because the surface area of a sphere is 2 41tr 47tr2, the total solid angle subtended by the sphere is Q = = 41T steradians 2 Now consider a point charge q surrounded by a closed surface of arbitrary shape. The total electric flux through this surface can be obtained by evaluating E.AA for each small area element AA and summing over all elements. The flux through each element is AA cose = = (Ecos O)AA = kq 2 where ris the distance from the charge to the area element, 0 is the angle between the electric field E and AA for the element. In Figure, we see that the projection of the area element perpendicular to the radius vector is AA cos0. Thus, the quantity (AA cos0)/r2 is equal to the solid angle AQ that the surface element AA subtends at the charge q. We also see that AQis equal to the solid angle subtended by the area element of a spherical surface of radius r. Because the total solid angle at a point is 47t steradians, the total flux through the closed surface is dA cos0 "E = kq f 2 kq f dQ = 411 k q Thus we have derived Gauss's law. Note that this result is independent of the shape of the closed surface and independent of the position of the charge within the surface. OR (1) According to this law, the total flux linked with a closed surface called Gaussian surface. (The surface need not be a real physical surface, it can also be an hypothetical one) is (1/80) times the charge enclosed by the closed surface i.e., = f • dA = (Qenc) o (2) Electric field in f E.dA is complete electric field. It may be partly due to charge with in the surface and partly due to charge outside the surface. However if there is no charge enclosed in the Gaussian surface, then f E. dA = 0 . (3) The electric field E is resulting from all charge, both those inside and those outside the Gaussian surface.
(6) If charge is kept at the centre of cube 1 "total = — • (Q) "face 680 (7) If charge is kept at the centre of a face: First we Gaussian surface (an identical imaginary cube) 4 (A) should enclosed O Q (B) (B) the charge by assuming a Total flux emerges from the system (Two cubes) Flux from given cube (i.e. from 5 face only) (8) If a charge is kept at the corner of a cube (A) "total — 280 For enclosing the charge seven more cubes are required so total flux from the 8 cube system is "-r . Flux from given cube . Flux from one face opposite to change, of the given cube cube 880 Q/ 880 Q (Because only three faces are seen). "face — 3 24 (9) A long straight wire of charge density penetrates a hollow body as shown. The flux emerges from the body is = x (Length of the wire inside the body) Exercise:- 1. Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 x 103 Nm2/C. (a) What is the net charge inside the box? (b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not? 2. A point charge +10 PC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in figure. What is the magnitude of the electric flux through the square? 3. A point charge of 2.0 PC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface? 4. A point charge causes an electric flux of —1.0 x 103 Nm2/C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge. (a) If the radius of the Gaussian surface were doubled, how 5 cm 10 cm much flux would pass through the surface? (b) What is the value of the point charge?
(Keep in mind, the electric field due to a charge outside the Gaussian surface contributes zero net flux through the surface, because as many lines due to that charge enter the surface as leave it). Flux from surface Sl = + Flux from surface S2 — , and flux from flux from surface S4 = O Application of Gauss's law: See flux emergence in the following cases (1) If a dipole is enclosed by a surface •.•Qenc (2) The net charge Qenc is the algebraic sum of all the enclosed positive, and negative charges. If Qenc is positive the net flux is outward; if Qenc is negative, the net flux is inward. 1 O (3) If a closed body (not enclosing any charge) is placed in an electric field (either uniform or non- uniform) total flux linked with it will be zero x a a (4) If a hemispherical body surface calculated as follows "Qurved + "Circular "Qzrved = —"arcular - (E x cosi800) = +ÆR2E Sphere is placed in n z uniform (B) "tn — "out = "T = O electric field then flux linked with the curved (5) If a hemispherical body is placed in non-uniform electric field as shown below, then flux linked with the circular surface calculated as follows "Circular = —"Cürved - (E x cos00) "Ocular -2ÆR2E n
5. The electric field components in figure are Ex = axl/2, Ey = Ez = O, in which u = 800 N/ C ml/2 Calculate (a) the flux through the cube, and (b) the charge within the cube. Assume that a = 0.1 m. 6. An electric field is uniform, and in the positive x direction for positive x, and uniform with the same magnitude but in the negative x direction for negative x. It is given that E = 200 N/ C for x > O and E = —200 N/ C for x < O. A right circular cylinder of length 20 cm and radius 5 cm has its centre at the origin and its axis along the x-axis so that one face is at x = +10 cm and the other is at x = —10 cm (figure). (a) What is the net outward flux through each flat _face? (b) What is the flux through the side of the cylinder? (c) What is the net outward flux through the cylinder? (d) What is the net charge inside the cylinder? 5 cm AS x = —10 cm 20 cm x cm 7. Four closed surfaces, Sl through S4, together with the charges -2Q, Q, and -Q are sketched in figure. 8. An infinitely long line charge having a uniform charge per unit length lies a distance d from point O as shown in figure. Determine the total electric flux through the surface of a sphere of radius R centered at O resulting from this line charge. Consider both cases, where R < d and R> d. 21 R2-d2 [O,
Electric field due to line of charge Consider an infinite line which has a linear charge density X. Using Gauss's law, let us find the electric field at a distance 'r' from the line charge. The cylindrical symmetry tells us that the field strength will be the same at all points at a fixed distance from the line. Thus, the field lines are directed radially outwards, perpendicular to the line charge. The appropriate choice of Gaussian surface is a cylinder of radius r and length L. On the flat end faces, S2 and S3, E is perpendicular to dS , which means flux is zero on them. On the curved surface Sl, E is parallel to dS , so that E.dS = Eds The charge enclosed by the cylinder is IL . Applying Gauss's law to the curved surface, we have EfdS = IT r L) Field Due to an Infinite Charged non-conducting Plane Sheet Let us consider a thin non-conducting charged plane sheet, infinite in extent, and having a surface charge density (charge per unit area) . Let P be a point, distance r from the sheet, at which the electric intensity is required. [Condition of infinite sheet:- r
(6) If charge is kept at the centre of cube 1 "total = — • (Q) "face 680 (7) If charge is kept at the centre of a face: First we Gaussian surface (an identical imaginary cube) 4 (A) should enclosed O Q (B) (B) the charge by assuming a Total flux emerges from the system (Two cubes) Flux from given cube (i.e. from 5 face only) (8) If a charge is kept at the corner of a cube (A) "total — 280 For enclosing the charge seven more cubes are required so total flux from the 8 cube system is "-r . Flux from given cube . Flux from one face opposite to change, of the given cube cube 880 Q/ 880 Q (Because only three faces are seen). "face — 3 24 (9) A long straight wire of charge density penetrates a hollow body as shown. The flux emerges from the body is = x (Length of the wire inside the body) Exercise:- 1. Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 x 103 Nm2/C. (a) What is the net charge inside the box? (b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not? 2. A point charge +10 PC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in figure. What is the magnitude of the electric flux through the square? 3. A point charge of 2.0 PC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface? 4. A point charge causes an electric flux of —1.0 x 103 Nm2/C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge. (a) If the radius of the Gaussian surface were doubled, how 5 cm 10 cm much flux would pass through the surface? (b) What is the value of the point charge?
1 Special case (i) If CA = 0B = o then Ep = ER = 0/80 and Eq = O (ii) If = o and 0B = — o then Ep = ER = O and EQ = 0/80 Electric Field Due to an Uniformly Charged Spherical Shell Using Gauss's law, let us find the intensity of the electric field due to a uniformly charged spherical shell or a solid conducting sphere Case I: At an external point. In case of an isolated charged spherical conductor any excess charge on it is distributed uniformly over its outer surface same as that of charged spherical shell or hollow sphere. Hence field lines must point radially outward. Also, the field strength will have the same value at all points on any imaginary spherical surface concentric with the imaginary spherical surface concentric with the charged conducting sphere or the shell. This symmetry leads us to choose the Gaussian surface to be a sphere of radius r>R. Any arbitrary element of area dS is parallel to the local field, so at all points on the surface. According to Gauss's law fÉ.dS=fEdS = r2) Q Therefore, E = 2 For points outside the charged conducting sphere or the charged spherical shell, the field is same as that of a point charge at the centre. Case 11 : At an Internal Point (r < R) The field still has the same symmetry and so we again pick a spherical Gaussian surface, but now with radius r less than R. Since the charge enclosed is zero, from Gauss's law we have E(4/T r 2) = 0 givesE = 0 Thus, we conclude that E = O at all points inside a uniformly charged conducting sphere or the charged spherical shell. Variation of E with the distance from the centre (r) Electric Field Due to an Uniformly Charged Sphere 2 4Tt€() R O r A Non-conducting uniformly charged sphere of radius R has a total charge Q uniformly distributed throughout its volume. Using the Gauss's Law, Let us find the field Case I: at an internal point (r < R) Positive charge Q is uniformly distributed throughout the volume of sphere of radius R. For finding the electric field at a distance (r < R) from the centre, we choose a spherical Gaussian surface of radius r, concentric with the charge distribution. From symmetry the magnitude E of the electric field has the same value at every point on the Gaussian surface, and the direction of is radial at every point on the surface. So, applying Gauss's law
Here f É.dS = ) 2 4 3 E(41T r 2) — Q r 4 3 3 3 where p is volume charge density 41TR3 Therefore orE= 3 3 The field increases linearly with distance form the centre. Case 11 : At an external point To find the electric field outside the charged sphere, we use a spherical Gaussian surface of radius (r >R). This surface encloses the entire charged sphere. So from Gauss's law, we have E(4/T r 2) — Q or E = 2 The field at points outside the sphere is a same as that of a point charge at the centre. Variation of E with the distance from the centre (r) 2 47t€() R O r Exercise:- An early model for an atom considered it to have a positively charged point nucleus of charge Ze, surrounded by a uniform density of negative charge up to a radius R. The atom as a whole is neutral. For this model, what is the electric field at a distance r from the nucleus?
5. The electric field components in figure are Ex = axl/2, Ey = Ez = O, in which u = 800 N/ C ml/2 Calculate (a) the flux through the cube, and (b) the charge within the cube. Assume that a = 0.1 m. 6. An electric field is uniform, and in the positive x direction for positive x, and uniform with the same magnitude but in the negative x direction for negative x. It is given that E = 200 N/ C for x > O and E = —200 N/ C for x < O. A right circular cylinder of length 20 cm and radius 5 cm has its centre at the origin and its axis along the x-axis so that one face is at x = +10 cm and the other is at x = —10 cm (figure). (a) What is the net outward flux through each flat _face? (b) What is the flux through the side of the cylinder? (c) What is the net outward flux through the cylinder? (d) What is the net charge inside the cylinder? 5 cm AS x = —10 cm 20 cm x cm 7. Four closed surfaces, Sl through S4, together with the charges -2Q, Q, and -Q are sketched in figure. 8. An infinitely long line charge having a uniform charge per unit length lies a distance d from point O as shown in figure. Determine the total electric flux through the surface of a sphere of radius R centered at O resulting from this line charge. Consider both cases, where R < d and R> d. 21 R2-d2 [O,
Electric field due to line of charge Consider an infinite line which has a linear charge density X. Using Gauss's law, let us find the electric field at a distance 'r' from the line charge. The cylindrical symmetry tells us that the field strength will be the same at all points at a fixed distance from the line. Thus, the field lines are directed radially outwards, perpendicular to the line charge. The appropriate choice of Gaussian surface is a cylinder of radius r and length L. On the flat end faces, S2 and S3, E is perpendicular to dS , which means flux is zero on them. On the curved surface Sl, E is parallel to dS , so that E.dS = Eds The charge enclosed by the cylinder is IL . Applying Gauss's law to the curved surface, we have EfdS = IT r L) Field Due to an Infinite Charged non-conducting Plane Sheet Let us consider a thin non-conducting charged plane sheet, infinite in extent, and having a surface charge density (charge per unit area) . Let P be a point, distance r from the sheet, at which the electric intensity is required. [Condition of infinite sheet:- r
Understanding Concept-I:- 1. A total charge of C is distributed uniformly throughout a 2.7 -cm radius sphere. The volume charge density is: A. 3.7 x 10-7 C/m3 D. 2.5 x 10-4 C/m3 2. Charge is placed on the B. 6.9 x 10-6 C/m3 E. 7.6 x 10-4 C/m3 C. 6.9 x 10-6 C/m2 surface of a 2.7 -cm radius isolated conducting sphere. The surface charge density is uniform and has the value 6.9 x 10-6 C/m2. The total charge on the sphere is: A. 5.6 x 10-10 C D. 6.3 x 10-8 C B. 2.1 10-8 c c. 4.7 x 10-8 c E. 9.5 x 10-3 C 3. A spherical shell has an inner radius of 3.7 cm and an outer radius of 4.5 cm. If charge is distributed uniformly throughout the shell with a volume density of 6. Ix 10-4 C/m3 the total charge is: A. 1.0 10-7 C D. 2.3 x 10-7 C B. 1.3 x 10-7 C E. 4.0 x 10-7 C c. 2.0 x 10-7 c 4. A cylinder has a radius of 2.1 cm and a length of 8.8 cm. Total charge 6.1 x 10-7 C is distributed uniformly throughout. The volume charge density is: A. 5.3 x 10-5 C/m3 B. 5.3 x 10-5 C/m2 D. 5.0 x 10-3 C/m3 E. 6.3 x 10-2 C/m3 C. 8.5 x 10-4 C/m3 5. When a piece of paper is held with one face perpendicular to a uniform electric field the flux through it is 25N•m2/C. When the paper is turned 250 with respect to the field the flux through it is: D. 23N • m2/C B. 12N • m2/C E. 25N • rn2/C C. 21N • m2/C 6. The flux of the electric field (24N/C)i + (30N/C)j + (16N/C)k through a 2.0m2 portion of the yz plane is: A. 32N • m2 /C D. 48N • m2 /C 7. Consider Gauss's law: fÉ.dÄ= q B. 34N • /C E. 60N • /C C. 42N • /C Which of the following is true? A. E must be the electric field due to the enclosed charge B. If q = O, then E = O everywhere on the Gaussian surface C. If the three particles inside have charges of +q, +q, and —2q, then the integral is zero D. on the surface E is everywhere parallel to dA E. If a charge is placed outside the surface, then it cannot affect E at any point on the surface 8. A charged point particle is placed at the center of a spherical Gaussian surface. The electric flux (I)E is changed if: A. the sphere is replaced by a cube of the same volume B. the sphere is replaced by a cube of one-tenth the volume C. the point charge is moved off center (but still inside the original sphere) D. the point charge is moved to just outside the sphere E. a second point charge is placed just outside the sphere 9. Choose the INCORRECT statement: A. Gauss' law can be derived from Coulomb's law B. Gauss' law states that the net number of lines crossing any closed surface in an outward direction is proportional to the net charge enclosed within the surface C. Coulomb's law can be derived from Gauss' law and symmetry D. Gauss' law applies to a closed surface of any shape E. According to Gauss' law, if a closed surface encloses no charge, then the electric field must vanish everywhere on the surface 10. The outer surface of the cardboard center of a paper towel roll: A. is a possible Gaussian surface B. cannot be a Gaussian surface because it encloses no charge C. cannot be a Gaussian surface since it is an insulator D. cannot be a Gaussian surface because it is not a closed surface E. none of the above 11. A physics instructor in an anteroom charges an electrostatic generator to 25 PC, then carries it into the lecture hall. The net electric flux in N • m2/C through the lecture hall walls is:
c. 2.2 x 105 B. 25 x 10-6 D. 2.8 x 106 E. can not tell unless the lecture hall dimensions are given 12. A point particle with charge q is placed inside the cube but not at its center. The electric flux through any one side of the cube: A. is zero C. is q/ 480 B. is q/ €0 D. is q/6eo E. cannot be computed using Gauss' law 13. A particle with charge 5.0-PC is placed at the corner of a cube. The total electric flux in N • m2 / C through all sides of the cube is: B. 7.1 x 104 D. 1.4 x 105 c. 9.4 x 104 E. 5.6 x 105 14. A point particle with charge q is at the center of a Gaussian surface in the form of a cube. The electric flux through any one face of the cube is: B. q/ 47t €0 E. q/ 12 €0 15. The table below gives the electric flux in N•m2/C through the ends and round surfaces of four Gaussian surfaces in the form of cylinders. Rank the cylinders according to the charge inside, from the most negative to the most positive. cylinder 1: cylinder 2: cylinder 9: cylinder 4: left end +2 X 10—9 +9 X 10—9 —2 X 10—9 +2 x 10-9 right end +4 X 10—9 —2 X 10—9 —5 X 10—9 -b x 10-9 rounded surface —6 X 10—9 +6 X 10—9 +3 X 10—9 -9 x 10-9 B. 4, 3,0, 1 16. A conducting sphere of radius 0.01m has a charge of 1.0 x 10-9 C deposited on it. The magnitude of the electric field in N/ C just outside the surface of the sphere is: B. 450 D. 4500 c. 900 E. 90, 000 17. A round wastepaper basket with a O. 15-m radius opening is in a uniform electric field of 300N [C, perpendicular to the opening. The total flux through the sides and bottom, in N • m2 C, is: c. 21 B. 4.2 D. 280 E. can not tell without knowing the areas of the sides and bottom 18. IOC of charge are placed on a spherical conducting shell. A particle with a charge of placed at the center of the cavity. The net charge on the inner surface of the shell is: c. oc 19. IOC of charge are placed on a spherical conducting shell. A particle with a charge of placed at the center of the cavity. The net charge on the outer surface of the shell is: c. oc -3C is -3C is 20. A 30-N/C uniform electric field points perpendicularly toward the left face of a large neutral conducting sheet. The surface charge density in C/m2 on the left and right faces, respectively, are: A. -2.7 x 10-9 C/m2•, +2.7 x 10-9 C/m2 B. +2.7 x 10-9 C/m2•, -2.7 x 10-9 C/m2 C. -5.3 x 10-9 C/m2•, +5.3 x 10-9 C/m2 D. +5.3 x 10-9 C/m2•, -5.3 x 10-9 C/m2
1 Special case (i) If CA = 0B = o then Ep = ER = 0/80 and Eq = O (ii) If = o and 0B = — o then Ep = ER = O and EQ = 0/80 Electric Field Due to an Uniformly Charged Spherical Shell Using Gauss's law, let us find the intensity of the electric field due to a uniformly charged spherical shell or a solid conducting sphere Case I: At an external point. In case of an isolated charged spherical conductor any excess charge on it is distributed uniformly over its outer surface same as that of charged spherical shell or hollow sphere. Hence field lines must point radially outward. Also, the field strength will have the same value at all points on any imaginary spherical surface concentric with the imaginary spherical surface concentric with the charged conducting sphere or the shell. This symmetry leads us to choose the Gaussian surface to be a sphere of radius r>R. Any arbitrary element of area dS is parallel to the local field, so at all points on the surface. According to Gauss's law fÉ.dS=fEdS = r2) Q Therefore, E = 2 For points outside the charged conducting sphere or the charged spherical shell, the field is same as that of a point charge at the centre. Case 11 : At an Internal Point (r < R) The field still has the same symmetry and so we again pick a spherical Gaussian surface, but now with radius r less than R. Since the charge enclosed is zero, from Gauss's law we have E(4/T r 2) = 0 givesE = 0 Thus, we conclude that E = O at all points inside a uniformly charged conducting sphere or the charged spherical shell. Variation of E with the distance from the centre (r) Electric Field Due to an Uniformly Charged Sphere 2 4Tt€() R O r A Non-conducting uniformly charged sphere of radius R has a total charge Q uniformly distributed throughout its volume. Using the Gauss's Law, Let us find the field Case I: at an internal point (r < R) Positive charge Q is uniformly distributed throughout the volume of sphere of radius R. For finding the electric field at a distance (r < R) from the centre, we choose a spherical Gaussian surface of radius r, concentric with the charge distribution. From symmetry the magnitude E of the electric field has the same value at every point on the Gaussian surface, and the direction of is radial at every point on the surface. So, applying Gauss's law
Here f É.dS = ) 2 4 3 E(41T r 2) — Q r 4 3 3 3 where p is volume charge density 41TR3 Therefore orE= 3 3 The field increases linearly with distance form the centre. Case 11 : At an external point To find the electric field outside the charged sphere, we use a spherical Gaussian surface of radius (r >R). This surface encloses the entire charged sphere. So from Gauss's law, we have E(4/T r 2) — Q or E = 2 The field at points outside the sphere is a same as that of a point charge at the centre. Variation of E with the distance from the centre (r) 2 47t€() R O r Exercise:- An early model for an atom considered it to have a positively charged point nucleus of charge Ze, surrounded by a uniform density of negative charge up to a radius R. The atom as a whole is neutral. For this model, what is the electric field at a distance r from the nucleus?
21. A solid insulating sphere of radius R contains positive charge that is distributed with a volume charge density that does not depend on angle but does increase with distance from the sphere center. Which of the graphs below might give the magnitude E of the electric field as a function of the distance r from the center of the sphere? distance from the center of a solid charged conducting sphere of radius R? .4 c 23. Charge Q is distributed uniformly throughout an insulating sphere of radius R. The magnitude of the electric field at a point R/ 2 from the center is: C. 3Q/4 E. none of these 24. Positive charge Q is distributed uniformly throughout an insulating sphere of radius R, centered at the origin. A particle with positive charge Q is placed at x = 2R on the x axis. The magnitude of the electric field at x = R/ 2 on the x axis is: C. Q/72 It E. none of these D. 17Q/72 25. Charge Q is distributed uniformly throughout a spherical insulating shell. The net electric flux in N • m2 / C through the inner surface of the shell is: C. 2Q/ 26. Charge Q is distributed uniformly throughout a spherical insulating shell. The net electric flux in N • m2 / C through the outer surface of the shell is: C. 2Q/ 27. A 3.5-cm radius hemisphere contains a total charge of 6.6 x 10-7 C. The flux through the rounded portion of the surface is 9.8 x 104 N • m2 / C. The flux through the flat base is: B. +2.3 x 104 N /C C. -2.3 x 104 N /C D. -9.8 x 104 N • /C E. +9.8 x 104 N /C 28. Charge is distributed uniformly along a long straight wire. The electric field 2 cm from the wire is 20N/C. The electric field 4 cm from the wire is: A. 120N/C C. 40N/C B. 80N/C D. ION/C
Understanding Concept-I:- 1. A total charge of C is distributed uniformly throughout a 2.7 -cm radius sphere. The volume charge density is: A. 3.7 x 10-7 C/m3 D. 2.5 x 10-4 C/m3 2. Charge is placed on the B. 6.9 x 10-6 C/m3 E. 7.6 x 10-4 C/m3 C. 6.9 x 10-6 C/m2 surface of a 2.7 -cm radius isolated conducting sphere. The surface charge density is uniform and has the value 6.9 x 10-6 C/m2. The total charge on the sphere is: A. 5.6 x 10-10 C D. 6.3 x 10-8 C B. 2.1 10-8 c c. 4.7 x 10-8 c E. 9.5 x 10-3 C 3. A spherical shell has an inner radius of 3.7 cm and an outer radius of 4.5 cm. If charge is distributed uniformly throughout the shell with a volume density of 6. Ix 10-4 C/m3 the total charge is: A. 1.0 10-7 C D. 2.3 x 10-7 C B. 1.3 x 10-7 C E. 4.0 x 10-7 C c. 2.0 x 10-7 c 4. A cylinder has a radius of 2.1 cm and a length of 8.8 cm. Total charge 6.1 x 10-7 C is distributed uniformly throughout. The volume charge density is: A. 5.3 x 10-5 C/m3 B. 5.3 x 10-5 C/m2 D. 5.0 x 10-3 C/m3 E. 6.3 x 10-2 C/m3 C. 8.5 x 10-4 C/m3 5. When a piece of paper is held with one face perpendicular to a uniform electric field the flux through it is 25N•m2/C. When the paper is turned 250 with respect to the field the flux through it is: D. 23N • m2/C B. 12N • m2/C E. 25N • rn2/C C. 21N • m2/C 6. The flux of the electric field (24N/C)i + (30N/C)j + (16N/C)k through a 2.0m2 portion of the yz plane is: A. 32N • m2 /C D. 48N • m2 /C 7. Consider Gauss's law: fÉ.dÄ= q B. 34N • /C E. 60N • /C C. 42N • /C Which of the following is true? A. E must be the electric field due to the enclosed charge B. If q = O, then E = O everywhere on the Gaussian surface C. If the three particles inside have charges of +q, +q, and —2q, then the integral is zero D. on the surface E is everywhere parallel to dA E. If a charge is placed outside the surface, then it cannot affect E at any point on the surface 8. A charged point particle is placed at the center of a spherical Gaussian surface. The electric flux (I)E is changed if: A. the sphere is replaced by a cube of the same volume B. the sphere is replaced by a cube of one-tenth the volume C. the point charge is moved off center (but still inside the original sphere) D. the point charge is moved to just outside the sphere E. a second point charge is placed just outside the sphere 9. Choose the INCORRECT statement: A. Gauss' law can be derived from Coulomb's law B. Gauss' law states that the net number of lines crossing any closed surface in an outward direction is proportional to the net charge enclosed within the surface C. Coulomb's law can be derived from Gauss' law and symmetry D. Gauss' law applies to a closed surface of any shape E. According to Gauss' law, if a closed surface encloses no charge, then the electric field must vanish everywhere on the surface 10. The outer surface of the cardboard center of a paper towel roll: A. is a possible Gaussian surface B. cannot be a Gaussian surface because it encloses no charge C. cannot be a Gaussian surface since it is an insulator D. cannot be a Gaussian surface because it is not a closed surface E. none of the above 11. A physics instructor in an anteroom charges an electrostatic generator to 25 PC, then carries it into the lecture hall. The net electric flux in N • m2/C through the lecture hall walls is:
E. 5N/C 29. Positive charge Q is placed on a conducting spherical shell with inner radius RI and outer radius R2. A particle with charge q is placed at the center of the cavity. The magnitude of the electric field at a point in the cavity, a distance r from the center, is: A. zero C. q/ 4 It eor2 E. (q + Q)/4 (R12 - r2) D. (q + Q)/ 4 Tt eor2 30. Positive charge Q is placed on a conducting spherical shell with inner radius RI and outer radius R2. A point charge q is placed at the center of the cavity. The magnitude of the electric field at a point outside the shell, a distance r from the center, is: A. zero C. q/ 4 Tt eor2 E. (q + Q)/4 (R12 - r2) 31. Positive charge Q is placed on a conducting spherical shell with inner radius RI and outer radius R2. A point charge q is placed at the center of the cavity. The magnitude of the electric field produced by the charge on the inner surface at a point in the interior of the conductor, a distance r from the center, is: D. q/ 4 Tt eor2 32. A long line of charge with Xl charge per unit length runs along the cylindrical axis of a cylindrical shell which carries a charge per unit length of Xc. The charge per unit length on the inner and outer surfaces of the shell, respectively are: A. Xl and Xc C. — Xl and Xc — Xc E. Xl — Xc and Xc + Xl B. - and Xc + D. Xl + Xc and Xc — Xl 33. Charge is distributed uniformly on the surface of a large flat plate. The electric field 2 cm from the plate is 30N/C. The electric field 4 cm from the plate is: A. 120N/C C. 30N/C E. 7.5N/C B. 80N/C D. 15N/C 34. Two large insulating parallel plates carry charge of equal magnitude, one positive and the other negative, that is distributed uniformly over their inner surfaces. Rank the points 1 through 5 according to the magnitude of the electric field at the points, least to greatest. B. 2, then 1, 3, and 4 tied, then 5 C. 1, 4, and 5 tie, then 2 and 3 tie D. 2 and 3 tie, then 1 and 4 tie, then 5 E. 2 and 3 tie, then 1, 4, and 5 tie 35. Two large parallel plates carry positive charge of equal magnitude that is distributed uniformly over their inner surfaces. Rank the points 1 through 5 according to the magnitude of the electric field at the points, least to greatest. C. 1, 4, and 5 tie, then 2 and 3 tie D. 2 and 3 tie, then 1 and 4 tie, then 5 E. 2 and 3 tie, then 1, 4, and 5 tie 1 1 23 29 36. A particle with charge Q is placed outside a large neutral conducting sheet. At any point in the interior of the sheet the electric field produced by charges on the surface is directed: A. toward the surface C. toward Q E. none of the above B. away from the surface D. away from Q 37. A hollow conductor is positively charged. A small uncharged metal ball is lowered by a silk thread through a small opening in the top of the conductor and allowed to touch its inner surface. After the ball is removed, it will have: A. a positive charge
c. 2.2 x 105 B. 25 x 10-6 D. 2.8 x 106 E. can not tell unless the lecture hall dimensions are given 12. A point particle with charge q is placed inside the cube but not at its center. The electric flux through any one side of the cube: A. is zero C. is q/ 480 B. is q/ €0 D. is q/6eo E. cannot be computed using Gauss' law 13. A particle with charge 5.0-PC is placed at the corner of a cube. The total electric flux in N • m2 / C through all sides of the cube is: B. 7.1 x 104 D. 1.4 x 105 c. 9.4 x 104 E. 5.6 x 105 14. A point particle with charge q is at the center of a Gaussian surface in the form of a cube. The electric flux through any one face of the cube is: B. q/ 47t €0 E. q/ 12 €0 15. The table below gives the electric flux in N•m2/C through the ends and round surfaces of four Gaussian surfaces in the form of cylinders. Rank the cylinders according to the charge inside, from the most negative to the most positive. cylinder 1: cylinder 2: cylinder 9: cylinder 4: left end +2 X 10—9 +9 X 10—9 —2 X 10—9 +2 x 10-9 right end +4 X 10—9 —2 X 10—9 —5 X 10—9 -b x 10-9 rounded surface —6 X 10—9 +6 X 10—9 +3 X 10—9 -9 x 10-9 B. 4, 3,0, 1 16. A conducting sphere of radius 0.01m has a charge of 1.0 x 10-9 C deposited on it. The magnitude of the electric field in N/ C just outside the surface of the sphere is: B. 450 D. 4500 c. 900 E. 90, 000 17. A round wastepaper basket with a O. 15-m radius opening is in a uniform electric field of 300N [C, perpendicular to the opening. The total flux through the sides and bottom, in N • m2 C, is: c. 21 B. 4.2 D. 280 E. can not tell without knowing the areas of the sides and bottom 18. IOC of charge are placed on a spherical conducting shell. A particle with a charge of placed at the center of the cavity. The net charge on the inner surface of the shell is: c. oc 19. IOC of charge are placed on a spherical conducting shell. A particle with a charge of placed at the center of the cavity. The net charge on the outer surface of the shell is: c. oc -3C is -3C is 20. A 30-N/C uniform electric field points perpendicularly toward the left face of a large neutral conducting sheet. The surface charge density in C/m2 on the left and right faces, respectively, are: A. -2.7 x 10-9 C/m2•, +2.7 x 10-9 C/m2 B. +2.7 x 10-9 C/m2•, -2.7 x 10-9 C/m2 C. -5.3 x 10-9 C/m2•, +5.3 x 10-9 C/m2 D. +5.3 x 10-9 C/m2•, -5.3 x 10-9 C/m2
B. a negative charge C. no appreciable charge D. a charge whose sign depends on what part of the inner surface it touched E. a charge whose sign depends on where the small hole is located in the conductor 38. A spherical conducting shell has charge Q. A particle with charge q is placed at the center of the cavity. The charge on the inner surface of the shell and the charge on the outer surface of the shell, respectively, are: 39. A charge Q is uniformly distributed over a large plastic plate. The electric field at a point P close to the centre of the plate is 10 V/ m. If the plastic plate is replaced by a copper plate of the same geometrical dimensions and carrying the same charge Q, the electric field at the point P will become (A) zero (C*) 10 V/m (D) 20 V/m 40. A metallic particle having no net charge is placed near a finite metal plate carrying a positive charge. The electric force on the particle will be (A*) towards the plate (B) away from the plate (C) parallel to the plate (D) zero 41. A thin, metallic spherical shell contains a charge Q on it. A point charge q is placed at the centre of the shell and another in fig. charge ql is placed outside it as shown All the three charges are positive. The force on the charge at the centre is (A) towards left (B) towards right (C) upward (D* ) zero 42. Consider the situation of the previous problem. The force on the centre charge due to the shell is (A) towards left (B*) towards right (C) upward (D) zero 43. Electric charges are distributed in a small volume. The flux of the electric field through a spherical surface of radius 10cm surrounding the total charge is 25 V-m. The flux over a concentric sphere of radius 20cm will be (A*) 25 V-m (B) 50 V-m (C) 100 V-m (D) 200 V-m 44. Fig. shows an imaginary cube of edge L/ 2. A uniformly charged rod of length L moves towards left at a small but constant speed v. At t = O, the left end just touches the centre of the face of the cube opposite it. Which of the graphs shown in fig. represents the flux of the electric filed through the cube at the rod goes through it? Flux d time q 45. A charge q is placed at the centre of the open end of a cylindrical vessel as shown in fig. The flux of the electric field through the surface of the vessel is - (A) zero
21. A solid insulating sphere of radius R contains positive charge that is distributed with a volume charge density that does not depend on angle but does increase with distance from the sphere center. Which of the graphs below might give the magnitude E of the electric field as a function of the distance r from the center of the sphere? distance from the center of a solid charged conducting sphere of radius R? .4 c 23. Charge Q is distributed uniformly throughout an insulating sphere of radius R. The magnitude of the electric field at a point R/ 2 from the center is: C. 3Q/4 E. none of these 24. Positive charge Q is distributed uniformly throughout an insulating sphere of radius R, centered at the origin. A particle with positive charge Q is placed at x = 2R on the x axis. The magnitude of the electric field at x = R/ 2 on the x axis is: C. Q/72 It E. none of these D. 17Q/72 25. Charge Q is distributed uniformly throughout a spherical insulating shell. The net electric flux in N • m2 / C through the inner surface of the shell is: C. 2Q/ 26. Charge Q is distributed uniformly throughout a spherical insulating shell. The net electric flux in N • m2 / C through the outer surface of the shell is: C. 2Q/ 27. A 3.5-cm radius hemisphere contains a total charge of 6.6 x 10-7 C. The flux through the rounded portion of the surface is 9.8 x 104 N • m2 / C. The flux through the flat base is: B. +2.3 x 104 N /C C. -2.3 x 104 N /C D. -9.8 x 104 N • /C E. +9.8 x 104 N /C 28. Charge is distributed uniformly along a long straight wire. The electric field 2 cm from the wire is 20N/C. The electric field 4 cm from the wire is: A. 120N/C C. 40N/C B. 80N/C D. ION/C
46. Mark the correct options: (A) Gauss's law is valid only for symmetrical charge distributions (B) Gauss's law is valid only for charges placed in vacuum (C) The electric field calculated by Gauss's law is the field due to the charges inside the Gaussian surface (D*) The flux of the electric field through a closed surface due to all the charge is equal to the flux due to the charges enclosed by the surface. 47. A positive point charge Q is brought near an isolated metal cube. (A) The cube becomes negatively charged (B) The cube becomes positively charged (C) The interior becomes positively charged and the surface becomes negatively charged. (D*) The interior becomes remains charged free and the surface gets non-uniform charge distribution. 48. A large non-conducting sheet M is given a uniform charge density. Two uncharged small metal rods A and B are placed near the (A*) M attracts A (B*) M attracts B sheet as shown in fig. (C*) A attracts B (D*) B attracts A 49. If flux of the electric field through a closed surface is zero, (A) the electric field must be zero everywhere on the surface (B*) the electric field may be zero everywhere in the surface (C*) the charge inside the surface must be zero (D) the charge in the vicinity of the surface must be zero. 50. An electric dipole is placed at the centre of a sphere, Mark the correct options. (A*) The flux of the electric filed through the sphere is zero (B) The electric field is zero at every point of the sphere (C*) The electric field is not zero anywhere on the sphere (D) The electric field is zero on a circle on the sphere. 51. Figure shows a charge q placed at the centre of a hemisphere. A second charge Q is placed at one of the positions A, B, C and D. In which position(s) of this second charge, the flux of the electric field through the hemisphere remains unchanged? c 52. A closed surface S is constructed around a conducting wire connected to a battery and a switch (fig.) As the switch is closed, the free electrons in the wire start moving along the wire. In any time interval, the number of electrons entering the closed surface S is equal to the number of electrons leaving it. On closing the switch, the flux of the electric field through the closed surface. (A) is increased (B) is decreased (C* ) remains unchanged (D* ) remains zero 53. Fig. shows a closed surface which intersects a conducting sphere. If a positive charged is placed at the point P, the flux of the 6 electric field through the closed surface (A) will remain zero (B*) will become positive (C) will become negative (D) will become undefined closed surface q battery conducting Shere 54. If the electric flux entering and leaving an enclosed surface respectively is (hand +2, the electric charge inside the surface will be: [AIEEE-2003] (1*) (+2 - +1)eo
ELECTROSTATICS-3 [Electric flux, Gauss's Law and its applications] ELECTRIC FLUX Electric flux is a measure of 'flow' of electric field through a surface. It is equal to the product of an area element and the perpendicular component of E integrated over a surface. (1) Flux of electric field E through any area A is defined as. = E.AcosO or (2) In case of variable electric field or curved area. = f E. dA 2 (3) It's S.I. Unit is (Volt x m) or 0 (4) For a closed body outward flux is taken to be positive while inward flux is taken to be negative. Negative flux Zero flux Positive flux Exercise:- 1. Consider a uniform electric field E = 3 x 103 i N/ C. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its plane makes a 600 angle with the x-axis? 2. What is the net flux of the uniform electric field of Exercise 1 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes? 3. A pyramid with horizontal square base, 6.00 m on each side, and a height of 4.00 m is placed in vertical electric field of 52.0 N/ C. Calculate the total electric flux through the pyramid's four slanted surfaces.[ +1.87 kN•m2/Cl 4. A cone with base radius R and height h is located on a horizontal table. A horizontal uniform field E penetrates the cone, as shown in Figure. Determine the electric flux that enters the left-hand side of the cone. [ERh] 5. A non-uniform electric field given by E = 3xi +4j pierces the Gaussian cube shown in figure. (E is in newtons per coulomb and x is in meters.) What is the electric flux through the right face, the left face, and the top face? [-12 N-m2/C, 36N- m2/C,16 N-m2/C] 6. An electric fleld given by E = 4i - 3(y2 + 2)j N/ C pierces a Gaussian cube of edge length 2.0 m and positioned as shown in figure same as for exercise 5. (The magnitude E is in newtons per coulomb and the position x is in meters.) What is the electric flux through the (a) top face, (b) bottom face, (c) left face, and (d) back face? (e) What is the net electric flux through the cube? Y 1.0m 3.0m Gaussian surface
E. 5N/C 29. Positive charge Q is placed on a conducting spherical shell with inner radius RI and outer radius R2. A particle with charge q is placed at the center of the cavity. The magnitude of the electric field at a point in the cavity, a distance r from the center, is: A. zero C. q/ 4 It eor2 E. (q + Q)/4 (R12 - r2) D. (q + Q)/ 4 Tt eor2 30. Positive charge Q is placed on a conducting spherical shell with inner radius RI and outer radius R2. A point charge q is placed at the center of the cavity. The magnitude of the electric field at a point outside the shell, a distance r from the center, is: A. zero C. q/ 4 Tt eor2 E. (q + Q)/4 (R12 - r2) 31. Positive charge Q is placed on a conducting spherical shell with inner radius RI and outer radius R2. A point charge q is placed at the center of the cavity. The magnitude of the electric field produced by the charge on the inner surface at a point in the interior of the conductor, a distance r from the center, is: D. q/ 4 Tt eor2 32. A long line of charge with Xl charge per unit length runs along the cylindrical axis of a cylindrical shell which carries a charge per unit length of Xc. The charge per unit length on the inner and outer surfaces of the shell, respectively are: A. Xl and Xc C. — Xl and Xc — Xc E. Xl — Xc and Xc + Xl B. - and Xc + D. Xl + Xc and Xc — Xl 33. Charge is distributed uniformly on the surface of a large flat plate. The electric field 2 cm from the plate is 30N/C. The electric field 4 cm from the plate is: A. 120N/C C. 30N/C E. 7.5N/C B. 80N/C D. 15N/C 34. Two large insulating parallel plates carry charge of equal magnitude, one positive and the other negative, that is distributed uniformly over their inner surfaces. Rank the points 1 through 5 according to the magnitude of the electric field at the points, least to greatest. B. 2, then 1, 3, and 4 tied, then 5 C. 1, 4, and 5 tie, then 2 and 3 tie D. 2 and 3 tie, then 1 and 4 tie, then 5 E. 2 and 3 tie, then 1, 4, and 5 tie 35. Two large parallel plates carry positive charge of equal magnitude that is distributed uniformly over their inner surfaces. Rank the points 1 through 5 according to the magnitude of the electric field at the points, least to greatest. C. 1, 4, and 5 tie, then 2 and 3 tie D. 2 and 3 tie, then 1 and 4 tie, then 5 E. 2 and 3 tie, then 1, 4, and 5 tie 1 1 23 29 36. A particle with charge Q is placed outside a large neutral conducting sheet. At any point in the interior of the sheet the electric field produced by charges on the surface is directed: A. toward the surface C. toward Q E. none of the above B. away from the surface D. away from Q 37. A hollow conductor is positively charged. A small uncharged metal ball is lowered by a silk thread through a small opening in the top of the conductor and allowed to touch its inner surface. After the ball is removed, it will have: A. a positive charge
Gauss's Law and it's Application Formal proof of Gauss's Law*** One way of deriving Gauss's law involves solid angles. Consider a spherical surface of radius r containing an area element AA . A closed surface of arbitrary shape surrounds a point charge q. The net electric flux through the surface is independent of the shape of the surface. The solid angle AQ subtended at the center of the sphere by this element is defined to be 2 From this equation, we see that AQhas no dimensions because AA and r 2 both have dimensions L2. The dimensionless unit of a solid angle is the steradian. Because the surface area of a sphere is 2 41tr 47tr2, the total solid angle subtended by the sphere is Q = = 41T steradians 2 Now consider a point charge q surrounded by a closed surface of arbitrary shape. The total electric flux through this surface can be obtained by evaluating E.AA for each small area element AA and summing over all elements. The flux through each element is AA cose = = (Ecos O)AA = kq 2 where ris the distance from the charge to the area element, 0 is the angle between the electric field E and AA for the element. In Figure, we see that the projection of the area element perpendicular to the radius vector is AA cos0. Thus, the quantity (AA cos0)/r2 is equal to the solid angle AQ that the surface element AA subtends at the charge q. We also see that AQis equal to the solid angle subtended by the area element of a spherical surface of radius r. Because the total solid angle at a point is 47t steradians, the total flux through the closed surface is dA cos0 "E = kq f 2 kq f dQ = 411 k q Thus we have derived Gauss's law. Note that this result is independent of the shape of the closed surface and independent of the position of the charge within the surface. OR (1) According to this law, the total flux linked with a closed surface called Gaussian surface. (The surface need not be a real physical surface, it can also be an hypothetical one) is (1/80) times the charge enclosed by the closed surface i.e., = f • dA = (Qenc) o (2) Electric field in f E.dA is complete electric field. It may be partly due to charge with in the surface and partly due to charge outside the surface. However if there is no charge enclosed in the Gaussian surface, then f E. dA = 0 . (3) The electric field E is resulting from all charge, both those inside and those outside the Gaussian surface.
55. A charged ball B hangs from a silk thread S, which makes an angle q with a large charged conducting sheet P, as shown in the figure. The surface density s of the sheet is proportional to: [AIEEE-2005] (1) sin 0 (2*) tan 0 (3) cos 0 (4) cot 0 56. Consider the charge configuration and a spherical Gaussian surface as shown in the figure. When calculating the flux of the electric / field over the spherical surface, the electric field will be due to charge [JEE 2004] (A) q2 (C) all the charges (B) only the positive charges (D) +ql and -ql 57. Three large parallel plates have uniform surface charge densities as shown in the figure. What is the electric field at P. [JEE- 2005 (Scr)] 40 Comprehension (Q.57 to Q.58) 20 20 (D) o s z=a z=—a z=—2a The nuclear charge (Ze) is non-uniformly distributed within a nucleus of radius R. The charge density p(r) [charge per unit volume] is dependent only on the radial distance r from the centre of the nucleus as shown in figure. The electric field is only along the radial p(r) d direction. 58. The electric field at r = R is: (A) independent of a (C) directly proportional to a2 [JEE 2008] (B) directly proportional to a (D) inversely proportional to a a 59. For a = O, the value of d (maximum value of r as shown in the figure) is . 3 (B) 3 4 Ze (C) 3 nR3 Ze (D) 60. The electric field within the nucleus is generally observed to be linearly dependent on r. This implies (D) a =2R/3 61. A disk of radius a / 4 having a uniformly distributed charge 6C is placed in the x-y plane with its centre at (—a / 2, O, O). A rod of length a carrying a uniformly distributed charge 8C is placed on the x-axis from x = a/ 4 to x = 5a/4. Two point charges —7 C and 3C are placed at (a/4, —a/ 4, O) and (—3a/4, 3a/4, O), respectively. Consider a cubical surface formed by six surfaces x = ± a/ 2, y — ± a/ 2, z = ± a/ 2. The electric flux through this cubical surface is o IOC (C) [JEE-2009] (B) o 12C (D) z (a,O,a) x (a,a,a) 62. Consider an electric field E = Eoi , where Eo is a constant. The flux through the shaded area (as shown in the figure) due to this field is: [JEE-2011] (A) 2Eoa2 (C) Eoa2 2 Eoa (D) x
B. a negative charge C. no appreciable charge D. a charge whose sign depends on what part of the inner surface it touched E. a charge whose sign depends on where the small hole is located in the conductor 38. A spherical conducting shell has charge Q. A particle with charge q is placed at the center of the cavity. The charge on the inner surface of the shell and the charge on the outer surface of the shell, respectively, are: 39. A charge Q is uniformly distributed over a large plastic plate. The electric field at a point P close to the centre of the plate is 10 V/ m. If the plastic plate is replaced by a copper plate of the same geometrical dimensions and carrying the same charge Q, the electric field at the point P will become (A) zero (C*) 10 V/m (D) 20 V/m 40. A metallic particle having no net charge is placed near a finite metal plate carrying a positive charge. The electric force on the particle will be (A*) towards the plate (B) away from the plate (C) parallel to the plate (D) zero 41. A thin, metallic spherical shell contains a charge Q on it. A point charge q is placed at the centre of the shell and another in fig. charge ql is placed outside it as shown All the three charges are positive. The force on the charge at the centre is (A) towards left (B) towards right (C) upward (D* ) zero 42. Consider the situation of the previous problem. The force on the centre charge due to the shell is (A) towards left (B*) towards right (C) upward (D) zero 43. Electric charges are distributed in a small volume. The flux of the electric field through a spherical surface of radius 10cm surrounding the total charge is 25 V-m. The flux over a concentric sphere of radius 20cm will be (A*) 25 V-m (B) 50 V-m (C) 100 V-m (D) 200 V-m 44. Fig. shows an imaginary cube of edge L/ 2. A uniformly charged rod of length L moves towards left at a small but constant speed v. At t = O, the left end just touches the centre of the face of the cube opposite it. Which of the graphs shown in fig. represents the flux of the electric filed through the cube at the rod goes through it? Flux d time q 45. A charge q is placed at the centre of the open end of a cylindrical vessel as shown in fig. The flux of the electric field through the surface of the vessel is - (A) zero
(Keep in mind, the electric field due to a charge outside the Gaussian surface contributes zero net flux through the surface, because as many lines due to that charge enter the surface as leave it). Flux from surface Sl = + Flux from surface S2 — , and flux from flux from surface S4 = O Application of Gauss's law: See flux emergence in the following cases (1) If a dipole is enclosed by a surface •.•Qenc (2) The net charge Qenc is the algebraic sum of all the enclosed positive, and negative charges. If Qenc is positive the net flux is outward; if Qenc is negative, the net flux is inward. 1 O (3) If a closed body (not enclosing any charge) is placed in an electric field (either uniform or non- uniform) total flux linked with it will be zero x a a (4) If a hemispherical body surface calculated as follows "Qurved + "Circular "Qzrved = —"arcular - (E x cosi800) = +ÆR2E Sphere is placed in n z uniform (B) "tn — "out = "T = O electric field then flux linked with the curved (5) If a hemispherical body is placed in non-uniform electric field as shown below, then flux linked with the circular surface calculated as follows "Circular = —"Cürved - (E x cos00) "Ocular -2ÆR2E n
46. Mark the correct options: (A) Gauss's law is valid only for symmetrical charge distributions (B) Gauss's law is valid only for charges placed in vacuum (C) The electric field calculated by Gauss's law is the field due to the charges inside the Gaussian surface (D*) The flux of the electric field through a closed surface due to all the charge is equal to the flux due to the charges enclosed by the surface. 47. A positive point charge Q is brought near an isolated metal cube. (A) The cube becomes negatively charged (B) The cube becomes positively charged (C) The interior becomes positively charged and the surface becomes negatively charged. (D*) The interior becomes remains charged free and the surface gets non-uniform charge distribution. 48. A large non-conducting sheet M is given a uniform charge density. Two uncharged small metal rods A and B are placed near the (A*) M attracts A (B*) M attracts B sheet as shown in fig. (C*) A attracts B (D*) B attracts A 49. If flux of the electric field through a closed surface is zero, (A) the electric field must be zero everywhere on the surface (B*) the electric field may be zero everywhere in the surface (C*) the charge inside the surface must be zero (D) the charge in the vicinity of the surface must be zero. 50. An electric dipole is placed at the centre of a sphere, Mark the correct options. (A*) The flux of the electric filed through the sphere is zero (B) The electric field is zero at every point of the sphere (C*) The electric field is not zero anywhere on the sphere (D) The electric field is zero on a circle on the sphere. 51. Figure shows a charge q placed at the centre of a hemisphere. A second charge Q is placed at one of the positions A, B, C and D. In which position(s) of this second charge, the flux of the electric field through the hemisphere remains unchanged? c 52. A closed surface S is constructed around a conducting wire connected to a battery and a switch (fig.) As the switch is closed, the free electrons in the wire start moving along the wire. In any time interval, the number of electrons entering the closed surface S is equal to the number of electrons leaving it. On closing the switch, the flux of the electric field through the closed surface. (A) is increased (B) is decreased (C* ) remains unchanged (D* ) remains zero 53. Fig. shows a closed surface which intersects a conducting sphere. If a positive charged is placed at the point P, the flux of the 6 electric field through the closed surface (A) will remain zero (B*) will become positive (C) will become negative (D) will become undefined closed surface q battery conducting Shere 54. If the electric flux entering and leaving an enclosed surface respectively is (hand +2, the electric charge inside the surface will be: [AIEEE-2003] (1*) (+2 - +1)eo
63. A cubical region of sid a has its centre at the origin. It encloses three fixed point charges, -q at (O,-a/4,O) , + 3q at (0,0,0) and -q at (O, + a/ 4,0). Choose the correct option(s) z a x (A) The net electric flux crossing the plane x = +a/2 is equal to the net electric flux crossing the plane x= -a/ 2. (B) The net electric flux crossing the plane y = +a/2 is more than the net electric flux crossing the plane y = -a/ 2. (C* ) The net electric flux crossing the entire region is q/ (D*) The net electric flux crossing the plane z = +a/2 is equal to the net electric flux crossing the plane + a/2. [JED-2012] 64. Two non-conducting solid spheres of radii R and 2R, having uniform volume charge densities PI and respectively, touch each other. The net electric field at a distance 2 R from the centre of the smaller sphere, along the line joining the centres of the spheres, is zero. The ratiopl/p2 can be (*B) -32/25 (C)32/25 (*D) 4 [JEE-2013 Al] 65. Let El(r), E2(r) and E3(r) be the respective electric fields at a distance r from a point charge Q, an infinitely long wire with constant linear charge density X, and an infinite plane with uniform surface charge density o. If E2(ro ) at a given distance ro, then [JEE-2014 Al] (A) Q— 40 IT r: (C*) E2(ro/2) (D) E2(ro/2) = 4E3(ro/2) 2 ITC 66. Charges Q, 2Q and 4Q are uniformly distributed in three dielectric solid spheres 1, 2 and 3 of radii R/ 2, R and 2R respectively, as shown in figure. If magnitudes of the electric fields at point P at a distance R from the centre of spheres 1, 2 and 3 are El, E2 and E3 respectively, then [JEE-2014 A2] 2 Sphere 1 (A) El > E2 > E3 (B) E3 > El > E2 2Q Sphere 2 67. An infinitely long uniform line charge distribution of charge per unit length lies parallel to the y-axis in the y-z plane at z = —a (see figure). If the magnitude of the flux of 2 the electric field through the rectangular surface ABCD lying in the x-y plane with its center at the origin is (eo o permittivity of free space), then the value of n is [JEE-2015 Al] 4Q Sphere 3 (D) E3 > E2 > El o x 2
(6) If charge is kept at the centre of cube 1 "total = — • (Q) "face 680 (7) If charge is kept at the centre of a face: First we Gaussian surface (an identical imaginary cube) 4 (A) should enclosed O Q (B) (B) the charge by assuming a Total flux emerges from the system (Two cubes) Flux from given cube (i.e. from 5 face only) (8) If a charge is kept at the corner of a cube (A) "total — 280 For enclosing the charge seven more cubes are required so total flux from the 8 cube system is "-r . Flux from given cube . Flux from one face opposite to change, of the given cube cube 880 Q/ 880 Q (Because only three faces are seen). "face — 3 24 (9) A long straight wire of charge density penetrates a hollow body as shown. The flux emerges from the body is = x (Length of the wire inside the body) Exercise:- 1. Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 x 103 Nm2/C. (a) What is the net charge inside the box? (b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not? 2. A point charge +10 PC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in figure. What is the magnitude of the electric flux through the square? 3. A point charge of 2.0 PC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface? 4. A point charge causes an electric flux of —1.0 x 103 Nm2/C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge. (a) If the radius of the Gaussian surface were doubled, how 5 cm 10 cm much flux would pass through the surface? (b) What is the value of the point charge?
55. A charged ball B hangs from a silk thread S, which makes an angle q with a large charged conducting sheet P, as shown in the figure. The surface density s of the sheet is proportional to: [AIEEE-2005] (1) sin 0 (2*) tan 0 (3) cos 0 (4) cot 0 56. Consider the charge configuration and a spherical Gaussian surface as shown in the figure. When calculating the flux of the electric / field over the spherical surface, the electric field will be due to charge [JEE 2004] (A) q2 (C) all the charges (B) only the positive charges (D) +ql and -ql 57. Three large parallel plates have uniform surface charge densities as shown in the figure. What is the electric field at P. [JEE- 2005 (Scr)] 40 Comprehension (Q.57 to Q.58) 20 20 (D) o s z=a z=—a z=—2a The nuclear charge (Ze) is non-uniformly distributed within a nucleus of radius R. The charge density p(r) [charge per unit volume] is dependent only on the radial distance r from the centre of the nucleus as shown in figure. The electric field is only along the radial p(r) d direction. 58. The electric field at r = R is: (A) independent of a (C) directly proportional to a2 [JEE 2008] (B) directly proportional to a (D) inversely proportional to a a 59. For a = O, the value of d (maximum value of r as shown in the figure) is . 3 (B) 3 4 Ze (C) 3 nR3 Ze (D) 60. The electric field within the nucleus is generally observed to be linearly dependent on r. This implies (D) a =2R/3 61. A disk of radius a / 4 having a uniformly distributed charge 6C is placed in the x-y plane with its centre at (—a / 2, O, O). A rod of length a carrying a uniformly distributed charge 8C is placed on the x-axis from x = a/ 4 to x = 5a/4. Two point charges —7 C and 3C are placed at (a/4, —a/ 4, O) and (—3a/4, 3a/4, O), respectively. Consider a cubical surface formed by six surfaces x = ± a/ 2, y — ± a/ 2, z = ± a/ 2. The electric flux through this cubical surface is o IOC (C) [JEE-2009] (B) o 12C (D) z (a,O,a) x (a,a,a) 62. Consider an electric field E = Eoi , where Eo is a constant. The flux through the shaded area (as shown in the figure) due to this field is: [JEE-2011] (A) 2Eoa2 (C) Eoa2 2 Eoa (D) x
5. The electric field components in figure are Ex = axl/2, Ey = Ez = O, in which u = 800 N/ C ml/2 Calculate (a) the flux through the cube, and (b) the charge within the cube. Assume that a = 0.1 m. 6. An electric field is uniform, and in the positive x direction for positive x, and uniform with the same magnitude but in the negative x direction for negative x. It is given that E = 200 N/ C for x > O and E = —200 N/ C for x < O. A right circular cylinder of length 20 cm and radius 5 cm has its centre at the origin and its axis along the x-axis so that one face is at x = +10 cm and the other is at x = —10 cm (figure). (a) What is the net outward flux through each flat _face? (b) What is the flux through the side of the cylinder? (c) What is the net outward flux through the cylinder? (d) What is the net charge inside the cylinder? 5 cm AS x = —10 cm 20 cm x cm 7. Four closed surfaces, Sl through S4, together with the charges -2Q, Q, and -Q are sketched in figure. 8. An infinitely long line charge having a uniform charge per unit length lies a distance d from point O as shown in figure. Determine the total electric flux through the surface of a sphere of radius R centered at O resulting from this line charge. Consider both cases, where R < d and R> d. 21 R2-d2 [O,
68. The figures below depict two situations in which two infinitely long static line charges of constant positive line charge density X are kept parallel to each other. In their resulting electric field, point charges q and —q are kept in equilibrium between them. The point charges are confined to move in the x direction only. If they are given a small displacement about their equilibrium positions, then the correct statement(s) is(are) [JEE-2015 Al] (A) Both charges execute simple harmonic motion. (B) Both charges will continue moving in the direction of their displacement. (C) Charge +q executes simple harmonic motion while charge -q continues moving in the direction of its displacement. (D) Charge -q executes simple harmonic motion while charge +q continues moving in the direction of its displacement. 69. The region between two concentric spheres of radii 'a' and 'b' respectively (see figure), has volume charge density p = A/ r, where A is a constant and r is the distance from the centre. At the centre of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant, is: [JED-M-2016] (1) 2 It b2—a 2 (3) 2 (2) 2 (4) 2 2 IT a 70. A point charge +Q is placed just outside an imaginary hemispherical surface of radius R as shown in the figure. Which of the following statements is/ are correct? [JEE-A2-2017] (A) Total flux through the curved and the flat surface is Q/ €0. (B) The component of the electric field normal to the flat surface is constant over the surface (C) The circumference of the flat surface is an equipotential (D) The electric flux passing through the curved surface of the Q 1 hemisphere is 71. An infinitely long thin non-conducting wire is parallel to the z-axis and carries a uniform line charge density X. It pierces a thin non-conducting spherical shell of radius R in such a way that the arc PQ subtends an angle 1200 at the centre O of the spherical shell, as shown in the figure. The permittivity of free space is €0. Which of the following statements is (are) true? [JEE-A2-2018] (A) The electric flux through the shell is NfiRX/eo (B) The z-component of the electric field is zero at all the points on the surface of the shell (C) The electric flux through the shell is URX/eo (D) The electric field is normal to the surface of the shell at all points a Q b o 1200
Electric field due to line of charge Consider an infinite line which has a linear charge density X. Using Gauss's law, let us find the electric field at a distance 'r' from the line charge. The cylindrical symmetry tells us that the field strength will be the same at all points at a fixed distance from the line. Thus, the field lines are directed radially outwards, perpendicular to the line charge. The appropriate choice of Gaussian surface is a cylinder of radius r and length L. On the flat end faces, S2 and S3, E is perpendicular to dS , which means flux is zero on them. On the curved surface Sl, E is parallel to dS , so that E.dS = Eds The charge enclosed by the cylinder is IL . Applying Gauss's law to the curved surface, we have EfdS = IT r L) Field Due to an Infinite Charged non-conducting Plane Sheet Let us consider a thin non-conducting charged plane sheet, infinite in extent, and having a surface charge density (charge per unit area) . Let P be a point, distance r from the sheet, at which the electric intensity is required. [Condition of infinite sheet:- r
63. A cubical region of sid a has its centre at the origin. It encloses three fixed point charges, -q at (O,-a/4,O) , + 3q at (0,0,0) and -q at (O, + a/ 4,0). Choose the correct option(s) z a x (A) The net electric flux crossing the plane x = +a/2 is equal to the net electric flux crossing the plane x= -a/ 2. (B) The net electric flux crossing the plane y = +a/2 is more than the net electric flux crossing the plane y = -a/ 2. (C* ) The net electric flux crossing the entire region is q/ (D*) The net electric flux crossing the plane z = +a/2 is equal to the net electric flux crossing the plane + a/2. [JED-2012] 64. Two non-conducting solid spheres of radii R and 2R, having uniform volume charge densities PI and respectively, touch each other. The net electric field at a distance 2 R from the centre of the smaller sphere, along the line joining the centres of the spheres, is zero. The ratiopl/p2 can be (*B) -32/25 (C)32/25 (*D) 4 [JEE-2013 Al] 65. Let El(r), E2(r) and E3(r) be the respective electric fields at a distance r from a point charge Q, an infinitely long wire with constant linear charge density X, and an infinite plane with uniform surface charge density o. If E2(ro ) at a given distance ro, then [JEE-2014 Al] (A) Q— 40 IT r: (C*) E2(ro/2) (D) E2(ro/2) = 4E3(ro/2) 2 ITC 66. Charges Q, 2Q and 4Q are uniformly distributed in three dielectric solid spheres 1, 2 and 3 of radii R/ 2, R and 2R respectively, as shown in figure. If magnitudes of the electric fields at point P at a distance R from the centre of spheres 1, 2 and 3 are El, E2 and E3 respectively, then [JEE-2014 A2] 2 Sphere 1 (A) El > E2 > E3 (B) E3 > El > E2 2Q Sphere 2 67. An infinitely long uniform line charge distribution of charge per unit length lies parallel to the y-axis in the y-z plane at z = —a (see figure). If the magnitude of the flux of 2 the electric field through the rectangular surface ABCD lying in the x-y plane with its center at the origin is (eo o permittivity of free space), then the value of n is [JEE-2015 Al] 4Q Sphere 3 (D) E3 > E2 > El o x 2
72. The electric field E is measured at a point P(O,O,d) generated due to various charge distributions and the dependence of E on d is found to be different for different charge distributions. List-I contains different relations between E and d. List-Il describes different electric charge distributions, along with their locations. Match the functions in List-I with the related char e distributions in List-Il. Q. R. s. List-I E is independent of d 1 d 1 1 JEE-A2-2018 List-Il 1. A point char e Q at the ori •n 2. A small dipole with point charges Q at and - Q at (O,o, - 0. Take 2 R then + = Q/€O C. If h > 2R and r = 4R/5 then + = Q/5€0 D. If h > 2R and r = 3R/5 then + = Q/5€O 75. Two infinite planes each with uniform surface charge density +0 are kept in such a way that the angle between them is 300. The electric field in the region shown between them is given by [JEE (Main)/MS/07-01-2020] [JEE-A1-2019] o (1) 280 o (2) 280 o (3) 280 o (4) 280 x 2 x 2 1- 2 2 2 2
1 Special case (i) If CA = 0B = o then Ep = ER = 0/80 and Eq = O (ii) If = o and 0B = — o then Ep = ER = O and EQ = 0/80 Electric Field Due to an Uniformly Charged Spherical Shell Using Gauss's law, let us find the intensity of the electric field due to a uniformly charged spherical shell or a solid conducting sphere Case I: At an external point. In case of an isolated charged spherical conductor any excess charge on it is distributed uniformly over its outer surface same as that of charged spherical shell or hollow sphere. Hence field lines must point radially outward. Also, the field strength will have the same value at all points on any imaginary spherical surface concentric with the imaginary spherical surface concentric with the charged conducting sphere or the shell. This symmetry leads us to choose the Gaussian surface to be a sphere of radius r>R. Any arbitrary element of area dS is parallel to the local field, so at all points on the surface. According to Gauss's law fÉ.dS=fEdS = r2) Q Therefore, E = 2 For points outside the charged conducting sphere or the charged spherical shell, the field is same as that of a point charge at the centre. Case 11 : At an Internal Point (r < R) The field still has the same symmetry and so we again pick a spherical Gaussian surface, but now with radius r less than R. Since the charge enclosed is zero, from Gauss's law we have E(4/T r 2) = 0 givesE = 0 Thus, we conclude that E = O at all points inside a uniformly charged conducting sphere or the charged spherical shell. Variation of E with the distance from the centre (r) Electric Field Due to an Uniformly Charged Sphere 2 4Tt€() R O r A Non-conducting uniformly charged sphere of radius R has a total charge Q uniformly distributed throughout its volume. Using the Gauss's Law, Let us find the field Case I: at an internal point (r < R) Positive charge Q is uniformly distributed throughout the volume of sphere of radius R. For finding the electric field at a distance (r < R) from the centre, we choose a spherical Gaussian surface of radius r, concentric with the charge distribution. From symmetry the magnitude E of the electric field has the same value at every point on the Gaussian surface, and the direction of is radial at every point on the surface. So, applying Gauss's law
(Y2 + 1b N/C passes 76 An electric field E = 4x i — through the box shown in figure. The flux of the electric field through surfaces ABCD and BCGF are marked as +1 and +11 respectively. The difference between (+1 — +11) is (in Nm2/C) [JEE (Main)/ES/09-01-2020] z c x 77 Two charged thin infinite plane sheets of uniform surface charge density 0+ and on, where | 0+ | > I I , intersect at right angle. Which of the following best represents the electric field lines for this system? (1) (2) [JEE (Main)/MS/04-09-2020] 0+ 0+ (3) (4) 0+ o_ 78 Consider the force F on a charge 'q' due to a uniformly charged spherical shell of radius R carrying charge Q distributed uniformly over it. Which one of the following statements is true for F, if 'q' is placed at distance r from the centre of the shell ? [JEE (Main)/ES/06-09-2020] (1) (2) (3) (4) — for all r 2 41-11% r > F > O for r < R 4111% R 2 for r > R 2 4111% r for r < R R 2 78 A circular disc of radius R carries surface charge density o(r) 1 where is a constant
68. The figures below depict two situations in which two infinitely long static line charges of constant positive line charge density X are kept parallel to each other. In their resulting electric field, point charges q and —q are kept in equilibrium between them. The point charges are confined to move in the x direction only. If they are given a small displacement about their equilibrium positions, then the correct statement(s) is(are) [JEE-2015 Al] (A) Both charges execute simple harmonic motion. (B) Both charges will continue moving in the direction of their displacement. (C) Charge +q executes simple harmonic motion while charge -q continues moving in the direction of its displacement. (D) Charge -q executes simple harmonic motion while charge +q continues moving in the direction of its displacement. 69. The region between two concentric spheres of radii 'a' and 'b' respectively (see figure), has volume charge density p = A/ r, where A is a constant and r is the distance from the centre. At the centre of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant, is: [JED-M-2016] (1) 2 It b2—a 2 (3) 2 (2) 2 (4) 2 2 IT a 70. A point charge +Q is placed just outside an imaginary hemispherical surface of radius R as shown in the figure. Which of the following statements is/ are correct? [JEE-A2-2017] (A) Total flux through the curved and the flat surface is Q/ €0. (B) The component of the electric field normal to the flat surface is constant over the surface (C) The circumference of the flat surface is an equipotential (D) The electric flux passing through the curved surface of the Q 1 hemisphere is 71. An infinitely long thin non-conducting wire is parallel to the z-axis and carries a uniform line charge density X. It pierces a thin non-conducting spherical shell of radius R in such a way that the arc PQ subtends an angle 1200 at the centre O of the spherical shell, as shown in the figure. The permittivity of free space is €0. Which of the following statements is (are) true? [JEE-A2-2018] (A) The electric flux through the shell is NfiRX/eo (B) The z-component of the electric field is zero at all the points on the surface of the shell (C) The electric flux through the shell is URX/eo (D) The electric field is normal to the surface of the shell at all points a Q b o 1200
Here f É.dS = ) 2 4 3 E(41T r 2) — Q r 4 3 3 3 where p is volume charge density 41TR3 Therefore orE= 3 3 The field increases linearly with distance form the centre. Case 11 : At an external point To find the electric field outside the charged sphere, we use a spherical Gaussian surface of radius (r >R). This surface encloses the entire charged sphere. So from Gauss's law, we have E(4/T r 2) — Q or E = 2 The field at points outside the sphere is a same as that of a point charge at the centre. Variation of E with the distance from the centre (r) 2 47t€() R O r Exercise:- An early model for an atom considered it to have a positively charged point nucleus of charge Ze, surrounded by a uniform density of negative charge up to a radius R. The atom as a whole is neutral. For this model, what is the electric field at a distance r from the nucleus?
and r is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is +0. Electric flux through another spherical surface of radius e— is R/4 and concentric with the disc is 4). Then the ratio [JEE-A1-2020] 79. A point charge of + 12gC is at a distance 6 cm vertically above the centre of a square of side 12 cm as shown in figure. The magnitude of the electric flux through the square will be x 103 Nm2/C. [JEE (Main)/ES/24-02-2021] 6 cm 80. The electric field in a region is given by E = 2021] 3 —Eoi+—E0j — 5 [JEE (Main)/MS/25-02- The ratio of flux of reported field through the rectangular surface of area 0.2 m2 (parallel to y — z plane) to that of the surface of area 0.3 m2 (parallel to x z plane) is a : b, where a = [Here, i, j, k are unit vectors along x, y and z-axes respectively] 81. A charge 'q' is placed at one corner of a cube as shown in figure. The flux of electrostatic field E through the shaded area is: [JEE (Main)/ES/25-02-2021] (1) (2) (3) 48 80 (4) 82. Given below are two statements : z q x Statement I : An electric dipole is placed at the centre of a hollow sphere. The flux of electric field through the sphere is zero but the electric field is not zero anywhere in the sphere. If R is the radius of a solid metallic sphere and Q be the total charge on it. The Statement Il electric field at any point on the spherical surface of radius r (< R) is zero but the electric flux passing through this closed spherical surface of radius r is not zero. In the light of the above statements, choose the correct answer from the options given below. (1) Statement I is false but Statement Il is true (2) Both Statement I and Statement Il are false (3) Statement I is true but Statement Il is false (4) Both Statement I and Statement Il are true [JEE-M-ES-26-2-2021] 83 The electric field in a region is given by E = —Eoi + —E oj with Eo = 4.0 x 103 N/ C. The flux of this field through a rectangular surface area 0.4 m2 parallel to the Y — Z plane is [JEE-M-ES- 17-03-2021] Nm2 c-l 84. The total charge enclosed in an incremental volume of 2 x 10-9 m3 located at the origin is 2 nC, if electric flux density of its field is found as D = e xsinyi — e-xcosyj + 2zk C/ m [JEE-M-ES-22-07-2021] 85. A solid metal sphere of radius R having charge q is enclosed inside the concentric spherical shell of inner radius a and outer radius b as shown in figure. The approximate variation electric field E as a function of distance r from centre O is given by : [JEE-M-MS-26-08-2021] b a (2) b R a b
72. The electric field E is measured at a point P(O,O,d) generated due to various charge distributions and the dependence of E on d is found to be different for different charge distributions. List-I contains different relations between E and d. List-Il describes different electric charge distributions, along with their locations. Match the functions in List-I with the related char e distributions in List-Il. Q. R. s. List-I E is independent of d 1 d 1 1 JEE-A2-2018 List-Il 1. A point char e Q at the ori •n 2. A small dipole with point charges Q at and - Q at (O,o, - 0. Take 2 R then + = Q/€O C. If h > 2R and r = 4R/5 then + = Q/5€0 D. If h > 2R and r = 3R/5 then + = Q/5€O 75. Two infinite planes each with uniform surface charge density +0 are kept in such a way that the angle between them is 300. The electric field in the region shown between them is given by [JEE (Main)/MS/07-01-2020] [JEE-A1-2019] o (1) 280 o (2) 280 o (3) 280 o (4) 280 x 2 x 2 1- 2 2 2 2
Understanding Concept-I:- 1. A total charge of C is distributed uniformly throughout a 2.7 -cm radius sphere. The volume charge density is: A. 3.7 x 10-7 C/m3 D. 2.5 x 10-4 C/m3 2. Charge is placed on the B. 6.9 x 10-6 C/m3 E. 7.6 x 10-4 C/m3 C. 6.9 x 10-6 C/m2 surface of a 2.7 -cm radius isolated conducting sphere. The surface charge density is uniform and has the value 6.9 x 10-6 C/m2. The total charge on the sphere is: A. 5.6 x 10-10 C D. 6.3 x 10-8 C B. 2.1 10-8 c c. 4.7 x 10-8 c E. 9.5 x 10-3 C 3. A spherical shell has an inner radius of 3.7 cm and an outer radius of 4.5 cm. If charge is distributed uniformly throughout the shell with a volume density of 6. Ix 10-4 C/m3 the total charge is: A. 1.0 10-7 C D. 2.3 x 10-7 C B. 1.3 x 10-7 C E. 4.0 x 10-7 C c. 2.0 x 10-7 c 4. A cylinder has a radius of 2.1 cm and a length of 8.8 cm. Total charge 6.1 x 10-7 C is distributed uniformly throughout. The volume charge density is: A. 5.3 x 10-5 C/m3 B. 5.3 x 10-5 C/m2 D. 5.0 x 10-3 C/m3 E. 6.3 x 10-2 C/m3 C. 8.5 x 10-4 C/m3 5. When a piece of paper is held with one face perpendicular to a uniform electric field the flux through it is 25N•m2/C. When the paper is turned 250 with respect to the field the flux through it is: D. 23N • m2/C B. 12N • m2/C E. 25N • rn2/C C. 21N • m2/C 6. The flux of the electric field (24N/C)i + (30N/C)j + (16N/C)k through a 2.0m2 portion of the yz plane is: A. 32N • m2 /C D. 48N • m2 /C 7. Consider Gauss's law: fÉ.dÄ= q B. 34N • /C E. 60N • /C C. 42N • /C Which of the following is true? A. E must be the electric field due to the enclosed charge B. If q = O, then E = O everywhere on the Gaussian surface C. If the three particles inside have charges of +q, +q, and —2q, then the integral is zero D. on the surface E is everywhere parallel to dA E. If a charge is placed outside the surface, then it cannot affect E at any point on the surface 8. A charged point particle is placed at the center of a spherical Gaussian surface. The electric flux (I)E is changed if: A. the sphere is replaced by a cube of the same volume B. the sphere is replaced by a cube of one-tenth the volume C. the point charge is moved off center (but still inside the original sphere) D. the point charge is moved to just outside the sphere E. a second point charge is placed just outside the sphere 9. Choose the INCORRECT statement: A. Gauss' law can be derived from Coulomb's law B. Gauss' law states that the net number of lines crossing any closed surface in an outward direction is proportional to the net charge enclosed within the surface C. Coulomb's law can be derived from Gauss' law and symmetry D. Gauss' law applies to a closed surface of any shape E. According to Gauss' law, if a closed surface encloses no charge, then the electric field must vanish everywhere on the surface 10. The outer surface of the cardboard center of a paper towel roll: A. is a possible Gaussian surface B. cannot be a Gaussian surface because it encloses no charge C. cannot be a Gaussian surface since it is an insulator D. cannot be a Gaussian surface because it is not a closed surface E. none of the above 11. A physics instructor in an anteroom charges an electrostatic generator to 25 PC, then carries it into the lecture hall. The net electric flux in N • m2/C through the lecture hall walls is:
(Y2 + 1b N/C passes 76 An electric field E = 4x i — through the box shown in figure. The flux of the electric field through surfaces ABCD and BCGF are marked as +1 and +11 respectively. The difference between (+1 — +11) is (in Nm2/C) [JEE (Main)/ES/09-01-2020] z c x 77 Two charged thin infinite plane sheets of uniform surface charge density 0+ and on, where | 0+ | > I I , intersect at right angle. Which of the following best represents the electric field lines for this system? (1) (2) [JEE (Main)/MS/04-09-2020] 0+ 0+ (3) (4) 0+ o_ 78 Consider the force F on a charge 'q' due to a uniformly charged spherical shell of radius R carrying charge Q distributed uniformly over it. Which one of the following statements is true for F, if 'q' is placed at distance r from the centre of the shell ? [JEE (Main)/ES/06-09-2020] (1) (2) (3) (4) — for all r 2 41-11% r > F > O for r < R 4111% R 2 for r > R 2 4111% r for r < R R 2 78 A circular disc of radius R carries surface charge density o(r) 1 where is a constant
(3) b 86. Choose the incorrect statement: b (a) The electric lines of force entering into a Gaussian surface provide negative flux. (b) A charge 'q' is placed at the centre of a cube. The flux through all the faces will be the same, (c) In a uniform electric field net flux through a closed Gaussian surface containing no net charge, is zero. (d) When electric field is parallel to a Gaussian surface, it provides a finite non-zero flux. Choose the most appropriate answer from the options given below: [JEE-M-ES-31-08-2021] (1) (d) Only (2) (c) and (d) Only (3) (a) and (c) Only (4) (b) and (d) Only 87. A cube is placed inside an electric field, E = 150 y j The side of the cube is 0.5 m and is placed in the field shown in the given figure. The charge inside the cube is: (1) 8.3 x 10-11 c (3) 8.3 x 10-12 c 88. If a charge q (B) q/2eo (C) q/4eo (D) q/ 2neo (2) 3.8 x 10-11 c (4) 3.8 x 10-12 c [JEE-M-MS-01-09-2021] as z is placed at the centre of a closed hemispherical non- conducting surface, the total flux passing through the flat surface would be: [JEE-ES-27-06-2022] 89. The volume charge density of a sphere of radius 6 m is 2 PC cm-3. The number of lines of force per unit surface area coming out from the surface of the sphere is x 1010 NC-I. [Given Permittivity of vacuum eo = 8.85 x 10-12 N-l - 90. A long cylindrical volume contains a uniformly distributed charge of density p Cm-3. The electric field inside the cylindrical volume at a distance x =2eo/p from its axis is Vm-l [JED-MS-27-07-2022] 91. A solid sphere is charged such that charge density p varies with radial distance r as P = PO I — for r < R. The electric field at a radial distance r, at point P is por [JED-MS-29-07-2022] 1 1 3 por
c. 2.2 x 105 B. 25 x 10-6 D. 2.8 x 106 E. can not tell unless the lecture hall dimensions are given 12. A point particle with charge q is placed inside the cube but not at its center. The electric flux through any one side of the cube: A. is zero C. is q/ 480 B. is q/ €0 D. is q/6eo E. cannot be computed using Gauss' law 13. A particle with charge 5.0-PC is placed at the corner of a cube. The total electric flux in N • m2 / C through all sides of the cube is: B. 7.1 x 104 D. 1.4 x 105 c. 9.4 x 104 E. 5.6 x 105 14. A point particle with charge q is at the center of a Gaussian surface in the form of a cube. The electric flux through any one face of the cube is: B. q/ 47t €0 E. q/ 12 €0 15. The table below gives the electric flux in N•m2/C through the ends and round surfaces of four Gaussian surfaces in the form of cylinders. Rank the cylinders according to the charge inside, from the most negative to the most positive. cylinder 1: cylinder 2: cylinder 9: cylinder 4: left end +2 X 10—9 +9 X 10—9 —2 X 10—9 +2 x 10-9 right end +4 X 10—9 —2 X 10—9 —5 X 10—9 -b x 10-9 rounded surface —6 X 10—9 +6 X 10—9 +3 X 10—9 -9 x 10-9 B. 4, 3,0, 1 16. A conducting sphere of radius 0.01m has a charge of 1.0 x 10-9 C deposited on it. The magnitude of the electric field in N/ C just outside the surface of the sphere is: B. 450 D. 4500 c. 900 E. 90, 000 17. A round wastepaper basket with a O. 15-m radius opening is in a uniform electric field of 300N [C, perpendicular to the opening. The total flux through the sides and bottom, in N • m2 C, is: c. 21 B. 4.2 D. 280 E. can not tell without knowing the areas of the sides and bottom 18. IOC of charge are placed on a spherical conducting shell. A particle with a charge of placed at the center of the cavity. The net charge on the inner surface of the shell is: c. oc 19. IOC of charge are placed on a spherical conducting shell. A particle with a charge of placed at the center of the cavity. The net charge on the outer surface of the shell is: c. oc -3C is -3C is 20. A 30-N/C uniform electric field points perpendicularly toward the left face of a large neutral conducting sheet. The surface charge density in C/m2 on the left and right faces, respectively, are: A. -2.7 x 10-9 C/m2•, +2.7 x 10-9 C/m2 B. +2.7 x 10-9 C/m2•, -2.7 x 10-9 C/m2 C. -5.3 x 10-9 C/m2•, +5.3 x 10-9 C/m2 D. +5.3 x 10-9 C/m2•, -5.3 x 10-9 C/m2
(1) (3) Por 1- (2) c arge is surrounded by ea rface consisting of an inverted 92. cone of height h and base radius R, and a hemisphere of radius R as shown in the figure. The electric flux through the conical surface is nq/6eo (in SI units). The value of n is [JEE-A2-2022] h
and r is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is +0. Electric flux through another spherical surface of radius e— is R/4 and concentric with the disc is 4). Then the ratio [JEE-A1-2020] 79. A point charge of + 12gC is at a distance 6 cm vertically above the centre of a square of side 12 cm as shown in figure. The magnitude of the electric flux through the square will be x 103 Nm2/C. [JEE (Main)/ES/24-02-2021] 6 cm 80. The electric field in a region is given by E = 2021] 3 —Eoi+—E0j — 5 [JEE (Main)/MS/25-02- The ratio of flux of reported field through the rectangular surface of area 0.2 m2 (parallel to y — z plane) to that of the surface of area 0.3 m2 (parallel to x z plane) is a : b, where a = [Here, i, j, k are unit vectors along x, y and z-axes respectively] 81. A charge 'q' is placed at one corner of a cube as shown in figure. The flux of electrostatic field E through the shaded area is: [JEE (Main)/ES/25-02-2021] (1) (2) (3) 48 80 (4) 82. Given below are two statements : z q x Statement I : An electric dipole is placed at the centre of a hollow sphere. The flux of electric field through the sphere is zero but the electric field is not zero anywhere in the sphere. If R is the radius of a solid metallic sphere and Q be the total charge on it. The Statement Il electric field at any point on the spherical surface of radius r (< R) is zero but the electric flux passing through this closed spherical surface of radius r is not zero. In the light of the above statements, choose the correct answer from the options given below. (1) Statement I is false but Statement Il is true (2) Both Statement I and Statement Il are false (3) Statement I is true but Statement Il is false (4) Both Statement I and Statement Il are true [JEE-M-ES-26-2-2021] 83 The electric field in a region is given by E = —Eoi + —E oj with Eo = 4.0 x 103 N/ C. The flux of this field through a rectangular surface area 0.4 m2 parallel to the Y — Z plane is [JEE-M-ES- 17-03-2021] Nm2 c-l 84. The total charge enclosed in an incremental volume of 2 x 10-9 m3 located at the origin is 2 nC, if electric flux density of its field is found as D = e xsinyi — e-xcosyj + 2zk C/ m [JEE-M-ES-22-07-2021] 85. A solid metal sphere of radius R having charge q is enclosed inside the concentric spherical shell of inner radius a and outer radius b as shown in figure. The approximate variation electric field E as a function of distance r from centre O is given by : [JEE-M-MS-26-08-2021] b a (2) b R a b
21. A solid insulating sphere of radius R contains positive charge that is distributed with a volume charge density that does not depend on angle but does increase with distance from the sphere center. Which of the graphs below might give the magnitude E of the electric field as a function of the distance r from the center of the sphere? distance from the center of a solid charged conducting sphere of radius R? .4 c 23. Charge Q is distributed uniformly throughout an insulating sphere of radius R. The magnitude of the electric field at a point R/ 2 from the center is: C. 3Q/4 E. none of these 24. Positive charge Q is distributed uniformly throughout an insulating sphere of radius R, centered at the origin. A particle with positive charge Q is placed at x = 2R on the x axis. The magnitude of the electric field at x = R/ 2 on the x axis is: C. Q/72 It E. none of these D. 17Q/72 25. Charge Q is distributed uniformly throughout a spherical insulating shell. The net electric flux in N • m2 / C through the inner surface of the shell is: C. 2Q/ 26. Charge Q is distributed uniformly throughout a spherical insulating shell. The net electric flux in N • m2 / C through the outer surface of the shell is: C. 2Q/ 27. A 3.5-cm radius hemisphere contains a total charge of 6.6 x 10-7 C. The flux through the rounded portion of the surface is 9.8 x 104 N • m2 / C. The flux through the flat base is: B. +2.3 x 104 N /C C. -2.3 x 104 N /C D. -9.8 x 104 N • /C E. +9.8 x 104 N /C 28. Charge is distributed uniformly along a long straight wire. The electric field 2 cm from the wire is 20N/C. The electric field 4 cm from the wire is: A. 120N/C C. 40N/C B. 80N/C D. ION/C
(3) b 86. Choose the incorrect statement: b (a) The electric lines of force entering into a Gaussian surface provide negative flux. (b) A charge 'q' is placed at the centre of a cube. The flux through all the faces will be the same, (c) In a uniform electric field net flux through a closed Gaussian surface containing no net charge, is zero. (d) When electric field is parallel to a Gaussian surface, it provides a finite non-zero flux. Choose the most appropriate answer from the options given below: [JEE-M-ES-31-08-2021] (1) (d) Only (2) (c) and (d) Only (3) (a) and (c) Only (4) (b) and (d) Only 87. A cube is placed inside an electric field, E = 150 y j The side of the cube is 0.5 m and is placed in the field shown in the given figure. The charge inside the cube is: (1) 8.3 x 10-11 c (3) 8.3 x 10-12 c 88. If a charge q (B) q/2eo (C) q/4eo (D) q/ 2neo (2) 3.8 x 10-11 c (4) 3.8 x 10-12 c [JEE-M-MS-01-09-2021] as z is placed at the centre of a closed hemispherical non- conducting surface, the total flux passing through the flat surface would be: [JEE-ES-27-06-2022] 89. The volume charge density of a sphere of radius 6 m is 2 PC cm-3. The number of lines of force per unit surface area coming out from the surface of the sphere is x 1010 NC-I. [Given Permittivity of vacuum eo = 8.85 x 10-12 N-l - 90. A long cylindrical volume contains a uniformly distributed charge of density p Cm-3. The electric field inside the cylindrical volume at a distance x =2eo/p from its axis is Vm-l [JED-MS-27-07-2022] 91. A solid sphere is charged such that charge density p varies with radial distance r as P = PO I — for r < R. The electric field at a radial distance r, at point P is por [JED-MS-29-07-2022] 1 1 3 por
E. 5N/C 29. Positive charge Q is placed on a conducting spherical shell with inner radius RI and outer radius R2. A particle with charge q is placed at the center of the cavity. The magnitude of the electric field at a point in the cavity, a distance r from the center, is: A. zero C. q/ 4 It eor2 E. (q + Q)/4 (R12 - r2) D. (q + Q)/ 4 Tt eor2 30. Positive charge Q is placed on a conducting spherical shell with inner radius RI and outer radius R2. A point charge q is placed at the center of the cavity. The magnitude of the electric field at a point outside the shell, a distance r from the center, is: A. zero C. q/ 4 Tt eor2 E. (q + Q)/4 (R12 - r2) 31. Positive charge Q is placed on a conducting spherical shell with inner radius RI and outer radius R2. A point charge q is placed at the center of the cavity. The magnitude of the electric field produced by the charge on the inner surface at a point in the interior of the conductor, a distance r from the center, is: D. q/ 4 Tt eor2 32. A long line of charge with Xl charge per unit length runs along the cylindrical axis of a cylindrical shell which carries a charge per unit length of Xc. The charge per unit length on the inner and outer surfaces of the shell, respectively are: A. Xl and Xc C. — Xl and Xc — Xc E. Xl — Xc and Xc + Xl B. - and Xc + D. Xl + Xc and Xc — Xl 33. Charge is distributed uniformly on the surface of a large flat plate. The electric field 2 cm from the plate is 30N/C. The electric field 4 cm from the plate is: A. 120N/C C. 30N/C E. 7.5N/C B. 80N/C D. 15N/C 34. Two large insulating parallel plates carry charge of equal magnitude, one positive and the other negative, that is distributed uniformly over their inner surfaces. Rank the points 1 through 5 according to the magnitude of the electric field at the points, least to greatest. B. 2, then 1, 3, and 4 tied, then 5 C. 1, 4, and 5 tie, then 2 and 3 tie D. 2 and 3 tie, then 1 and 4 tie, then 5 E. 2 and 3 tie, then 1, 4, and 5 tie 35. Two large parallel plates carry positive charge of equal magnitude that is distributed uniformly over their inner surfaces. Rank the points 1 through 5 according to the magnitude of the electric field at the points, least to greatest. C. 1, 4, and 5 tie, then 2 and 3 tie D. 2 and 3 tie, then 1 and 4 tie, then 5 E. 2 and 3 tie, then 1, 4, and 5 tie 1 1 23 29 36. A particle with charge Q is placed outside a large neutral conducting sheet. At any point in the interior of the sheet the electric field produced by charges on the surface is directed: A. toward the surface C. toward Q E. none of the above B. away from the surface D. away from Q 37. A hollow conductor is positively charged. A small uncharged metal ball is lowered by a silk thread through a small opening in the top of the conductor and allowed to touch its inner surface. After the ball is removed, it will have: A. a positive charge
(1) (3) Por 1- (2) c arge is surrounded by ea rface consisting of an inverted 92. cone of height h and base radius R, and a hemisphere of radius R as shown in the figure. The electric flux through the conical surface is nq/6eo (in SI units). The value of n is [JEE-A2-2022] h
B. a negative charge C. no appreciable charge D. a charge whose sign depends on what part of the inner surface it touched E. a charge whose sign depends on where the small hole is located in the conductor 38. A spherical conducting shell has charge Q. A particle with charge q is placed at the center of the cavity. The charge on the inner surface of the shell and the charge on the outer surface of the shell, respectively, are: 39. A charge Q is uniformly distributed over a large plastic plate. The electric field at a point P close to the centre of the plate is 10 V/ m. If the plastic plate is replaced by a copper plate of the same geometrical dimensions and carrying the same charge Q, the electric field at the point P will become (A) zero (C*) 10 V/m (D) 20 V/m 40. A metallic particle having no net charge is placed near a finite metal plate carrying a positive charge. The electric force on the particle will be (A*) towards the plate (B) away from the plate (C) parallel to the plate (D) zero 41. A thin, metallic spherical shell contains a charge Q on it. A point charge q is placed at the centre of the shell and another in fig. charge ql is placed outside it as shown All the three charges are positive. The force on the charge at the centre is (A) towards left (B) towards right (C) upward (D* ) zero 42. Consider the situation of the previous problem. The force on the centre charge due to the shell is (A) towards left (B*) towards right (C) upward (D) zero 43. Electric charges are distributed in a small volume. The flux of the electric field through a spherical surface of radius 10cm surrounding the total charge is 25 V-m. The flux over a concentric sphere of radius 20cm will be (A*) 25 V-m (B) 50 V-m (C) 100 V-m (D) 200 V-m 44. Fig. shows an imaginary cube of edge L/ 2. A uniformly charged rod of length L moves towards left at a small but constant speed v. At t = O, the left end just touches the centre of the face of the cube opposite it. Which of the graphs shown in fig. represents the flux of the electric filed through the cube at the rod goes through it? Flux d time q 45. A charge q is placed at the centre of the open end of a cylindrical vessel as shown in fig. The flux of the electric field through the surface of the vessel is - (A) zero
46. Mark the correct options: (A) Gauss's law is valid only for symmetrical charge distributions (B) Gauss's law is valid only for charges placed in vacuum (C) The electric field calculated by Gauss's law is the field due to the charges inside the Gaussian surface (D*) The flux of the electric field through a closed surface due to all the charge is equal to the flux due to the charges enclosed by the surface. 47. A positive point charge Q is brought near an isolated metal cube. (A) The cube becomes negatively charged (B) The cube becomes positively charged (C) The interior becomes positively charged and the surface becomes negatively charged. (D*) The interior becomes remains charged free and the surface gets non-uniform charge distribution. 48. A large non-conducting sheet M is given a uniform charge density. Two uncharged small metal rods A and B are placed near the (A*) M attracts A (B*) M attracts B sheet as shown in fig. (C*) A attracts B (D*) B attracts A 49. If flux of the electric field through a closed surface is zero, (A) the electric field must be zero everywhere on the surface (B*) the electric field may be zero everywhere in the surface (C*) the charge inside the surface must be zero (D) the charge in the vicinity of the surface must be zero. 50. An electric dipole is placed at the centre of a sphere, Mark the correct options. (A*) The flux of the electric filed through the sphere is zero (B) The electric field is zero at every point of the sphere (C*) The electric field is not zero anywhere on the sphere (D) The electric field is zero on a circle on the sphere. 51. Figure shows a charge q placed at the centre of a hemisphere. A second charge Q is placed at one of the positions A, B, C and D. In which position(s) of this second charge, the flux of the electric field through the hemisphere remains unchanged? c 52. A closed surface S is constructed around a conducting wire connected to a battery and a switch (fig.) As the switch is closed, the free electrons in the wire start moving along the wire. In any time interval, the number of electrons entering the closed surface S is equal to the number of electrons leaving it. On closing the switch, the flux of the electric field through the closed surface. (A) is increased (B) is decreased (C* ) remains unchanged (D* ) remains zero 53. Fig. shows a closed surface which intersects a conducting sphere. If a positive charged is placed at the point P, the flux of the 6 electric field through the closed surface (A) will remain zero (B*) will become positive (C) will become negative (D) will become undefined closed surface q battery conducting Shere 54. If the electric flux entering and leaving an enclosed surface respectively is (hand +2, the electric charge inside the surface will be: [AIEEE-2003] (1*) (+2 - +1)eo
55. A charged ball B hangs from a silk thread S, which makes an angle q with a large charged conducting sheet P, as shown in the figure. The surface density s of the sheet is proportional to: [AIEEE-2005] (1) sin 0 (2*) tan 0 (3) cos 0 (4) cot 0 56. Consider the charge configuration and a spherical Gaussian surface as shown in the figure. When calculating the flux of the electric / field over the spherical surface, the electric field will be due to charge [JEE 2004] (A) q2 (C) all the charges (B) only the positive charges (D) +ql and -ql 57. Three large parallel plates have uniform surface charge densities as shown in the figure. What is the electric field at P. [JEE- 2005 (Scr)] 40 Comprehension (Q.57 to Q.58) 20 20 (D) o s z=a z=—a z=—2a The nuclear charge (Ze) is non-uniformly distributed within a nucleus of radius R. The charge density p(r) [charge per unit volume] is dependent only on the radial distance r from the centre of the nucleus as shown in figure. The electric field is only along the radial p(r) d direction. 58. The electric field at r = R is: (A) independent of a (C) directly proportional to a2 [JEE 2008] (B) directly proportional to a (D) inversely proportional to a a 59. For a = O, the value of d (maximum value of r as shown in the figure) is . 3 (B) 3 4 Ze (C) 3 nR3 Ze (D) 60. The electric field within the nucleus is generally observed to be linearly dependent on r. This implies (D) a =2R/3 61. A disk of radius a / 4 having a uniformly distributed charge 6C is placed in the x-y plane with its centre at (—a / 2, O, O). A rod of length a carrying a uniformly distributed charge 8C is placed on the x-axis from x = a/ 4 to x = 5a/4. Two point charges —7 C and 3C are placed at (a/4, —a/ 4, O) and (—3a/4, 3a/4, O), respectively. Consider a cubical surface formed by six surfaces x = ± a/ 2, y — ± a/ 2, z = ± a/ 2. The electric flux through this cubical surface is o IOC (C) [JEE-2009] (B) o 12C (D) z (a,O,a) x (a,a,a) 62. Consider an electric field E = Eoi , where Eo is a constant. The flux through the shaded area (as shown in the figure) due to this field is: [JEE-2011] (A) 2Eoa2 (C) Eoa2 2 Eoa (D) x
63. A cubical region of sid a has its centre at the origin. It encloses three fixed point charges, -q at (O,-a/4,O) , + 3q at (0,0,0) and -q at (O, + a/ 4,0). Choose the correct option(s) z a x (A) The net electric flux crossing the plane x = +a/2 is equal to the net electric flux crossing the plane x= -a/ 2. (B) The net electric flux crossing the plane y = +a/2 is more than the net electric flux crossing the plane y = -a/ 2. (C* ) The net electric flux crossing the entire region is q/ (D*) The net electric flux crossing the plane z = +a/2 is equal to the net electric flux crossing the plane + a/2. [JED-2012] 64. Two non-conducting solid spheres of radii R and 2R, having uniform volume charge densities PI and respectively, touch each other. The net electric field at a distance 2 R from the centre of the smaller sphere, along the line joining the centres of the spheres, is zero. The ratiopl/p2 can be (*B) -32/25 (C)32/25 (*D) 4 [JEE-2013 Al] 65. Let El(r), E2(r) and E3(r) be the respective electric fields at a distance r from a point charge Q, an infinitely long wire with constant linear charge density X, and an infinite plane with uniform surface charge density o. If E2(ro ) at a given distance ro, then [JEE-2014 Al] (A) Q— 40 IT r: (C*) E2(ro/2) (D) E2(ro/2) = 4E3(ro/2) 2 ITC 66. Charges Q, 2Q and 4Q are uniformly distributed in three dielectric solid spheres 1, 2 and 3 of radii R/ 2, R and 2R respectively, as shown in figure. If magnitudes of the electric fields at point P at a distance R from the centre of spheres 1, 2 and 3 are El, E2 and E3 respectively, then [JEE-2014 A2] 2 Sphere 1 (A) El > E2 > E3 (B) E3 > El > E2 2Q Sphere 2 67. An infinitely long uniform line charge distribution of charge per unit length lies parallel to the y-axis in the y-z plane at z = —a (see figure). If the magnitude of the flux of 2 the electric field through the rectangular surface ABCD lying in the x-y plane with its center at the origin is (eo o permittivity of free space), then the value of n is [JEE-2015 Al] 4Q Sphere 3 (D) E3 > E2 > El o x 2
68. The figures below depict two situations in which two infinitely long static line charges of constant positive line charge density X are kept parallel to each other. In their resulting electric field, point charges q and —q are kept in equilibrium between them. The point charges are confined to move in the x direction only. If they are given a small displacement about their equilibrium positions, then the correct statement(s) is(are) [JEE-2015 Al] (A) Both charges execute simple harmonic motion. (B) Both charges will continue moving in the direction of their displacement. (C) Charge +q executes simple harmonic motion while charge -q continues moving in the direction of its displacement. (D) Charge -q executes simple harmonic motion while charge +q continues moving in the direction of its displacement. 69. The region between two concentric spheres of radii 'a' and 'b' respectively (see figure), has volume charge density p = A/ r, where A is a constant and r is the distance from the centre. At the centre of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant, is: [JED-M-2016] (1) 2 It b2—a 2 (3) 2 (2) 2 (4) 2 2 IT a 70. A point charge +Q is placed just outside an imaginary hemispherical surface of radius R as shown in the figure. Which of the following statements is/ are correct? [JEE-A2-2017] (A) Total flux through the curved and the flat surface is Q/ €0. (B) The component of the electric field normal to the flat surface is constant over the surface (C) The circumference of the flat surface is an equipotential (D) The electric flux passing through the curved surface of the Q 1 hemisphere is 71. An infinitely long thin non-conducting wire is parallel to the z-axis and carries a uniform line charge density X. It pierces a thin non-conducting spherical shell of radius R in such a way that the arc PQ subtends an angle 1200 at the centre O of the spherical shell, as shown in the figure. The permittivity of free space is €0. Which of the following statements is (are) true? [JEE-A2-2018] (A) The electric flux through the shell is NfiRX/eo (B) The z-component of the electric field is zero at all the points on the surface of the shell (C) The electric flux through the shell is URX/eo (D) The electric field is normal to the surface of the shell at all points a Q b o 1200
72. The electric field E is measured at a point P(O,O,d) generated due to various charge distributions and the dependence of E on d is found to be different for different charge distributions. List-I contains different relations between E and d. List-Il describes different electric charge distributions, along with their locations. Match the functions in List-I with the related char e distributions in List-Il. Q. R. s. List-I E is independent of d 1 d 1 1 JEE-A2-2018 List-Il 1. A point char e Q at the ori •n 2. A small dipole with point charges Q at and - Q at (O,o, - 0. Take 2 R then + = Q/€O C. If h > 2R and r = 4R/5 then + = Q/5€0 D. If h > 2R and r = 3R/5 then + = Q/5€O 75. Two infinite planes each with uniform surface charge density +0 are kept in such a way that the angle between them is 300. The electric field in the region shown between them is given by [JEE (Main)/MS/07-01-2020] [JEE-A1-2019] o (1) 280 o (2) 280 o (3) 280 o (4) 280 x 2 x 2 1- 2 2 2 2
(Y2 + 1b N/C passes 76 An electric field E = 4x i — through the box shown in figure. The flux of the electric field through surfaces ABCD and BCGF are marked as +1 and +11 respectively. The difference between (+1 — +11) is (in Nm2/C) [JEE (Main)/ES/09-01-2020] z c x 77 Two charged thin infinite plane sheets of uniform surface charge density 0+ and on, where | 0+ | > I I , intersect at right angle. Which of the following best represents the electric field lines for this system? (1) (2) [JEE (Main)/MS/04-09-2020] 0+ 0+ (3) (4) 0+ o_ 78 Consider the force F on a charge 'q' due to a uniformly charged spherical shell of radius R carrying charge Q distributed uniformly over it. Which one of the following statements is true for F, if 'q' is placed at distance r from the centre of the shell ? [JEE (Main)/ES/06-09-2020] (1) (2) (3) (4) — for all r 2 41-11% r > F > O for r < R 4111% R 2 for r > R 2 4111% r for r < R R 2 78 A circular disc of radius R carries surface charge density o(r) 1 where is a constant
and r is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is +0. Electric flux through another spherical surface of radius e— is R/4 and concentric with the disc is 4). Then the ratio [JEE-A1-2020] 79. A point charge of + 12gC is at a distance 6 cm vertically above the centre of a square of side 12 cm as shown in figure. The magnitude of the electric flux through the square will be x 103 Nm2/C. [JEE (Main)/ES/24-02-2021] 6 cm 80. The electric field in a region is given by E = 2021] 3 —Eoi+—E0j — 5 [JEE (Main)/MS/25-02- The ratio of flux of reported field through the rectangular surface of area 0.2 m2 (parallel to y — z plane) to that of the surface of area 0.3 m2 (parallel to x z plane) is a : b, where a = [Here, i, j, k are unit vectors along x, y and z-axes respectively] 81. A charge 'q' is placed at one corner of a cube as shown in figure. The flux of electrostatic field E through the shaded area is: [JEE (Main)/ES/25-02-2021] (1) (2) (3) 48 80 (4) 82. Given below are two statements : z q x Statement I : An electric dipole is placed at the centre of a hollow sphere. The flux of electric field through the sphere is zero but the electric field is not zero anywhere in the sphere. If R is the radius of a solid metallic sphere and Q be the total charge on it. The Statement Il electric field at any point on the spherical surface of radius r (< R) is zero but the electric flux passing through this closed spherical surface of radius r is not zero. In the light of the above statements, choose the correct answer from the options given below. (1) Statement I is false but Statement Il is true (2) Both Statement I and Statement Il are false (3) Statement I is true but Statement Il is false (4) Both Statement I and Statement Il are true [JEE-M-ES-26-2-2021] 83 The electric field in a region is given by E = —Eoi + —E oj with Eo = 4.0 x 103 N/ C. The flux of this field through a rectangular surface area 0.4 m2 parallel to the Y — Z plane is [JEE-M-ES- 17-03-2021] Nm2 c-l 84. The total charge enclosed in an incremental volume of 2 x 10-9 m3 located at the origin is 2 nC, if electric flux density of its field is found as D = e xsinyi — e-xcosyj + 2zk C/ m [JEE-M-ES-22-07-2021] 85. A solid metal sphere of radius R having charge q is enclosed inside the concentric spherical shell of inner radius a and outer radius b as shown in figure. The approximate variation electric field E as a function of distance r from centre O is given by : [JEE-M-MS-26-08-2021] b a (2) b R a b
(3) b 86. Choose the incorrect statement: b (a) The electric lines of force entering into a Gaussian surface provide negative flux. (b) A charge 'q' is placed at the centre of a cube. The flux through all the faces will be the same, (c) In a uniform electric field net flux through a closed Gaussian surface containing no net charge, is zero. (d) When electric field is parallel to a Gaussian surface, it provides a finite non-zero flux. Choose the most appropriate answer from the options given below: [JEE-M-ES-31-08-2021] (1) (d) Only (2) (c) and (d) Only (3) (a) and (c) Only (4) (b) and (d) Only 87. A cube is placed inside an electric field, E = 150 y j The side of the cube is 0.5 m and is placed in the field shown in the given figure. The charge inside the cube is: (1) 8.3 x 10-11 c (3) 8.3 x 10-12 c 88. If a charge q (B) q/2eo (C) q/4eo (D) q/ 2neo (2) 3.8 x 10-11 c (4) 3.8 x 10-12 c [JEE-M-MS-01-09-2021] as z is placed at the centre of a closed hemispherical non- conducting surface, the total flux passing through the flat surface would be: [JEE-ES-27-06-2022] 89. The volume charge density of a sphere of radius 6 m is 2 PC cm-3. The number of lines of force per unit surface area coming out from the surface of the sphere is x 1010 NC-I. [Given Permittivity of vacuum eo = 8.85 x 10-12 N-l - 90. A long cylindrical volume contains a uniformly distributed charge of density p Cm-3. The electric field inside the cylindrical volume at a distance x =2eo/p from its axis is Vm-l [JED-MS-27-07-2022] 91. A solid sphere is charged such that charge density p varies with radial distance r as P = PO I — for r < R. The electric field at a radial distance r, at point P is por [JED-MS-29-07-2022] 1 1 3 por
(1) (3) Por 1- (2) c arge is surrounded by ea rface consisting of an inverted 92. cone of height h and base radius R, and a hemisphere of radius R as shown in the figure. The electric flux through the conical surface is nq/6eo (in SI units). The value of n is [JEE-A2-2022] h

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Note On Gauss's Law

Naveen B / Chandigarh

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