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Potential Class XII

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Published in: IIT JEE Mains
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Notes On Potential.

Akhilesh K / Lucknow

4 years of teaching experience

Qualification: M.Sc (NIT Rourkela - 2019)

Teaches: All Subjects, English, Mathematics, Science, Chemistry, Physics, Algebra, IIT JEE Mains, AIPMT, NEET

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  1. ELECTROSTATIC POTENTIAL ELECTRIC FIELD IS CONSERVATIVE In an electric field work done by the electric field in moving a unit positive charge from one point to the other, depends only on the position of those two points and does not depend on the path joining them. ELECTROSTATIC POTENTIAL Electrostatic potential is defined as "Work required to be done against the force by electric field in bringing a unit positive charge from infinite distance to the given point in the electric field us called the electrostatic potential (V) at that point" According to above definition the electric potential at point P is given by the formula Electric potential is scalar quantity. Sl units (J/C) called as volt (V) POTENTIAL AT APOINT DUE TO A POINT CHARGE Consider a point charge positive Qat the origin. To deter let P be the point at a distance 'r' from origin of coordinate axis . Since work done in electric field is independent of path, we will consider radial path as shown in figure. According to definition of electric potential we can use the equation p dr 1 1 Q 1 And electric field E is given by 4780 r -1 ELECTRIC POTENTIAL DUE TO GROUP OF POINT CHARGE The potential at any point due to group of point charges is the algebraic sum of the potentials contributed at the same point by all the individual point charges V = VI +V2 + V3 +
  2. ELECTRIC POTENTIAL DIFFERENCE Electric potential difference is defined as "Work required to be done to take a unit positive charge from one point (say P) to another point (say Q) against the electric field According to formula for potential at point P Thus potential at point Q is given by From above formula potential difference between points Q and P is given by E • dr Or work done in moving charge from point P to point Q Sl unit of potential is 'V and dimensional formula [M LT A ] ELECTROSTATIC POTENTIAL ENERGY The electric potential energy is defined as "The work required to be done against the electric field in bringing a given charge (q), from infinite distance to the given point in the electric field motion without acceleration is called the electric potential energy of that charge at that point." From definition of electric potential energy and the electric potential we can write electric potential energy of charge q at point P, as qÉ. dr = q E. dr = qVp
  3. The absolute value of the electric potential energy is not at all important, only the difference in its value is important. Here, in moving a charge q, from point P to Q, without acceleration, the work required to be done by the external force, shows the difference In the electric potential energies (UQ - Up) of this charge q, at those two points. [JQ — Up = —q Éi dr Electric potential energy is of the entire system of the sources producing the field and the charge, 'for some configuration, and when the configuration changes the electric potential energy of the system also changes. POTENTIAL ENERGY OF A SYSTEM OF TWO POINT CHARGES The potential energy possessed by a system of two point charges qi and q2 separated by a distance 'r' is 'the work done required to bring them to this arrangements from infinity. This electrostatic potential energy is given by 4TtEor ELECTRIC POTENTIAL ENERGY OF A SYSTEM OF POINT CHARGES The electric potential energy of such a system is the work done in assembling this system starting from infinite separation between any two-point charges. For a system of point charges qi, Q, q3 .... qn the potential energy is 47804 It simply means that we have to consider all the pairs that are possible Important points regarding electrostatic potential energy (i) Work done required by an external agency to move a charge q from A to B in an electric field with constant speed (ii) When a charge q is let free in an electric field, it loses potential energy and gains kinetic energy, if it goes from A to B, then loss in potential energy = gain in kinetic energy Or ELECTRIC POTENTIAL DUE TO CONTINUOUS CHARGE DISTRIBUTION The electric potential due to continuous charge distribution is the sum of potential of all the infinitesimal charge elements in which the distribution may be divided dq v = fdv=J—— 4rteor
  4. ELECTRIC POTENTIAL DUE TO A CHARGED RING A charge Q is uniformly distributed over the circumference of a ring. Let us calculate the electric potential at an axial point at a distance r from the centre of the ring. The electric potential at P due to the charge element dq of the ring is given by dq Hence electric potential at P due to the uniformly charged ring is given by 1 dq 1 dq ELECTRIC POTENTIAL DUE TO A CHARGED DISC AT A POINT ON THE AXIS A non-conducting disc of radius 'R' has a uniform surface charge density o C/m2to calculate the potential at a point on the axis of the disc at a distance from its centre. Consider a circular element of disc of radius x' and thickness dx. All points on this ring are at the same distance Z , from the point P. The charge on the ring is dq = OA o(2nxdx) and so the potential due to the ring is 1 c(2zxdx) 4760 Z 4760 x2 + r Since potential is scalar, there are no components. The potential due to the whole disc is given by 28 280 280
  5. For large distance R/r
  6. (b) At an internal point At point inside the shell, E = O. So work done in bringing a unit positive charge from a point on the surface to any point inside the shell is zero. Thus, the potential has a fixed value at all points within the spherical shell and is equal to the potential at the surface. 4760 R 4rco R Above results hold for a conducting sphere also whose charge lies entirely on the outer surface. ELECTRIC POTENTIAL DUE TO A NON-CONDUCTING CHARGED SPHERE A charge Q is uniformly distributed through a non-conducting volume of radius R. (a) Electric potential at external point is given by equation. 'r' is the distance of point from the center of the sphere 1 (b) Electric potential at an internal point is given by equation 1 Q r 3R2_r21 2R3 L Here R is the radius of the sphere and r is the distance of point from the centre RELATION BETWEEN THE ELECTRIC FIELD AND ELECTRIC POTENTIAL We know that electric potential from electric field is given by And potential difference between two points is given by If points P and Q are very close to each other, then for such a small displacementdr, integration is not required and only term E. dr can be kept thus dV = - E. dr
  7. I) If dr, is the direction of electric field E , E. dr = Edrcos9 = Edr dV = - Edr This equation gives the magnitude of electric field in the direction of displacement dr . = potential difference per unit distance. It is called the potential gradient. It unit Here is Vm-l, which is equivalent to N/C Il) If dr, is not in the direction of E , but in some other direction, the — would give us Idr the component of electric field in the direction of that displacement If electric field is in X direction and displacement is in any direction ( in three dimensions) then E = Ext and dr = dxt + dyj + dznc dV = - (Exi + dyj + dzk) = -Exdx dV dx Similarly, if the electric field is Y and only in Z direction respectively, we would get dV dy and dV dz Now if the electric field also have three ( x,y,z) components then ax And E = — , — , — shows the partial differentiation of V(x, y, z) with respect to x, y, z Here ax ay Dz respectively. Moreover, the potential differentiation of V(x, y, z) with respect to x means the differentiation of V with respect to x only, by taking y and z in the formula of V as constant EQUIPOTENTIAL SURFACE An equipotential surface is that surface on which the electric potentials at all points are equal Important points regarding equipotential surfaces (i) The line of forces are always normal to equipotential surface. (ii) The net work done in taking a charge from A to B is zero, if A and B are on same equipotential surface.
  8. Suppose a unit positive charge is given a small displacement dl on the equipotential surface from a given point. In this process the work required to be done against the electric field is dW = -E. dl = potential difference between those two point But the potential difference on the equipotential surface = O dl = O — E dl cos0 = O, where 0 = angle between E and dl But E and dl COS 0 = 0 = Tt /2 But dl is along this surface. Hence electric field is normal to the equipotential surface at that point (iii) Equipotential surface never intersect each other. If they intersect then electric field lines will also intersect which is not possible. ELECTRIC POTENTIAL DUE TO DIPOLE o p B Let a dipole consisting of equal and opposite charge q separated by a distance 2a. Let zero of coordinate system is at centre of the dipole as shown in figure. Let p be any point in x-y plane. Let PO = r. AP = r- BP = Let r makes angle of 0 with the axis of dipole. Potential at point P is the sum of potential due at point P due to —q and +q charges. BM is perpendicular on OP and AN is perpendicular on ON 1 q = acosO r+ 4780 r 1 1 From figure r — r + ON but ON — r + acosO Similarly r = r. + 0M but 0M = acose = + a cosO = r — acose r- - = 2acosO Substituting the values we get
  9. q 1 p cos 2a cos 0 (r +a cos e) (r — a cos d) 2a cos (r2 —a2 cos2 e) 2qa cos (r 2 —az cos2 e) 2 Since r >>2a Case l) Potential on the axis : For point on the axis of the dipole 0 or TT — 4TtEo r2 From the given point, if the nearer charge is +q, then we get V as positive. And if it is —q, then we get V as negative Case Il) Potential on the equator From a point on the equator 0 =n/2 V= 0 POTENTIAL ENERGY OF DIPOLE When a dipole is placed in external uniform electric field E , a torque i = x E acts on it. If we rotate the dipole through a small angle dB, the work done by torque is dW = T dB or dW = -PEsin6 d9 The work is negative as the rotation d0 is opposite to the torque. 'The change in electric potential energy of the dipole is therefore dlJ -PEsine dO
  10. If dipole is rotated from angle 61 to 62, then J dU PE sin O dO U ( g ) — U (q) = pE[— cos (J (02 ) — U (q) = —PE (cos — cos q) Work done by external force = U(02) — IJ(OI) OR Wext = cos 91 - cos 92) Work done by electric force - Wext = COS 02 - COS 01) Welectric force — PERIODIC TIME OF DIPOLE When a dipole is suspended in a uniform electric field, it will align itself parallel to the field. Now if it is given a small angular displacement 0 about its equilibrium, the restoring couple will be C = -pEsin0 Or C = -PEO [as for small 0, sino = 0] Also couple d20 Thus = -PEO d20 9 Comparing above equation with standard equation for SHM we get 2 Thus periodic time 1 T = 27T
  11. a) EFFECT OF EXTERNAL FIELD ON CONDUCTOR Iln a metallic conductor there are positive ions situated at the lattice points and the free electrons are moving randomly between these ions. They are free to move within the metal but not free to come out of the metal. When such a conductor is placed in an external electric field, the free electrons move under the effect of the force in the direction opposite to the direction of electric field and get deposited on the surface of one end of the conductor, and equal amount of positive charge can be considered as deposited on the other end. These induced charges produce an electric field inside the conductor, in the direction opposite to external electric field, When these two electric fields become equal in magnitude, the resultant net electric field inside the conductor becomes zero. Now the motion of charges in the conductor stops and the charges become steady on the end surfaces Thus in the case of metallic conductor, placed in an external electric field (1) A steady electric charge distribution is induced on the surface of the conductor (2) The net electric field inside the conductor is zero (3) The net electric charge inside the conductor is zero (4) On the outer surface of the conductor, the electric field at every point is locally normal to the surface. If the electric field were not normal (perpendicular) a component of electric field parallel to the surface would exist and due to it the charge would move on the surface. But now the motion is stopped and the charges have become steady. Thus the component of electric field parallel to the surface would be zero, and hence the electric field would be normal to the surface (5) Since E at every point inside the conductor, the electric potential everywhere inside the conductor and equal to the value of potential on the surface CAVITY INSIDE A CONDUCTOR Gaussian Surface Consider a +qo suspended in a cavity in a conductor. Consider. Consider a Gaussian surface just outside the cavity and inside the conductor E on this Gaussian surface as it is inside the conductor form Gauss's law
  12. E. ds we have This concludes that a charge of —q must reside on the metal surface on the cavity so that the sum of this induced charge —q and the original charge +q within the Gaussian surface is zero. In other words, a charge q suspended inside a cavity in a conductor is electrically neutral, a charge +q is induced on the outer surface of the conductor. As field inside the conductor, as shown in figure ELECTROSTATIC SHIELDING When a conductor with a cavity is placed inside the electric field, Charge induces on the surface of the conductor. These induce charges produce electric field inside the conductor such that net electric field inside the conductor and inside the cavity is zero. Thus electric field everywhere inside the cavity is zero. This fact is called electrostatic shielding EFFECT PRODUCED BY PUTTING CHARGE ON THE CONDUCTOR The charge placed on a conductor is always distributed only on the outer surface of the conductor. We can understand this by the fact that the electric field inside a conductor is zero. Consider a Gaussian Surface shown by the dots inside the surface and very close to it. Every point on it is inside the surface and not on the surface of conductor. Hence the electric field at every point on this surface is zero. Hence according to Gauss's theorem the charge enclosed by the surface is also zero Consider a Gaussian surface of a pill-box of extremely Gaussian small length and extremely small cross-section ds. A Surface condu ctor fraction of it is inside the surface and the remaining part is outside the surface . The total charge enclosed by this pill-box is q=cds, where = surface charge density of the charge on the conductor. At every point on the surface of the conductor E is perpendicular to the local surface element. Hence it is parallel to the surface vector ds But inside the surface E is zero. Hence flux coming out from the cross-section of pillbox inside the surface is zero. The flux coming out from the cross-section of pill-box outside the surface is E ids = E ds q According to Gauss's theorem E ds — In vector form cds

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