Page 1 :
CHAPTER 2, ELECTRICAL POTENTIAL &, CAPACITANCE, 1. Define electric potential at a, point., Ans: Electric potential at a point, is defined as the work done to, bring a unit positive charge from, infinity to that point., , V, , W, q, , 2. Derive an expression for electric, potential at a point due to a point, charge., Ans:, , r, , =, , , , =, , q, 40, , , , , , 1 q, . dx, 40 x 2, , , , r, , , , 1, dx, x2, , q  1 , =, , 4 0 ,  x , , =, , q 1, , 4 0 , x, , r, , , , r, , , , q 1 1 , , 40  r  , q, 1, ., =, 4 0 r, , =, , q 1, ie, electric potential, V = 4 . r, 0, Note: Potential is a scalar quantity.SI, unit of electric potential is J/C or volt, (V), 3. Is electric potential a vector or a, scalar?, Ans: Scalar, , By definition electric potential at, a point P is the work done to bring a, +1C charge from infinity to the point, P., Let the +1C charge is at A; The work, done to move it from A to B, dW= E.(-dx) [ Since displacement is, against force], = -Edx …………………..(1), But E =, , 4. Draw the graph showing the, variation of ‘V’ and ‘E’ with distance, r., 1, 1, and V 2, r, r, 1, Since E  2 it decreases suddenly, r ,, , Ans: E, , with distance., , q, 1, . 2 …… …….(2), 40 x, , Substituting (2) in (1), 1, q, dW =  4 . x 2 dx, 0, The total work done to bring the, +1C charge from ∞ to P, W=, , , , r, , , , dw, , SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut, , 1

Page 2 :
5[P]. Two charges 5 x 10-8C and -, , 1, q, 1, q, = 4 . r  4 . r, 0, 1, 0, 2, , 3x10-8C are located 16cm apart. At, what point(s) on the line joining the, two charges is the electric potential, zero., 6[P]. A regular hexagon of side, 10cm has a charge 5µC at each of its, , 9. Derive the relation between, electric field and potential., Ans: We know that the potential, difference between A and B is the, work done, to move +1C charge from, A to B., , vertices. Calculate the potential at the, centre of the hexagon., 7[P]. A cube of side ‘a’ has a charge, ‘q’ at each of its vertices. Determine, the potential and electric field at the, , dV = dw = -Edr, i.e, dV = - Edr, dV, dr, , centre of the cube?, , E = , , 8., Define electric potential, difference., Ans:, Potential, difference, , ie, electric field is the negative, gradient of electrostatic potential., , between two points is defined as, the work done to bring a unit +ve, charge, , from, , one, , point, , to, , another., , 10. Derive an expression for the, potential due to an electric dipole., Ans: Consider an electric dipole, having a dipole moment p  q  2ap, .We have to find the electric potential, at a point P, distant ‘r’ from the mid –, point of the dipole., , Potential at A,, V1 =, , q, 1, ., 4 0 r1, , Potential at B,, V2 =, , q, 1, ., 4 0 r2, ,  Potential difference between A and, B, = V 1 – V2, , Electric potential at P due to the +q, charge, V+ =, , 1 q, 4 0 r1, , SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut, , 2

Page 3 :
Electric potential at P due to the –q, charge, V_ =, , 1 q, 4 0 r2, , Total electric potential at P,, V=, , 1, q, 1 q, --------- (1), 40 r1 4 0 r2, , From figure, r1 =r – OC, = r – a cos, From figure, r2 =r +OD, = r + a cos, Substituting in equation (1), V=, , q, 40, , , , 1, 1, , , ,  r  a cos  r  a cos  , , =, , q  r  a cos   (r  a cos ) , , , 40 , r 2  a 2 cos 2 , , , =, , q, 40, ,  2a cos  ,  2 2, 2 ,  r  a cos  , , 11. Define one Electron Volt (eV)., Give its relation with joule., Ans: Electron volt is a smaller unit of, energy., 1eV is defined as the energy acquired, by an electron, when it is accelerated, through a p.d. of 1V ., W = qV => 1eV = 1.6 × 10-19 × 1J, = 1.6 × 10-19J, 1eV = 1.6 × 10-19 Joule, 12. Define potential energy of a, system of charges., Ans: Potential energy of a system, of charges is the work done to, bring the charges from infinity to, their present positions., , If r2>> a2, a2 can be neglected., V=, , q  2a cos  , , 4 o , r2, , , , V, , 1 P cos , 40 r 2, , Special cases, Potential at a point on the axial, , 13. Derive expressions for potential, energy of (i) a single charge (ii) a two, charge system in an external electric, field., Ans:, Potential energy of a single charge, , line, Put  = 00, V=, =, , 1 P cos 0, 40 r 2, , 1 p 1, 1 p, , 40 r 2, 40 r 2, , Potential at a point on the, equatorial line, Put  = 900, V=, =, , 1 P cos 90, 40, r2, , Let V(r) the potential at a point due to, an external e.f. E ., The potential energy of q at that point,, PE = W = qV(r), , 1 p0, 0, 40 r 2, , Note: The equatorial plane of the, dipole is an equipotential surface, having a potential zero., , SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut, , 3

Page 4 :
PE of a system of two charges in, , P.E. of a system of three charges, , an external e.f., , Total potential energy of this system, =, PE of the system of charges is the total, work done to assemble the charges, from infinity., Work done to bring q1 = q1V(r1), Work done to bring q2 =q2V(r2)+, 1 q1q 2, 4 0 r12, , PE of the system=q1V(r) + q2V (r2) +, 1 q1q 2, 40 r12, , 14. Derive expressions for potential, energy of a (i) two charges system (ii), three charge system, in the absence, of external electric field., Ans:, , q 2 q3, 1 q1 q 2, 1, 1 q1 q 3, , , 40 r12, 40 r23, 40 r13, , 1  q1 q 2 q 2 q3 q1 q3 , , , , , 40  r12, r23, r13 , , PE =, , 15. Derive an expression for the, work done in rotating a dipole in an, external electric field., Ans:, Consider a dipole placed in a uniform, electric field at an angle ‘’ with the, electric field., , P.E. of a system of two charges, , Work done to bring q1= 0, The work done to bring q2 to the point, B from infinity in presence of q1 is, = Potential at B due to q1 charge × q2, =, , q1, 1,  q2, 4 0 r, , W=, , 1 q1q 2, 4 0, r, , 1 q1q 2, P.E. =, 40 r, , Work done in rotating the dipole, through an angle d, dW = τ d, But τ = PE Sin , dw = PE sin .d, The work done for rotating the, dipole from an angle  to an angle , W=, , , , 2, , 1, , d, , 2, , =, ,  PE sin  d, , 1, , 2, , =, , PE  sin  d  PE   cos , , 2, 1, , 1, , = - PE cos 2, , , 1, , SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut, , 4

Page 5 :
= -PE [cos θ2 – cos θ1], = PE [cos θ1 – cos θ2], If the dipole is rotated by an angle θ, from stable equilibrium position, θ1 =, 0 and θ2 = θ, , W = PE [1- cosθ], Special cases:, Case I :- when θ = 00, Work done w = PE [1 – cos0], = PE [1 – 1], = PE × 0 = 0, Case II:- When θ = 900, Work done W = PE [1 – cos90], = PE [1 – 0], W = PE, Case III:- When θ = 1800, Work done W = PE[1-cos1800], = PE [1 – (-1)], = PE [2], W = 2PE, This is the maximum work done and, also the maximum potential energy., 16. Derive an expression for the, potential energy of an electric dipole, in an electric field., Ans: Let – PE be the initial potential, energy of the dipole when it is in stable, equilibrium (for convenience)., The total potential energy, when the, dipole is rotated by an angle θ0., U = U0 + W, = -PE + PE(1-cosθ), = -PE + PE –PE cos θ, = -PE cosθ, = - P.E, , Case I, When θ = 00 (Stable equilibrium), U = -PE cos0 = -PE, U = -PE (minimum), Case II, When θ = 900, U = -PE cos900 = -PE × 0, U=0, Case III, When θ = 1800, = -PE × -1, U = PE, [Maximum potential energy], Therefore, unstable equilibrium., CAPACITORS, 17. What is the use capacitor?, Define capacitance., Ans: It is a device used to store, electric charge., Capacitance or capacity (C), It is the ability to capacitance to store, electric charge, Capacitance C =, , Q, V, , Q – charge, V – potential, 18. What is the SI unit capacitance?, Ans: SI unit of capacitance is C/V or, farad (F), 19. Define one farad, Ans: Capacitance of a capacitor is, said to be one farad if one, coulomb, , of, , charge, , raises, , its, , potential by one volt., , U = -PE cosθ, , SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut, , 5

Page 6 :
20. Explain the principle of a, parallel plate capacitor., Ans:, , =, , , d, 0, , We know, capacitance C =, =, , Consider a positively charged, plate P1. If another plate P2 with no, charge, is brought near to P1 (and, placed without touching), then on the, inside of the plate negative charges are, induced and on the outside positive, charges are induced. If the second plate, is earthed all the positive charges, will, flow to earth. Now due to the presence, of negative charges on the plate P2, the, potential (V) of P1 decreases., Therefore, by equation C =, , Q, V, , ,  A, A,  A  0  0, d, d,  d , , ,  0 , , This is the expression for capacitance, of a capacitor with air as the medium, between the plates., 22. What happens if a dielectric, material is introduced between the, plates of the capacitor?, Ans:, , Q, V, , When potential decreases capacitance, increases. This is the principle of, capacitor., 21. Derive an expression for the, capacitance of a parallel plate, capacitor., Ans:, Consider a parallel plate air capacitor, having plate area ‘A’ and charge, density σ ., Charge on a plate, Q = σA, V = Ed, , If a dielectric material is, introduced between the plates of the, capacitor, the capacitance becomes, , Cdielectric =, , k.0 A, d, , k- dielectric, , const., When the dielectric is introduced, in the region between the plates, the, capacitance increases k times, , SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut, , 6

Page 7 :
Cdielectric =k.Cair, , k = εr, , (ii) Dielectric medium avoids, sparking between the plates., , 23[P]. A parallel plate capacitor with, air between the plates has a, capacitance of 8pF (1pF=10-12F)., What will be the capacitance if the, distance between the plates is, reduced by half, and the space, between them is filled with a, substance of dielectric constant 6?, 24. Give the expression for, capacitance of a parallel plate, capacitor partially filled with a, dielectric slab., , Ans:, When a dielectric of relative, permittivity εr of thickness‘t’is, introduced partially between the plates, of the capacitor., Then capacity, C =, , 26. What are the different uses of a, capacitor?, Ans: (i) To store charge, (ii) To generate electromagnetic, radiation, (iii) To tune radio circuits, (iv)To reduce voltage fluctuation, in power supply, 27. Derive expressions for effective, capacitance when capacitors are, connected in (i) series and (ii), parallel., Ans:, (i), Series:, Consider 3, capacitors of capacitances, C1, C2, C3 connected in series, with a voltage V., In a series circuit the charge stored in, each of the capacitors is the same but, the voltages across them are different., , 0 A, (d  t) , , t, r, , t = thickness of dielectric slab, d = distance between the plates of, capacitor, 25. What are the advantages of, introducing dielectric slab between, the plates of a capacitor., Ans:, (i) The capacitance increases, 𝜀𝑟 times, , Applied voltage,, V = V1 + V2 + V3 …………. (1), q, V, , We have C =, V=, , But V =, , q, C, , q, Cs, , Cs effective capacitance (in series), V1 =, , q, C1, , ,, , V2 =, , q, C2, , V3 =, , q, C3, , Eqn. (1) gives, SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut, , 7

Page 9 :
35. Derive an expression for the, , 31[P]. You are given two capacitors of, 2μF and 3μF. What are the maximum, and minimum values of capacitance, , the capacitor combination given, , energy stored in a capacitor., Ans: Consider a capacitor of a, capacitance ‘C’; it has given a voltage, ‘V’. Let at any instant the charge in the, capacitor be ‘q’. Now the work done to, increase the charge by an amount ‘dq’, is given by, , below., , dw = Vdq, , that can be obtained by combining, them?, 32[P]. Calculate effective capacity of, , But V =, , between A and B., , W, q, , W = Vq, , q, .dq, C, , dw=, , 33[P]. Calculate the effective capacity, , q, C, , V=, ,  the total work done to increase the, charge from O to Q is given by, Q, , W=, ,  dW, 0, , Q, , =, , q, ,  C dq, 0, , Q, , =, , 1, q.dq, C 0, , =, , 1  q2 ,  , C  2 0, , =, , 1  Q 2 02 ,  , , C 2, 2, , =, , , 1  Q2,  0, , C 2, , , =, , 1  Q2 ,  , C 2 , , Q, , 34[P]. Obtain the equivalent, capacitance of the network in figure, below. For a 300V supply, determine, the charge and voltage across each, capacitor., , W=, , Q2, 2C, , But Q = CV, , SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut, , 9

Page 10 :
W=, =, =, , 38[P]. A 12pF capacitor is connected, , (CV) 2, 2C, , to a 50V battery. How much, , C2 V 2, 2C, CV, 2, , electrostatic energy is stored in the, , 2, , capacitor?, , W = ½ CV2, This work done is stored, as the energy of the capacitor., U = ½ CV2, , 39[P]. In an experiment with a, capacitor,, , the charge which, , was stored is measured for different, values of p.d. The results are, tabulated as follows:, Charge, , 36. Derive an expression for energy, density of a parallel plate capacitor., Ans: We have the expression for, energy of a capacitor as, U = ½CV2, , , 1  0 A , 1 0 AE d, (Ed) 2 , 2  d , 2, d, 2, , 2, , , , 1, 0 AE 2 d, 2, , Energy, Volume, 1, 0 AE 2 d,  2, Ad, 1,  0 E 2, 2, , Energy Density(u) , , 1, u  0 E 2, 2, 37. If you connect the plates of a, parallel plate capacitor by a copper, wire, what happens to the capacitor?, Justify your answer., Ans: Sparking is produced. A part of, the energy in the capacitor is wasted in, the form of heat, sound and, electromagnetic radiations., , stored/µC, , 7.5 30 60, , 75 90, , pd/ V, , 1, , 10 12, , 4, , 8, , a) Plot a graph with charge on y-axis, and p.d on x-axis, b) Using the graph, calculate the, capacitance of the capacitor., c) Determine the energy stored in, the capacitor., 40. Derive an expression for the lost, energy due to sharing of Capacitors., Ans: Let two capacitors C1 and C2, having charged to potentials V1 and V2,, connected in parallel. Let V be the, common potential., Now we have (C1 + C2) V = C1V1 +, C2V2, Common potential V =, , C1V1  C2 V2, C1  C2, , C1+C2 = total capacitance, Energy after sharing,, U= ½ CV2, C V  C V , 1, = (C1  C2 )  1 1 2 2 , 2,  C1  C2 , , 2, , SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut, , 10

Page 11 :
C1V1  C2 V2 , 1, (C1  C2 ), 2, 2, C1  C2 , , 2, , =, =, , 1  C1V1  C2 V2 , 2, C1  C2, , V=, , 1, q, 4 0 R, , Potential inside the shell (r < R), , 2, , V=, , Total energy before sharing, U1+U2 = ½ C1V12+ ½C2V22, Loss of energy = (U1 + U2) – U, = ½C1V12 + ½ C2V22 -, , 1  C1V1  C2 V2 , 2, C1  C2, , 1, q, 4 0 R, , , constant, , Van de Graaff Generator, 2, , 44. What is the use of a Van De, , On simplification we get, , 1  C1C2 , , Loss of energy ΔU = 2  C  C  (V1  V2 ),  1 2, , 2, , 41[P]. A 600pF capacitor is charged, by a 200V supply. It is then, disconnected from the supply and, then connected to another uncharged, 600pF capacitor. How much, electrostatic energy is lost in the, process?, 42[P]. A 4µF capacitor is charged by, 200V supply. It is then disconnected, from the supply, and is connected to, another uncharged 2µF capacitor., How much electrostatic energy of the, , Graaff Generator? Give its, principle. Explain its construction, and working, Ans:, Use:- It is a device used to create very, high electrostatic potential of the order, of a few million volts., This high voltage is used to supply the, high energy needed for particle, accelerators., Principle, Van de Graaff generator works on the, following two principles., 1., , Discharging action of sharp, , points:- electric discharge takes, place in air or gases readily at, pointed conductors., 2., , If a charged conductor is, , brought, , into, , internal, , contact, , first capacitor is lost in the form of, , with a hollow conductor, all the, , heat and electromagnetic radiation?, , charges are transferred to the, , 43. Write the expressions for the, potential due to a shell., Ans: Potential outside the shell (r > R), V=, , 1 q, 4 0 r, , Potential on the surface of the shell, (r = R), , surface of the hollow conductor, irrespective of the potential of, the hollow conductor., , Explanation, Consider a large spherical shell of, radius R and charge Q. Let us suppose, we introduce a small sphere of radius, , SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut, , 11

Page 12 :
‘r’ carrying a charge q into the large, one, and place it at the centre., , Now the potential at the surface, of the large sphere, V(R) =, , 1 Q q , , 40  R R , , Potential at the surface of the smaller, sphere V(r) =, , 1 Q q, , 40  R r , , V(r) – V(R) =, , 1 q q , , 40  r R , , =, , q 1 1 , , 40  r R , , , which is always, , positive (Assume that q is +ve). Thus, the smaller sphere is always at a higher, potential. So charges are transferred, from it to the larger sphere., Construction, It consists of a large conducting shell, supported on an insulator column of, several meters height. There is an, insulating belt wound around two, pulleys, moving continuously by a, driven motor.The spray comb is, connected to a high tension (10kV), battery. The collector comb is, connected to the shell., , the belt moves up the charges reach the, upper pulley. A similar discharge takes, place at the collector comb and finally, charges are transferred to the, conducting shell, raising its potential to, a few million volts., ELCTRIC AND DIELECTRIC, POLARIZATIONS, , 45. Distinguish between polar and, non-polar molecules., Ans: In certain molecules, the centre, of gravity of positive charges and, centre of gravity of negative charges, do not coincide. These molecules are, called polar molecules., Eg: HCl, H2O, NH3, etc., , Working, The high electric field applied to the, spray comb ionizes the air near to it., The positive charges produced in air, are repelled and get deposited on the, moving belt, by a corona discharge. As, , SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut, , 12

Page 13 :
But in some other molecules the, centre of gravity of positive charges, and centre of gravity of negative, charges coincide. These molecules are, called non-polar molecules., Eg: O2, N2, H2, CO2, etc., , than the external electric field. Thus, , 46. What are dielectrics?, Dielectrics are non-conducting, substances or insulators. But they, allow electric field to pass through, them., , Ans: Infinity, , 47. What is the difference in the, behavior of a conductor and, dielectric in an external electric, field?, Ans:, , When a conductor is placed in an, external electric field ( E 0 ) the free, charge, carries, (electrons), are, redistributed in such a way that an, equal and opposite electric field, , the dielectric only reduces the, external field. Here E 0 + E in ≠ 0, 48. What is the value of dielectric, constant for a metal?, , 49. Explain the polarization in nonpolar molecules., Ans: In the absence of external e.f.,, non-polar, molecules, have, no, permanent dipole moment. In an, external e.f., the positive and negative, centres of the non-polar molecule are, displaced in the opposite directions., Thus the molecule develops an, induced dipole moment. Then the, dielectric is said to be polarized. The, induced dipole moments of different, molecules add up giving a net dipole, moment of the dielectric in the, presence of external electric field., , ( E in ) is set up inside the conductor. So, net electrostatic field is zero, inside the conductor. E0  Ein  0, 50. Explain the polarization in polar, molecules., Ans:, , But when a dielectric is placed in, an external electric field ( E 0 ), the, molecular dipoles are arranged in such, a way that an opposite electric field, ( E in ) is set up inside the dielectric., But this electric field is always less, SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut, , 13

Page 14 :
In a polar dielectric, each, molecule has permanent dipole, moment but in the absence of external, e.f., the dipoles are arranged randomly, due to thermal agitation; so the total, dipole moment is zero., When an external e.f. is applied, the, individual dipoles tend to align with, the field. Then a net dipole moment is, developed., 51. Define Polarization and electric, susceptibility., Whether polar or non-polar, a, dielectric develops a net dipole, moment in the presence of an external, electric field., The dipole moment per unit volume of, the dielectric is called polarization ( P )., For linear isotropic dielectrics,, , P  e E, e is called electric susceptibility of the, dielectric medium., 52. How does external electric field, is reduced in a polarized dielectric?, Ans: Consider two parallel plates, having charge densities + and - and, a dielectric slab placed between them., Due to polarization of the dielectric in, the external field (E0), the charge, densities of plates P1 and P2 are, reduced to P and -P. We, know that, electric field E =, , , 0, , between two sheets of opposite charge, densities (+and). But because of, the polarization of dielectric slab,, charge densities are reduced so electric, field is reduced to E =, ,   P, 0, , ., , ELECTROSTATICS OF, CONDUCTORS, , 53. Explain the main points of, electrostatics of conductors., Ans: The following are the important, results regarding the electrostatics of, conductors:, 1. Inside, , a, , conductor,, , electrostatic field is zero., , Inside a conductor (neutral or, charged) the electrostatic field is zero., This is true even in the presence of an, external field., Reason: In the static situation, the free, charge carriers are so distributed, themselves that the e.f is zero, everywhere inside., 2. At the surface of a charged, conductor, electric field must, be normal to the surface at, every point., , Reason: If E were not normal to the, surface, it would have some non-zero, component along the surface. Free, charges on the surface of the conductor, would then experience force and move., 3. The interior of a conductor can, have no excess charge in static, situation., , Reason: A neutral conductor has equal, amounts of positive and negative, charges. When the conductor is, charged the excess charge can reside, only on the surface in the static, situation., 4. Electrostatic, , potential, , is, , constant through the volume of, the conductor and has the same, value (as inside) on its surface., , SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut, , 14

Page 15 :
Reason: Since E = 0 inside the, conductor and has no work is done in, moving a small test charge within the, conductor and on its surface. That is, there is no potential difference, between any two points inside or on, the surface of the conductor., 5. Electric field at the surface of, a charged conductor E =, , , n̂, 0, , is the surface charge density and n̂, is a unit vector normal to the surface in, the outward direction., If  is –ve, electric field is normal to, the surface inward., 6. Electrostatic shielding., , Electric field is zero inside the cavity, of a conductor of any shape., , Properties, surfaces:-, , of, , equipotential, , 1. The work done to move a, , charge, , from, , one, , point, , to, , another on an equipotential, surface is zero., 2. Two equipotential surfaces will, , never intersect., 3. Electric, , lines of force pass, , normal to an equipotential, , equipotential surface, 54. What is an equipotential surface?, Give examples. Write some properties, of it., , Ans: It is a surface having same, , surface., , More Examples, The figure below shows, equipotential surfaces due to, (i) a dipole, , the, , potential at all points., , Example 1: Concentric spheres with a, point charge at the centre are, equipotential surfaces., (ii) two positive charges, , Example 2: In a uniform e.f parallel, planes perpendicular to the electric, lines of force are equipotential, surfaces., , SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut, , 15

Page 16 :
55[P]. Two charges 2µCand -2µC, are placed at points A and B, 6cm, apart., a) Identify the equipotential surface, of the system., b) What is the direction of the, electric field at every point on this, surface?, , SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut, , 16