XII_Physics_Chapter 2- Electric Potential_Capacitance_saju_hsslive

CHAPTER 2 = r 1 . q dx ELECTRICAL POTENTIAL &  CAPACITANCE  40 x2 1. Define electric potential at a q r 1 point. = 40 dx Ans: Electric potential at a point  x2 is defined as the work done to q  1r = 40  x   bring a unit positive charge from q  1 r infinity to that point. = 40  x   V W = q  1  1  q 40  r   2. Derive an expression for electric = 1 .q potential at a point due to a point 40 r charge. Ans: q1 . ie, electric potential, V = 40 r 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 4. Draw the graph showing the variation of ‘V’ and ‘E’ with distance a point P is the work done to bring a r. +1C charge from infinity to the point P. Ans: E 1 and V 1 r2 Let the +1C charge is at A; The work r done to move it from A to B Since E  1 it decreases suddenly r2 , dW= E.(-dx) [ Since displacement is against force] with distance. = -Edx …………………..(1) 1q …… …….(2) But E = . 40 x2 Substituting (2) in (1) dW = 1 q dx 40 . x2 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

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

Electric potential at P due to the –q 11. Define one Electron Volt (eV). charge Give its relation with joule. Ans: Electron volt is a smaller unit of 1 q energy. 1eV is defined as the energy acquired V_ = 40 r2 by an electron, when it is accelerated Total electric potential at P, through a p.d. of 1V . W = qV => 1eV = 1.6 × 10-19 × 1J 1q 1 q = 1.6 × 10-19J V= - --------- (1) 1eV = 1.6 × 10-19 Joule 40 40 r2 r1 12. Define potential energy of a system of charges. From figure, r1 =r – OC Ans: Potential energy of a system of charges is the work done to = r – a cos bring the charges from infinity to their present positions. From figure, r2 =r +OD 13. Derive expressions for potential = r + a cos energy of (i) a single charge (ii) a two charge system in an external electric Substituting in equation (1) field. Ans: V= q  1  1 Potential energy of a single charge 40   a cos    r r  a cos   Let V(r) the potential at a point due to an external e.f. E . = q  r  a cos   (r  a cos )  The potential energy of q at that point,   PE = W = qV(r) 40  r2  a2 cos2   = q  2a cos     40  r 2  a2 cos2   If r2>> a2, a2 can be neglected. V= q  2a cos   4o  r2  V  1 P cos  40 r2 Special cases Potential at a point on the axial line Put  = 00 V = 1 P cos 0 40 r2 = 1 p1  1 p 40 r2 40 r2 Potential at a point on the equatorial line Put  = 900 V = 1 P cos 90 40 r2 = 1 p0  0 40 r2 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

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 1 q1q2  1 q2 q3  1 q1q3 work done to assemble the charges 40 r12 40 r23 40 r13 from infinity. Work done to bring q1 = q1V(r1) PE = 1  q1q 2  q2 q3  q1q3  Work done to bring q2 =q2V(r2)+ 40    r12 r23 r13  1 q1q2 40 r12 15. Derive an expression for the work done in rotating a dipole in an PE of the system=q1V(r) + q2V (r2) + external electric field. Ans: 1 q1q2 Consider a dipole placed in a uniform 40 r12 electric field at an angle ‘’ with the electric field. 14. Derive expressions for potential energy of a (i) two charges system (ii) three charge system, in the absence of external electric field. Ans: P.E. of a system of two charges Work done in rotating the dipole through an angle d dW = τ d But τ = PE Sin  Work done to bring q1= 0 dw = PE sin .d The work done to bring q2 to the point B from infinity in presence of q1 is The work done for rotating the = Potential at B due to q1 charge × q2 dipole from an angle  to an angle  = 1 q1  q2 W = 2 d 40 r 1 W = 1 q1q2 2 40 r =  PE sin d 1 2 sin  d  PE  cos  2 1 P.E. = 1 q1q2 = PE 40 r 1 = - PE cos  2 1 SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut 4

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

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

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

q qq q Where Cp is the effective Cs C1 C2 C3 capacitance when the three capacitors are connected in parallel. q  q  1  1  1  q1 = C1V q2 = C2V q3 = C3V CS  C1 C2 C3    (1) CpV = C1V + C2V + C3V CpV = V(C1+ C2 + C3) If there are ‘n’ capacitors If there are ‘n’ capacitors connected in parallel connected in series Cp = C1 + C2 + C3 + ………. + Cn 1  1  1  1  ...........  1 28[P]. Three capacitors each of CS C1 C2 C3 Cn capacitance, 9pF are connected in series. In the case of two capacitors a) What is the capacitance of the 1  1  1  C2  C1 combination? CS C1 C2 C1C2 b) What is the potential difference 1  C1  C2 across each capacitor if the CS C1C2 combination is connected to a 120V supply? (ii)In Parallel 29[P]. Three capacitors of Consider 3 capacitors C1, C2, C3 capacitances 2pF, 3pF and 4pF are connected in parallel with a voltage connected in parallel. ‘V’. a) What is the capacitance of the In a parallel circuit, the voltage is combination? the same but the charges stored in the capacitors are different. b) Determine the charge on each capacitor if the combination is Here the total charge connected to a 100V supply. q = q1 + q2 + q3 …………. (1) But C = q q = CV 30[P]. Find the effective capacitance of the capacitors given in the V network. q = CpV SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut 8

35. Derive an expression for the 31[P]. You are given two capacitors of energy stored in a capacitor. 2μF and 3μF. What are the maximum and minimum values of capacitance Ans: Consider a capacitor of a that can be obtained by combining capacitance ‘C’; it has given a voltage them? ‘V’. Let at any instant the charge in the 32[P]. Calculate effective capacity of capacitor be ‘q’. Now the work done to the capacitor combination given increase the charge by an amount ‘dq’ below. is given by 33[P]. Calculate the effective capacity between A and B. dw = Vdq W 34[P]. Obtain the equivalent V= q capacitance of the network in figure below. For a 300V supply, determine But V = q W = Vq the charge and voltage across each capacitor. C dw= q .dq C  the total work done to increase the charge from O to Q is given by Q W =  dW 0 Q q = dq 0C 1Q q.dq = C0 = 1  q2 Q   C  2 0 = 1  Q2  02  C  2   2  = 1  Q2  C   0  2  = 1  Q2    C  2  W = Q2 2C But Q = CV SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut 9

W = (CV)2 38[P]. A 12pF capacitor is connected to a 50V battery. How much 2C electrostatic energy is stored in the capacitor? = C2V2 39[P]. In an experiment with a 2C capacitor, the charge which = CV2 2 W = ½ CV2, This work done is stored as the energy of the capacitor. U = ½ CV2 was stored is measured for different values of p.d. The results are tabulated as follows: Charge stored/µC 7.5 30 60 75 90 pd/ V 1 4 8 10 12 36. Derive an expression for energy a) Plot a graph with charge on y-axis density of a parallel plate capacitor. and p.d on x-axis Ans: We have the expression for b) Using the graph, calculate the energy of a capacitor as, U = ½CV2 capacitance of the capacitor.  1  0A  (Ed)2  1 0AE2d2  1 0 AE2d 2  d  2d 2 Energy Density(u)  Energy c) Determine the energy stored in Volume the capacitor. 1 0AE 2d 2  40. Derive an expression for the lost Ad energy due to sharing of Capacitors. 1  2 0E2 Ans: Let two capacitors C1 and C2 having charged to potentials V1 and V2, u  1 0E2 connected in parallel. Let V be the 2 common potential. 37. If you connect the plates of a Now we have (C1 + C2) V = C1V1 + parallel plate capacitor by a copper wire, what happens to the capacitor? C2V2 Justify your answer. Ans: Sparking is produced. A part of Common potential V = C1V1  C2V2 the energy in the capacitor is wasted in the form of heat, sound and C1  C2 electromagnetic radiations. C1+C2 = total capacitance Energy after sharing, U= ½ CV2 = 1 (C1  C2 )  C1V1  C2 V2 2 2  C1  C2    SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut 10

 = 1 2 V= 1 q  2 (C1  C2 ) C1V1  C2V2 40 R C1  C2 2 Potential inside the shell (r < R) =  1 C1V1  C2V2 2 V = 1 q , constant 2 C1  C2 40 R Total energy before sharing Van de Graaff Generator U1+U2 = ½ C1V12+ ½C2V22 44. What is the use of a Van De Loss of energy = (U1 + U2) – U Graaff Generator? Give its principle. Explain its construction = ½C1V12 + ½ C2V22 -  1 C1V1  C2V2 2 and working Ans: 2 C1  C2 Use:- It is a device used to create very high electrostatic potential of the order On simplification we get of a few million volts. This high voltage is used to supply the Loss of energy ΔU = 1  C1C2  (V1  V2 )2 high energy needed for particle 2  C1  C2  accelerators.   Principle Van de Graaff generator works on the 41[P]. A 600pF capacitor is charged following two principles. by a 200V supply. It is then 1. Discharging action of sharp disconnected from the supply and then connected to another uncharged points:- electric discharge takes 600pF capacitor. How much electrostatic energy is lost in the place in air or gases readily at process? pointed conductors. 42[P]. A 4µF capacitor is charged by 200V supply. It is then disconnected 2. If a charged conductor is from the supply, and is connected to another uncharged 2µF capacitor. brought into internal contact How much electrostatic energy of the 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. surface of the hollow conductor Ans: Potential outside the shell (r > R) V= 1 q irrespective of the potential of 40 r the hollow conductor. Explanation Potential on the surface of the shell Consider a large spherical shell of (r = R) 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

‘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  the belt moves up the charges reach the 40  R r  upper pulley. A similar discharge takes place at the collector comb and finally V(r) – V(R) = 1 q  q  charges are transferred to the 40  r R  conducting shell, raising its potential to a few million volts. = q  1  1  , which is always 40  r R  ELCTRIC AND DIELECTRIC POLARIZATIONS positive (Assume that q is +ve). Thus 45. Distinguish between polar and the smaller sphere is always at a higher non-polar molecules. Ans: In certain molecules, the centre potential. So charges are transferred of gravity of positive charges and centre of gravity of negative charges from it to the larger sphere. do not coincide. These molecules are called polar molecules. Construction Eg: HCl, H2O, NH3, etc. 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. 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

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

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

Reason: Since E = 0 inside the Properties of equipotential conductor and has no work is done in surfaces:- moving a small test charge within the conductor and on its surface. That is 1. The work done to move a there is no potential difference between any two points inside or on charge from one point to the surface of the conductor. 5.Electric field at the surface of another on an equipotential a charged conductor E =  nˆ surface is zero. 0 2. Two equipotential surfaces will is the surface charge density and nˆ never intersect. is a unit vector normal to the surface in the outward direction. 3. Electric lines of force pass If  is –ve, electric field is normal to the surface inward. normal to an equipotential 6.Electrostatic shielding. Electric field is zero inside the cavity surface. the of a conductor of any shape. More Examples equipotential surface The figure below shows equipotential surfaces due to 54. What is an equipotential surface? (i) a dipole Give examples. Write some properties of it. Ans: It is a surface having same 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

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