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Electric Field

Home > Study Notes > Electric Field
Published in: AIEEE | IIT JEE Mains | Physics
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Notes On Electric Field.

Akhilesh K / Lucknow

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Qualification: M.Sc (NIT Rourkela - 2019)

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  1. ELECTRIC FIELD An electric field is defined as a region in which there should be a force on a charge brought into that region. Whenever a charge is being placed in an electric field, it experiences a force Electric fields are usually produced by different types of charged bodies, point charges, charged plates, charged spheres etc If two point charges are placed as shown in figure, we can describe the force on them in two ways (i) The charge q2 is in the electric field of charge qi. Thus the electric field of charge qi exerts force on Q. (ii) The charge qi is in the electric field of charge Q. Hence the electric field of charge q2 exerts a force on qi Electric field intensity or Electric field Strength (E) The electric field intensity at a point in an electric field is the force experienced by a unit positive charge placed at that point, it is being assumed that the unit charge does not affect the field. Thus, if a positive test charge qo experiences a force F at a point in an electric field, then the electric field intensity E at a point is given by Important points regarding electric field intensity (i) It is vector quantity. The direction of the electric field intensity at a point inside the electric field is the direction in which the electric field exerts force on the unit positive charge. (ii) Direction of electric field due to positive charge is outward while direction of electric field due to negative charge is inward (iii) Dimensions of electric field intensity E = [MLT 3A 1] S.l. unit of Electric field is C/C or V/m as N C m
  2. Force exerted by a field on a charge inside it By definition E qo If qo is positive charge force F on it is in the direction of E If qo is negative charge force F on it is opposite to the direction of E G) Electric field due to point charge Let a positive test charge qo be placed at a distance r from a point charge q. The magnitude of force acting on qo is given by Coulomb's law. 1 qoq 4TtEo r2 'The magnitude of the electric field at the site of the charge is 2 qo 47TEo r Electric field in vector form is 3 4TTEo r Here F is the position vector of point The direction of E is the same s the direction of F , along a radial line q, point outward if q is positive and inward is q is negative. The figure given shows the direction of the electric field E at various points near a positive point charge. Note length of arrow is more where electric field is more and point near to charge have more electric field 2 3 p 1 1
  3. Electric field intensity due to a group of point charges Since the principle of linear superposition is valid for Coulomb's law, it is also valid for the electric field. To calculate the electric field at a point due to a group of N point charges. We find the individual field strength El due to QI, E2 due to Q2 and so on.. 'The resultant field strength is the vector sum of individual field strengths É=É1+É2+É3+... Electric field at a point depends on the charge and position of point qn with position vectors Fl, F2, F3, —-rn relative Consider a system of charges qi, Q, CB to some origin O. Electric field at point P having position vector F . For this purpose place a very small test charge qo at that point and use the superposition principle. Electric field at point P due to qi is given by Electric field at point P due to q2 is given by IF-F213 Electric field at point P due to q3 is given by qo Same way, electric field at point P due to charge qn is Ir—rnl 122) According to superposition principle, net electric field at a point P is q2 (F — 722) + + k k r Ir — ri13 IF — 72213 IF— Fjl IF — Fn13 Here m, Q, q3 Are the sources of electric field Phvsical significance of electric field a) Equation for electric field is given by E-SL
  4. Equation of force acting on a unit positive charges at point once EF is kown, we do not have to worry about the source of electric field. In this since, the electric field itself is a special representation of the system of charges producing electric field, as far as the effect on other charges are concerned. Once the representation is done, the force acting on charge q kept at that point in the electric field can be determined using following equation FF = qEF b) True significance of electric field is when there is an accelerated motion of charge qi and Q. For effect of qi and q2 there will be a time delay between 'the force on q2 and the cause (motion of qi) . The field picture is this: the accelerated motion of charge qi produces electromagnetic waves, which then propagate with the speed of light and reaches q2 and cause a force on Q. The notation of field elegantly accounts for the time delay. Thus even though electric and magnetic fields can be detected only by their effect ( force) on charge, they are regarded as physical entity. Electric and magnetic field transport energy. Thus, a source of time-dependent electromagnetic fields, turned on briefly and switch off, leaves behind propagating electromagnetic field transporting energy ELECTRIC CHARGE DUE TO CHARGE UNIFORMALY CHARGED RING AT A POINT ON AXIS OF RING— x z dEsin O dEsin 2 d Ecos O x Let us consider a charge Q distributed uniformly on a thin, circular, non-conducting ring of radius a. We have to find electric field E at the point P on the axis of the ring, at a distance x from the centre. From symmetry we observe that every element dq be paired with a similar element on the opposite side of the ring. Every component dEsin0 perpendicular to the x-axis is thus cancelled by a component dEsin0 in the opposite direction. In a summation process, all the perpendicular components add to zero. Thus we only add dEx components
  5. x Now r2 = a2 + x2 and cose a2+x 2 1 dq 2 4TtEo r 1 4TEo (d2 + x2) IHence, the resultant electric field at P is given by dEx - dEcos9 1 dq x 41TEo (d2 + x2) a2 + x2 1 x dq 41TEo (d2 + x2)3/2 As we integrate around the ring, all the terms remain constant and dq — q 1 xQ 47TEo (d2 + X2)3/2 Electric field due to a uniformly charged disc at a point on the axis of the disc x Let us consider a flat, circular, non-conducting thin disc of radius R having a uniform surface charge density aC/m2. We have to find the electric field intensity at an axial point at a distance x from the disc. Let O be the centre of a uniformly charged disc of radius R and surface charge density o. Let P be an axial point, distance x from O, at which electric field intensity is required. From the circular symmetry of the disc, we imagine the disc to be made up of large number of concentric rings and consider one such ring of radius 'r' and infinitesimally small width dr. The area of the elemental ring = Circumference X width = (2Ttr dr) The charge dq on the elemental ring = (2Ttr dr) o Therefore, the electric field intensity at P due to the elementary ring is given by 1 x(2Ttr dr) o 4TTEo (r2 + X2)3/2 And is directed along x —axis. Hence, the electric intensity E due to the whole disc is
  6. xo 2E0 xo r dr o (r2 + x2)3/2 —1 (r 2 + X 2) 1/2 —1 1 2Eo (R2 + X2)1/2 X 1 (R2 + x2)1/2 Electric lines of Forces The properties of electric lens of forces are the following (i) The electric lines of force are continuous curves in an electric field starting from a positively charged body and ending on a negatively charged body. (ii) The tangent to the curve at any point gives the direction of the electric field intensity at that point (iii) Electric field lines of forces do not pass but leave or end on a charged conductor normally. Suppose the Iline of forces is not perpendicular to the conductor surface. In this situation, the component of electric field parallel to the surface cause the electrons to move and hence conductor will not remain equipotential which is absurd as electrostatics conductor is an equipotential surface. (iv) Electric field line of forces never intersects since if they cross at a point, electric field intensity at that point will have two directions, which is not possible. (v) The number of electric lines of force that originate from or terminate on a charge is proportional to the magnitude of the charge. (vi) As number of lines of force per unit area normal to the area at point represents magnitude of intensity, crowded lines represent strong field while distant lines weak field. Further, if the lines of force are equidistant straight lines, the field is uniform. Electric Dipole A system of equal and opposite charges, separated by a finite distance is called as an electric dipole. -q
  7. where p is the unit vector along the dipole axis ( from —q to +q) Electric field at point P due to positive charge is given by As shown in figure, the two electric charges of electric dipole are +q and —q and distance between them is 2a. Electric dipole moment (P) of the system can be defined as follows P = q(2d) Importatnt points regarding electric dipole (i) The Sl unit of electric dipole is coulomb metre ( C m) (ii) Electric dipole is a vector quantity and its direction is from negative charge (-q) to positive charge (+q) (iii)The net electric charge on an electric dipole is zero but its electric field is not zero, since the position of the two charges is different . If lim oo and 0 in P = q(2d), then electric dipole is called point dipole. Electric field of a dipole The electric field of the pair of charges (-q and +q) at any point in space can be found out from Coulombs law and the superposition principle. The resultants are simple for the following two cases: (i) When the point is on the dipole axis (ii) When it is in the equatorial plane (i) Electric field due to dipole for point on the axis Let the point P be at distance r from the centre of the dipole on the side of the charge q as shown in figure, then G) -q 2a Elctric field due to negative charge at point P is given by 47TEo (r2 — a2)2 q q 4TtEo(r + Total electric field at P E— q 41TEo But 2qa = P For r a 1 47TE (r 1 2pr 1 47TEo
  8. (ii) Electric field due to dipole for point on the equatorial plane Let the point P be at distance r from the centre of the dipole on the equatorial plane as shown in figure. Then Magnitude of electric field at point P due to positive charge is given by 1 q +q — 47TE0 r 2 + a 2 Magnitude of electric field at point P due to negative charge p is given by -q 1 q Both magnitude are equal . Directions of E-q and E+q are as Shown in fig. Clearly componants perpendicular to axis cancel away. The componant along the dipole axis add up. The total electric field is opposite to p . We have E = (E+q + E-q)cos0P 2 q cosØP 47TEo r2 + a2 From figure a cos9 — r2 + a2 2 q a 4TtE r2 +a2 r2 + a2 2qa 4TtEo(r2 + a2)3/2 At large distance ( r>>a), this reduces to 2qa 47TE (r)3 1 4TTEo (r)3 Important points (i) The electric field at large distances falls of as 1/r3 (ii) The magnitude and direction of the dipole field depends not only on the distance r but also on the angle between the position vector r and dipole moment p (iii) Electric field of dipole is cylindrically symmetrical.
  9. Dipole in uniform external electric field As shown in figure an electric dipole of magnitude p = q(2a) is kept in a uniform electric {G) q field Let 0 be the angle between dipole momentum F and electric field E . The force qE and -qE are acting on the charges q and —q respectively. These forces are equal but opposite in direction. The resultant force being zero keeps dipole in translational equilibrium. But, the two forces have different line of action, hence the dipole will experience a torque. When the net force is zero, the torque (couple) is independent of the origin. Its magnitude is equal to the magnitude of the force multiplied by the arm of the couple ( perpendicular distance between the two antiparallel forces) Magnitude of torque = qEX2asin0 = 2qaEsin0 but q(2a) = p Thus Direction of torque is perpendicular to paper, coming out. The torque rotates the dipole in such a way that the angle 0 reduces, (i) when the dipole aligns itself along the direction of electric field, the torque becomes zero. This is the normal position of dipole in electric field. (ii)lf the dipole is to be rotated by an angle 0 from this position, work is required to be done against torque. This work is equal to the change in potential energy of dipole. Behaviour of electric dipole in non-uniform electric field If the electric field is not uniform, the intensity of electric field will be different at different points as a result electric force acting on the positive charge and negative charge of the dipole will also be different. In this situation the net force and torque are acting on the dipole. As a result dipole experiences a linear displacement in addition to roatation. This rotation will continue only till the dipole aligns in the direction of the electric field. But linear motion will continue. When charged comb is brough near the piece of parer, the non-uniform electric field is produced by the comb. Electric dipole is induced along the direction of non-uniform eletric field in small piece of paper. Now non uniform electric field exters a net force on piece of paer and paper moves in the direction of comb.
  10. Flux of an electric field or Electric flux Let us consider a plane surface of area S placed in an electric fied E. Electric flux through an elementary area dS is defind as the scalar product of dS and E .i.e d E . dS, where dS is the area vector, whose magnitude is the area dS of the element and whose direction is along the outward normal to the elementary area. Hence, the electric flux through the entire surface is given by . or = f E dS cose 0 is the angle between area vector and electic field Area vector is always outward for closed surface. While for other surface it can be considered as inward outward. If the electric field is uniform, then = E cos0f dS When the electric flux through a closed surface is required, we use a small circular sign on the integration symbol Thus general defination of electic flux can be given as " The flux linked with any surface is the surface integration of the electric field over the given surface" Unit of flux N-m/C Importatnt points regarding electric flux (i) The number of lines of force passing normally to the given area gives the measure of flux of electric field over the given area. (ii) It is a real scalar physical quantity with units ( volt xm) (iii) It will be maximum when cose = max = 1. i.e 0 = 00, i.e. electric field is normal to the surface with (d +E)max = E(dS) (iv) It will be minimum when Icos01 = min = 0, = 900 i.e. field parallel the area with (v) For closed surface, is positive if the lines of force point outward everywhere (E will be outward everywhere, e < 900 and E . dS will be positive) and negative if they point inward (E will be inwards everywhere, e > 900 and E . dS will be negative) Positive flux G) Negative flux + < O
  11. Gauss's Law This law gives a relation between the electric flux through any closed hypothetical surface ( called gaussian surface) and the charge enclosed by the surface. It states "The electric 1 flux through any closed surface is equal to — times the /net' charge enclosed by the surface" That is Where q denotes the algebric sum of all the charges enclosed by the surface. If there are several charges +ql, +q2, +q3 , -q4, -qs ...etc inside the Gaussian surface then It is clear form above equation that flux inked with a closed body is independent of the sphape and size of the body and position of charge inside it The law implies that the total elecric flux through a closed surface is zero, if no net charge is enclosed by the surface Important points rega rding law (i) Gaus's law is true for any closed surface, no matter what its shape or size (ii) The term q on the right side of the eqation , include the sum of all charges enclosed by the surface. The charges may be located any where inside the surface (iii) The electric field appearing on the left hand side of the equation(l) is the electric field produced due to a system of charges, whether enclosed by the surface or outside it (iv) The surface that we choose for application of Gauss's law is called Gaussian surface (v) Gauss law is useful towards a much easier calculation of electric field when system has some symmetry. Applications of Gauss's Law Gauss's law is useful when there is symmetry in the charge distribution, as in the case of uniformly charged sphere, long cylinders and flat surface over which the surface integral gives by equation (1) can be easily evaluated These are steps to apply the Gauss's law (i) Use symmetry of the charge distribution to determine the pattern of the lines (ii) Choose a Gaussian surface for which E is either parallel to dS or perpendicular to dS (iii) If E is parallel to dS , then the magnitude of E should be constant over this part of the surface. The integral then reduces to sum over area elements. Field due to an infinite line of charge Consider an infinite line of charge has a linear charge density X. Using Gauss's law, let us find the electric field at a distance 'r' from the line. The cylindrical symmetry tells us that the field strength will be the same at all points at a fixed distance r from the line.
  12. Since the line is infinite and uniform, for every charge element on the other side the dl p p dl component along the line of the fields due to all such element cancels in pairs. Thus the field lines are directed radially outward, perpendicular to the line charge. Also perpendicular distance from line is same magnitude at all points PI,P2 will be same The appropriate choice of Gaussian surface is a cylinder of radius r and length L. On the flat faces S2 and S3 , E is perpendicular to dS, which means no flux cross them. On curved surface Sl, E is parallel to dS so that = EdS. The charge enclosed by the cylinder is Q = XL dS 1 Applying Gauss's law to the curved surface, we have EfdS = E(2nrL) — dS dS Vectorially , E 1 2nrE() = here F is the radial unit vector in 2TtE0 r the plane normal to the wire passing through the point. 31
  13. Field due to an infinite plane sheet of charge (5 Let us consider a thin non-conducting plane sheet of charge, infinite in extent and having a surface charge density o C/m2 . Let point P be a point at distance r from the sheet, at which the electric intensity is required. Let us choose a point P' symmetrical with P, on the other side of sheet. Let us now draw a Gaussian surface to be rectangular parallelepiped of cross sectional area A, as shown in figure By symmetry, the electric field at all points on either side near the sheet will be perpendicular to the sheet, directed outward ( if sheet is positively charged). 'Thus E is perpendicular to the plane ends contain point P and P' .Also magnitude of E will be same at P and P'. Therefore, the flux passing through the points containing P and P' É.ä3+ EdS + EdS ØE = 2EA The flux through remaining surface is zero because E is perpendicular to dS and do not contribute to the total flux The charge enclosed by the Gaussian surface shown by shaded area. q = OA Applying Gauss's law, we have 2EA = OA Electric field intensity is independent of the distance from an infinite sheet of charges
  14. Electric field due to a uniformly charged spherical shell Using Gauss's law, we can find the intensity of the electric field due to a uniformly charged dS spherical shell or a solid conducting sphere at Case(l): At an external point In an isolated charged spherical conductor an excess charge on it is distributed uniformly + ix---—Gaussian over its surface. Since charge lines is radially Surface 32 outward. Also, field strength will have the same value at all points on any imaginary spherical surface concentric with the charged conducting sphere or 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 E , so E. dS = EdS at all points on the surface According to Gauss's law E. dS EdS=E dS = E(4Ttr2) - 47TEo r2 For all points outside the charged conducting sphere or the charge spherical shell, the field is same as that appoint charge at the centre
  15. Case(ii) At an internal point (r
  16. electric field has the same value at every point on the Gaussian surface, and the direction of E is radial at every point on the surface So, applying Gauss's law Here EdS - 4 E dS = E(4nr2) — 4 — Ttr 3 3 — ma 4TR3 3 E (4mr2) 3 3 47TEo R3 Or 380 Case (ii) At an external point ( r ) To find the electric field outside the charged sphere, we use a spherical Gaussian surface of radius r ( r>R). This surface encloses the entire charged sphere, So, from Gauss's law, we have E (4m-2) 4TCEor2 Or 3r2Eo Variation of E with the distance from centre (r) 2
  17. N/A

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