{"id":118351,"date":"2022-04-25T18:42:10","date_gmt":"2022-04-25T13:12:10","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=118351"},"modified":"2022-04-25T18:42:10","modified_gmt":"2022-04-25T13:12:10","slug":"chapter-2-electrostatic-potential-and-capacitance-questions-and-answers-ncert-solutions-for-class-12-physics","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/education\/chapter-2-electrostatic-potential-and-capacitance-questions-and-answers-ncert-solutions-for-class-12-physics","title":{"rendered":"Chapter 2 – Electrostatic Potential And Capacitance Questions and Answers: NCERT Solutions for Class 12 Physics"},"content":{"rendered":"

1. Two charges
\n5\u00d710\u22128C5\u00d710\u22128C
\nand
\n\u22123\u00d710\u22128C\u22123\u00d710\u22128C
\nare located
\n16 cm16 cm
\napart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.<\/h2>\n

Ans: It is provided that,
\nFirst charge, q1=5\u00d710\u22128Cq1=5\u00d710\u22128C
\nSecond charge,
\nq2=\u22123\u00d710\u22128Cq2=\u22123\u00d710\u22128C
\nDistance between the two given charges, d=16cm=0.16md=16cm=0.16m
\nCase 1. When point P is inside the system of two charges.
\nConsider a point named P on the line connecting the two charges.
\nr is the distance of point P from q1q1.
\nPotential at point P will be,
\nV=q14\u03c0\u03b5or+q24\u03c0\u03b5o(d\u2212r)V=q14\u03c0\u03b5or+q24\u03c0\u03b5o(d\u2212r)
\nWhere, \u03b5o\u03b5ois the Permittivity of free space
\nBut V=0V=0so,
\n0=q14\u03c0\u03b5or+q24\u03c0\u03b5o(d\u2212r)0=q14\u03c0\u03b5or+q24\u03c0\u03b5o(d\u2212r)
\n\u21d2q14\u03c0\u03b5or=\u2212q24\u03c0\u03b5o(d\u2212r)\u21d2q14\u03c0\u03b5or=\u2212q24\u03c0\u03b5o(d\u2212r)
\n\u21d2q1r=\u2212q2(d\u2212r)\u21d2q1r=\u2212q2(d\u2212r)
\n\u21d25\u00d710\u22128r=\u2212\u22123\u00d710\u22128(0.16\u2212r)\u21d25\u00d710\u22128r=\u2212\u22123\u00d710\u22128(0.16\u2212r)
\n\u21d20.16r=85\u21d20.16r=85
\nWe get,
\nr=0.1m=10cmr=0.1m=10cm
\nTherefore, the potential is zero at
\n10 cm10 cm
\ndistance from the positive charge.
\nCase 2. When point P is outside the system of two charges.
\nPotential at point P will be,
\nV=q14\u03c0\u03b5os+q24\u03c0\u03b5o(s\u2212d)V=q14\u03c0\u03b5os+q24\u03c0\u03b5o(s\u2212d)
\nWhere, \u03b5o\u03b5ois the Permittivity of free space
\nBut V=0V=0so,
\n0=q14\u03c0\u03b5os+q24\u03c0\u03b5o(s\u2212d)0=q14\u03c0\u03b5os+q24\u03c0\u03b5o(s\u2212d)
\n\u21d2q14\u03c0\u03b5os=\u2212q24\u03c0\u03b5o(s\u2212d)\u21d2q14\u03c0\u03b5os=\u2212q24\u03c0\u03b5o(s\u2212d)
\n\u21d2q1s=\u2212q2(s\u2212d)\u21d2q1s=\u2212q2(s\u2212d)
\n\u21d25\u00d710\u22128s=\u2212\u22123\u00d710\u22128(s\u22120.16)\u21d25\u00d710\u22128s=\u2212\u22123\u00d710\u22128(s\u22120.16)
\n\u21d20.16s=25\u21d20.16s=25
\nWe get,
\ns=0.4m=40cms=0.4m=40cm
\nTherefore, the potential is zero at
\n40 cm40 cm
\ndistance from the positive charge.<\/h3>\n

2. A regular hexagon of side
\n10 cm10 cm
\nhas a charge
\n5\u03bcC5\u03bcC
\nat each of its vertices. Calculate the potential at the centre of the hexagon.<\/h2>\n

Ans: The given figure represents six equal charges, q=5\u00d710\u22126Cq=5\u00d710\u22126C, at the hexagon’s vertices.
\nSides of the hexagon, AB=BC=CD=DE=EF=FA=10cmAB=BC=CD=DE=EF=FA=10cm
\nThe distance of O from each vertex, d=10cmd=10cm
\nElectric potential at point O,
\nV=6q4\u03c0\u03b5odV=6q4\u03c0\u03b5od
\nWhere, \u03b5o\u03b5ois the Permittivity of free space
\nValue of
\n14\u03c0\u03b5o=9\u00d7109NC\u22122m\u2212214\u03c0\u03b5o=9\u00d7109NC\u22122m\u22122
\n\u21d2V=6\u00d79\u00d7109\u00d75\u00d710\u221260.1\u21d2V=6\u00d79\u00d7109\u00d75\u00d710\u221260.1
\n\u21d2V=2.7\u00d7106V\u21d2V=2.7\u00d7106V
\nClearly, the potential at the hexagon\u2019s centre is 2.7\u00d7106V2.7\u00d7106V.<\/h3>\n

3. Two charges
\n2\u03bcC2\u03bcC
\nand
\n\u22122\u03bcC\u22122\u03bcC
\nare placed at points A and B,
\n6 cm6 cm
\napart.
\na) Identify an equipotential surface of the system.<\/h2>\n

Ans: The given figure represents two charges.
\nAn equipotential surface is defined as that plane on which electric potential is equal at every point. One such plane is normal to line AB. The plane is placed at the mid-point of line AB because the magnitude of charges is equal.<\/h3>\n

b) What is the direction of the electric field at every point on this surface?<\/h2>\n

Ans: The electric field’s direction is perpendicular to the plane in the line AB direction at every location on this surface.<\/h3>\n

4. A spherical conductor of radius
\n12 cm12 cm
\nhas a charge of 1.6\u00d710\u22127C1.6\u00d710\u22127C distributed uniformly on its surface. What is the electric field,
\na) inside the sphere?<\/h2>\n

Ans: It is provided that,
\nSpherical conductor\u2019s radius,
\nr=12cm=0.12mr=12cm=0.12m
\nThe charge is evenly distributed across the conductor. The electric field within a spherical conductor is zero because the total net charge within a conductor is zero.<\/h3>\n

b) just outside the sphere?<\/h2>\n

Ans: Just outside the conductor, Electric field E is given by
\nE=q4\u03c0\u03b5or2E=q4\u03c0\u03b5or2
\nWhere, \u03b5o\u03b5ois the Permittivity of free space
\nValue of
\n14\u03c0\u03b5o=9\u00d7109NC\u22122m\u2212214\u03c0\u03b5o=9\u00d7109NC\u22122m\u22122
\n\u21d2E=1.6\u00d710\u22127\u00d79\u00d7109(0.12)2\u21d2E=1.6\u00d710\u22127\u00d79\u00d7109(0.12)2
\n\u21d2E=105NC\u22121\u21d2E=105NC\u22121
\nClearly, the electric field just outside the sphere is 105NC\u22121105NC\u22121.<\/h3>\n

c) at a point
\n18 cm18 cm
\nfrom the centre of the sphere?<\/h2>\n

Ans: Let electric field at a given point which is
\n18 cm18 cm
\nfrom the sphere centre = E1E1
\nDistance of the given point from the centre, d=18\u00a0cm=0.18md=18\u00a0cm=0.18m
\nThe formula for electric field is given by,
\nE1=q4\u03c0\u03b5od2E1=q4\u03c0\u03b5od2
\nWhere, \u03b5o\u03b5ois the Permittivity of free space
\nValue of
\n14\u03c0\u03b5o=9\u00d7109NC\u22122m\u2212214\u03c0\u03b5o=9\u00d7109NC\u22122m\u22122
\n\u21d2E=1.6\u00d710\u22127\u00d79\u00d7109(0.18)2\u21d2E=1.6\u00d710\u22127\u00d79\u00d7109(0.18)2
\n\u21d2E=4.4\u00d7104NC\u22121\u21d2E=4.4\u00d7104NC\u22121
\nTherefore, the electric field at a given point
\n18 cm18 cm
\nfrom the sphere centre is 4.4\u00d7104NC\u221214.4\u00d7104NC\u22121.<\/h3>\n

5. A parallel plate capacitor with air between the plates has a capacitance of
\n8pF 8pF
\n. 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
\n66
\n?<\/h2>\n

Ans: It is provided that,
\nCapacitance between the capacitor\u2019s parallel plates,
\nC=8pFC=8pF
\nOriginally, the distance separating the parallel plates was d, and the air was filled in it.
\nDielectric constant of air,
\nk=1k=1
\nThe formula for Capacitance is given by:
\nC=k\u03b5oAdC=k\u03b5oAd
\nHere,
\nk=1k=1
\n, so,
\nC=\u03b5oAdC=\u03b5oAd
\nWhere, \u03b5o\u03b5ois the Permittivity of free space
\nA is the area of each plate.
\nIf the distance separating the plates is decreased to half and the substance has a dielectric constant of
\n66
\nfilled in between the plates.
\nThen,
\nk\u2032=6,d\u2032=d2k\u2032=6,d\u2032=d2
\nHence, capacitor\u2019s capacitance becomes,
\nC\u2032=k\u2032\u03b5oAd\u2032C\u2032=k\u2032\u03b5oAd\u2032
\n\u21d2C\u2032=6\u03b5oAd2\u21d2C\u2032=6\u03b5oAd2
\n\u21d2C\u2032=12C\u21d2C\u2032=12C
\n\u21d2C\u2032=12\u00d78=96pF\u21d2C\u2032=12\u00d78=96pF
\nTherefore, the capacitance when the substance of dielectric constant 66 is filled between the plates is
\n96 pF96 pF<\/h3>\n

6. Three capacitors each of capacitance
\n9 pF9 pF
\nare connected in series.
\na) What is the total capacitance of the combination?<\/h2>\n

Ans: It is provided that,
\nCapacitance of each three capacitors,
\nC=9pFC=9pF
\nThe formula for equivalent capacitance (C\u2032)(C\u2032) of the capacitors\u2019 series combination is given by
\n1C\u2032=1C+1C+1C1C\u2032=1C+1C+1C
\n\u21d21C\u2032=19+19+19\u21d21C\u2032=19+19+19
\n\u21d21C\u2032=39\u21d21C\u2032=39
\n\u21d2C\u2032=3pF\u21d2C\u2032=3pF
\nClearly, total capacitance of the combination of the capacitors is
\n3pF3pF<\/h3>\n

b) What is the potential difference across each capacitor if the combination is connected to a
\n120 V120 V
\nsupply?<\/h2>\n

Ans: Provided that,
\nSupply voltage,
\nV=120 VV=120 V
\nPotential difference
\n(V\u2032)(V\u2032)
\nacross each capacitor will be one-third of the supply voltage.
\nV\u2032=120\/3=40VV\u2032=120\/3=40V
\nClearly, the potential difference across each capacitor is
\n40 V40 V<\/h3>\n

7. Three capacitors of capacitance
\n2pF, 3 pF and 4pF2pF, 3 pF and 4pF
\nare connected in parallel.
\na) What is the total capacitance of the combination?<\/h2>\n

Ans: Provided that,
\nCapacitances of the given capacitors are,
\nC1=2pF ; C2=3pF ; C3=4pFC1=2pF ; C2=3pF ; C3=4pF
\nThe formula for equivalent capacitance (C\u2032)(C\u2032) of the capacitors\u2019 parallel combination is given by
\nC\u2032=C1+C2+C3C\u2032=C1+C2+C3
\n\u21d2C\u2032=2+3+4=9pF\u21d2C\u2032=2+3+4=9pF
\nTherefore, total capacitance of the combination is
\n9pF.9pF.<\/h3>\n

b) Determine the charge on each capacitor if the combination is connected to a
\n100 V100 V
\nsupply.<\/h2>\n

Ans: We have,
\nSupply voltage,
\nV=100 VV=100 V
\nCharge on a capacitor with capacitance C and potential difference V is given by,
\nq=CVq=CV\u2026\u2026(i)
\nFor C=2pFC=2pF,
\nCharge =VC=100\u00d72=200pF=VC=100\u00d72=200pF
\nFor C=3pFC=3pF,
\nCharge =VC=100\u00d73=300pF=VC=100\u00d73=300pF
\nFor C=4pFC=4pF,
\nCharge =VC=100\u00d74=400pF=VC=100\u00d74=400pF<\/h3>\n

8. In a parallel plate capacitor with air between the plates, each plate has an area of 6\u00d710\u22123m26\u00d710\u22123m2 and the distance between the plates is
\n3 mm3 mm Calculate the capacitance of the capacitor. If this capacitor is connected to a
\n100 V100 V
\nsupply, what is the charge on each plate of the capacitor?<\/h2>\n

Ans: It is provided that,
\nArea of parallel plate capacitor\u2019s each plate, A=6\u00d710\u22123m2A=6\u00d710\u22123m2
\nDistance separating the plates,
\nd=3mm=3\u00d710\u22123md=3mm=3\u00d710\u22123m
\nSupply voltage,
\nV=100 VV=100 V
\nThe formula for parallel plate capacitor\u2019s Capacitance is given by,
\nC=\u03b5oAdC=\u03b5oAd
\nWhere, \u03b5o\u03b5ois the Permittivity of free space
\n\u03b5o=8.854\u00d710\u221212C2N\u22121m\u22122\u03b5o=8.854\u00d710\u221212C2N\u22121m\u22122
\n\u21d2C=8.854\u00d710\u221212\u00d76\u00d710\u221233\u00d710\u22123\u21d2C=8.854\u00d710\u221212\u00d76\u00d710\u221233\u00d710\u22123
\n\u21d2C=17.71\u00d710\u221212F\u21d2C=17.71\u00d710\u221212F
\n\u21d2C=17.71pF\u21d2C=17.71pF
\nThe formula for Potential V is related with charge q and capacitance C is given by,
\nV=qCV=qC
\n\u21d2q=CV=100\u00d717.71\u00d710\u221212\u21d2q=CV=100\u00d717.71\u00d710\u221212
\n\u21d2q=1.771\u00d710\u22129C\u21d2q=1.771\u00d710\u22129C
\nClearly, the capacitor’s capacitance is
\n17.71 pF17.71 pF
\nand charge on each plate is 1.771\u00d710\u22129C1.771\u00d710\u22129C.<\/h3>\n

9. Explain what would happen if in the capacitor given in Exercise 8, a
\n3 mm3 mm
\nthick mica sheet (of dielectric constant
\n=6=6
\n) were inserted between the plates,
\na) while the voltage supply remained connected.<\/h2>\n

Ans: It is provided that,
\nMica sheet\u2019s Dielectric constant, k = 6
\nInitial capacitance,
\nC=17.71\u00d710\u221212FC=17.71\u00d710\u221212F
\nNew capacitance,
\nC\u2032=kC=6\u00d717.71\u00d710\u221212=106pFC\u2032=kC=6\u00d717.71\u00d710\u221212=106pF
\nSupply voltage,
\nV=100VV=100V
\nNew charge, q\u2032=C\u2032V\u2032=106\u00d7100pC=1.06\u00d710\u22128Cq\u2032=C\u2032V\u2032=106\u00d7100pC=1.06\u00d710\u22128C
\nPotential across the plates will remain
\n100 V100 V<\/h3>\n

c) after the supply was disconnected.<\/h2>\n

Ans: It is provided that,
\nMica sheet\u2019s Dielectric constant, k = 6
\nInitial capacitance,
\nC=17.71\u00d710\u221212FC=17.71\u00d710\u221212F
\nNew capacitance,
\nC\u2032=kC=6\u00d717.71\u00d710\u221212=106pFC\u2032=kC=6\u00d717.71\u00d710\u221212=106pF
\nIf supply voltage is disconnected, then there will be no influence on the charge amount on the plates.
\nThe formula for potential across the plates is given by,
\nV\u2032=qC\u2032V\u2032=qC\u2032
\nV\u2032=1.771\u00d710\u22129106\u00d710\u221212=16.7VV\u2032=1.771\u00d710\u22129106\u00d710\u221212=16.7V
\nThe potential across the plates when the supply was removed is 16.7V16.7V.<\/h3>\n

10. A
\n12pF12pF
\ncapacitor is connected to a
\n50 V50 V
\nbattery. How much electrostatic energy is stored in the capacitor?<\/h2>\n

Ans: It is provided that,
\nCapacitance of the capacitor,
\nC=12\u00d710\u221212FC=12\u00d710\u221212F
\nPotential difference,
\nV=50 VV=50 V
\nThe formula for stored electrostatic energy in the capacitor is given by,
\nE=12CV2E=12CV2
\n\u21d2E=12\u00d712\u00d710\u221212\u00d7502\u21d2E=12\u00d712\u00d710\u221212\u00d7502
\n\u21d2E=1.5\u00d710\u22128J\u21d2E=1.5\u00d710\u22128J
\nTherefore, the stored electrostatic energy in the capacitor is 1.5\u00d710\u22128J1.5\u00d710\u22128J.<\/h3>\n

11. A
\n600 pF600 pF
\ncapacitor is charged by a
\n200 V200 V
\nsupply. It is then disconnected from the supply and is connected to another uncharged
\n600 pF600 pF
\ncapacitor. How much electrostatic energy is lost in the process?<\/h2>\n

Ans: It is provided that,
\nCapacitance of the capacitor,
\nC=600 pFC=600 pF
\nPotential difference,
\nV=200 VV=200 V
\nThe formula for stored electrostatic energy in the capacitor is given by,
\nE=12CV2E=12CV2
\n\u21d2E=12\u00d7600\u00d710\u221212\u00d72002\u21d2E=12\u00d7600\u00d710\u221212\u00d72002
\n\u21d2E=1.2\u00d710\u22125J\u21d2E=1.2\u00d710\u22125J
\nIf supply is removed from the capacitor and another capacitor of capacitance
\nC=600 pFC=600 pF
\nis joined to it, then equivalent capacitance (C\u2032)(C\u2032) of the series combination is given by
\n1C\u2032=1C+1C1C\u2032=1C+1C
\n\u21d21C\u2032=1600+1600\u21d21C\u2032=1600+1600
\n\u21d21C\u2032=2600\u21d21C\u2032=2600
\n\u21d2C\u2032=300pF\u21d2C\u2032=300pF
\nNew electrostatic energy will be,
\nE\u2032=12C\u2032V2E\u2032=12C\u2032V2
\n\u21d2E\u2032=12\u00d7300\u00d710\u221212\u00d72002\u21d2E\u2032=12\u00d7300\u00d710\u221212\u00d72002
\n\u21d2E\u2032=0.6\u00d710\u22125J\u21d2E\u2032=0.6\u00d710\u22125J
\nLoss in electrostatic energy =E\u2212E\u2032=E\u2212E\u2032
\n\u21d2E\u2212E\u2032=1.2\u00d710\u22125\u22120.6\u00d710\u22125=0.6\u00d710\u22125J\u21d2E\u2212E\u2032=1.2\u00d710\u22125\u22120.6\u00d710\u22125=0.6\u00d710\u22125J
\n\u21d2E\u2212E\u2032=6\u00d710\u22126J\u21d2E\u2212E\u2032=6\u00d710\u22126J
\nClearly, the lost electrostatic energy in the process is 6\u00d710\u22126J6\u00d710\u22126J.<\/h3>\n

12. A charge of
\n8 mC8 mC
\nis located at the origin. Calculate the work done in taking a small charge of \u22122\u00d710\u22129C\u22122\u00d710\u22129C from a point
\nP(0,0,3 cm)P(0,0,3 cm)
\nto a point
\nQ(0,4 cm,0)Q(0,4 cm,0)
\n, via a point
\nR(0,6 cm,9 cm)R(0,6 cm,9 cm)<\/h2>\n

Ans: Charge located at the origin, q=8\u00a0mC=8\u00d710\u22123Cq=8\u00a0mC=8\u00d710\u22123C
\nA small charge is moved from a point P to point R to point Q, q1=\u22122\u00d710\u22129Cq1=\u22122\u00d710\u22129C
\nThe figure given below represents all points.
\nPotential at point P,
\nV1=q4\u03c0\u03b5od1V1=q4\u03c0\u03b5od1
\nWhere d1=3cm=0.03md1=3cm=0.03m
\nPotential at point Q,
\nV2=q4\u03c0\u03b5od2V2=q4\u03c0\u03b5od2
\nWhere d2=4cm=0.04md2=4cm=0.04m
\nWork done by the electrostatic force does not depend on the path.
\nW=q1[V2\u2212V1]W=q1[V2\u2212V1]
\n\u21d2W=q1[q4\u03c0\u03b5od2\u2212q4\u03c0\u03b5od1]\u21d2W=q1[q4\u03c0\u03b5od2\u2212q4\u03c0\u03b5od1]
\n\u21d2W=q1q4\u03c0\u03b5o[1d2\u22121d1]\u21d2W=q1q4\u03c0\u03b5o[1d2\u22121d1]
\nValue of
\n14\u03c0\u03b5o=9\u00d7109NC\u22122m\u2212214\u03c0\u03b5o=9\u00d7109NC\u22122m\u22122
\n\u21d2W=9\u00d7109\u00d78\u00d710\u22123\u00d7(\u22122\u00d710\u22129)[10.04\u221210.03]\u21d2W=9\u00d7109\u00d78\u00d710\u22123\u00d7(\u22122\u00d710\u22129)[10.04\u221210.03]
\n\u21d2W=1.27J\u21d2W=1.27J
\nClearly, work done during the process is 1.27J1.27J.<\/h3>\n

13. A cube of side
\nbb
\nhas a charge
\nqq
\nat each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.<\/h2>\n

Ans: Length of each cube\u2019s side =l=l
\nCharge at each of cube\u2019s vertices =q=q
\nThis figure shows the cube of side b.
\nd is the length of the diagonal of face of the cube.
\nd=b2+b2\u2212\u2212\u2212\u2212\u2212\u2212\u221a=2b2\u2212\u2212\u2212\u221ad=b2+b2=2b2
\nd=2\u2013\u221abd=2b
\nl is the length of the diagonal of the cube.
\nl=3\u2013\u221abl=3b
\nr is the distance between the cube\u2019s centre and one of the eight vertices of the cube.
\nr=3\u221ab2r=3b2
\nThe electric potential (V) at the cube\u2019s centre is due to the presence of eight charges present at the vertices of the cube is given by,
\nV=8q4\u03c0\u03b5orV=8q4\u03c0\u03b5or
\n\u21d2V=8q4\u03c0\u03b5o(3\u221ab2)\u21d2V=8q4\u03c0\u03b5o(3b2)
\n\u21d2V=4q3\u221a\u03c0\u03b5ob\u21d2V=4q3\u03c0\u03b5ob
\nClearly, the potential at the cube\u2019s centre is 4q3\u221a\u03c0\u03b5ob4q3\u03c0\u03b5ob.
\nThe electric field at the cube’s centre is due to the eight charges, which get cancelled because the charges are divided symmetrically concerning the cube’s centre. Therefore, the electric field is zero at the centre.<\/h3>\n

14. Two tiny spheres carrying charges
\n1.5\u03bcC1.5\u03bcC
\nand
\n2.5\u03bcC2.5\u03bcC
\nare located
\n30 cm30 cm
\napart. Find the potential and electric field:
\na) at the mid-point of the line joining the two charges, and<\/h2>\n

Ans: Two charges placed at points A and B are represented in the given figure.
\nO is the mid-point of the line connecting the two charges.
\n(Image will Be Updated Soon)
\nMagnitude of charge located at A,
\nq1=1.5\u03bcCq1=1.5\u03bcC
\nMagnitude of charge located at B,
\nq2=2.5\u03bcCq2=2.5\u03bcC
\nDistance between the two charges,
\nd=30cm=0.3md=30cm=0.3m
\nLet
\nV1V1
\nand E1E1 are the electric potential and electric field respectively at O point.
\nV1V1
\nis the sum of potential due to charge at A and potential due to charge at B.
\nV1=q14\u03c0\u03b5o(d2)+q24\u03c0\u03b5o(d2)=14\u03c0\u03b5o(d2)(q1+q2)V1=q14\u03c0\u03b5o(d2)+q24\u03c0\u03b5o(d2)=14\u03c0\u03b5o(d2)(q1+q2)
\nWhere, \u03b5o\u03b5ois the Permittivity of free space
\n\u03b5o=8.854\u00d710\u221212C2N\u22121m\u22122\u03b5o=8.854\u00d710\u221212C2N\u22121m\u22122
\nValue of
\n14\u03c0\u03b5o=9\u00d7109NC\u22122m\u2212214\u03c0\u03b5o=9\u00d7109NC\u22122m\u22122
\n\u21d2V1=9\u00d7109\u00d710\u22126(0.32)(1.5+2.5)=2.4\u00d7105V\u21d2V1=9\u00d7109\u00d710\u22126(0.32)(1.5+2.5)=2.4\u00d7105V
\nAnd, E1E1is the electric field due to
\nq2q2
\n\u2212\u2212 electric field due to
\nq1q1
\nE1=q24\u03c0\u03b5o(d2)2\u2212q14\u03c0\u03b5o(d2)2=14\u03c0\u03b5o(d2)2(q2\u2212q1)E1=q24\u03c0\u03b5o(d2)2\u2212q14\u03c0\u03b5o(d2)2=14\u03c0\u03b5o(d2)2(q2\u2212q1)
\n\u21d2E1=9\u00d7109\u00d710\u22126(0.32)2(2.50\u22121.5)=4\u00d7105Vm\u22121\u21d2E1=9\u00d7109\u00d710\u22126(0.32)2(2.50\u22121.5)=4\u00d7105Vm\u22121
\nTherefore, the potential at mid-point is
\n2.4\u00d7105V2.4\u00d7105V
\nand the electric field at mid-point is
\n4\u00d7105Vm\u221214\u00d7105Vm\u22121
\n. The field is pointed from the greater charge to the smaller charge.<\/h3>\n

b) at a point
\n10 m10 cm
\nfrom this midpoint in a plane normal to the line and passing through the mid-point.<\/h2>\n

Ans: Consider a point Z such that
\nOZ=10 cm=0.1 mOZ=10 cm=0.1 m
\n, as shown in figure,
\nV2V2
\nand E2E2 are the electric potential and electric field respectively at Z point.
\nBZ=AZ=0.12+0.152\u2212\u221a=0.18mBZ=AZ=0.12+0.152=0.18m
\nV2V2
\nis the sum of potential due to charge at A and potential due to charge at B.
\nV2=q14\u03c0\u03b5o(AZ)+q24\u03c0\u03b5o(BZ)=V2=q14\u03c0\u03b5o(AZ)+q24\u03c0\u03b5o(BZ)=
\nWhere, \u03b5o\u03b5ois the Permittivity of free space
\n\u03b5o=8.854\u00d710\u221212C2N\u22121m\u22122\u03b5o=8.854\u00d710\u221212C2N\u22121m\u22122
\nValue of
\n14\u03c0\u03b5o=9\u00d7109NC\u22122m\u2212214\u03c0\u03b5o=9\u00d7109NC\u22122m\u22122
\n\u21d2V2=9\u00d7109\u00d710\u221260.18(1.5+2.5)=2\u00d7105V\u21d2V2=9\u00d7109\u00d710\u221260.18(1.5+2.5)=2\u00d7105V
\nElectric field due to
\nq1q1
\nat Z,
\nEA=q14\u03c0\u03b5o(AZ)2EA=q14\u03c0\u03b5o(AZ)2
\n\u21d2EA=9\u00d7109\u00d71.5\u00d710\u22126(0.18)2=0.416\u00d7106Vm\u22121\u21d2EA=9\u00d7109\u00d71.5\u00d710\u22126(0.18)2=0.416\u00d7106Vm\u22121
\nalong AZ
\nElectric field due to
\nq2q2
\nat Z,
\nEB=q24\u03c0\u03b5o(BZ)2EB=q24\u03c0\u03b5o(BZ)2
\n\u21d2EB=9\u00d7109\u00d72.5\u00d710\u22126(0.18)2=0.69\u00d7106Vm\u22121\u21d2EB=9\u00d7109\u00d72.5\u00d710\u22126(0.18)2=0.69\u00d7106Vm\u22121
\nalong BZ.
\nThe resultant field intensity at Z,
\nE=E2A+E2B+2EAEBcos2\u03b8\u2212\u221aE=EA2+EB2+2EAEBcos\u20612\u03b8
\nFrom figure,
\ncos\u03b8=0.100.18=0.5556cos\u2061\u03b8=0.100.18=0.5556
\n\u21d2\u03b8=cos\u22121(0.5556)=56.25\u2218\u21d2\u03b8=cos\u22121(0.5556)=56.25\u2218
\n\u21d22\u03b8=112.5\u2218\u21d22\u03b8=112.5\u2218
\n\u21d2cos2\u03b8=\u22120.38\u21d2cos\u20612\u03b8=\u22120.38
\nNow,
\nE=(0.416\u00d7106)2+(0.69\u00d7106)2+2\u00d70.416\u00d7106\u00d70.416\u00d7106\u00d7(\u22120.38)\u2212\u2212\u221aE=(0.416\u00d7106)2+(0.69\u00d7106)2+2\u00d70.416\u00d7106\u00d70.416\u00d7106\u00d7(\u22120.38)
\nE=6.6\u00d7105Vm\u22121E=6.6\u00d7105Vm\u22121
\nTherefore, the potential at a point
\n10 cm10 cm
\n(perpendicular to the mid-point) is
\n2\u00d7105V2\u00d7105V
\nand electric field is
\n6.6\u00d7105Vm\u221216.6\u00d7105Vm\u22121<\/h3>\n

15. A spherical conducting shell of inner radius r1r1 and outer radius r2r2 has a charge Q.
\na) A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell?<\/h2>\n

Ans: Provided that,
\nCharge located at the centre of the shell is
\n+q+q
\n. Hence, a charge of magnitude
\n\u2212q\u2212q
\nwill be induced to the inner surface of the shell. Therefore, net charge on the shell\u2019s inner surface is
\n\u2212q\u2212q
\nSurface charge density at the shell\u2019s inner surface is given by the relation,
\n\u03c31=\u2212q4\u03c0r21\u03c31=\u2212q4\u03c0r12
\nA charge of
\n+q+q
\nis induced on the outer surface of the shell. A charge of magnitude Q is placed on the outer surface of the shell. Therefore, the total charge on the outer surface of the shell is Q+qQ+q. Surface charge density at the shell\u2019s outer surface is,
\n\u03c32=\u2212q4\u03c0r22\u03c32=\u2212q4\u03c0r22
\nb) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.
\nAns: Yes, the electric field intensity is zero inside a cavity, even if the shell is not spherical and has any random shape. Take a closed circle such that a part of it is inside the hole along a field line while the rest is within the conductor. The network performed by the field in taking a test charge over a closed circuit is zero because the field is zero inside the conductor. Hence, the electric field is zero, whatever the frame is.<\/h3>\n

16.
\na) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by
\n(E\u20d7 2\u2212E\u20d7 1)\u22c5n^=\u03c3\u03b5o(E\u21922\u2212E\u21921)\u22c5n^=\u03c3\u03b5o
\nWhere
\nn^n^
\nis a unit vector normal to the surface at a point and
\n\u03c3\u03c3
\nis the surface charge density at that point (The direction of
\nn^n^
\nis from side 1 to side 2). Hence show that just outside a conductor, the electric field is
\nn^\u03c3\u03b5on^\u03c3\u03b5o
\n.<\/h2>\n

Ans: Electric field on the charged body\u2019s one side is E1E1 and the electric field on the same\u00a0 \u00a0body\u2019s other side is E2E2.
\nIf infinite plane body has a uniform thickness, then electric field due to one surface is given by,
\nE\u20d7 1=\u2212\u03c32\u03b5on^E\u21921=\u2212\u03c32\u03b5on^
\n\u2026\u2026(1)
\nWhere,
\nn^n^
\nis the unit vector normal to the surface at a point
\n\u03c3\u03c3
\nis the surface charge density at that point
\nElectric field due to the other surface of the charged body,
\nE\u20d7 2=\u2212\u03c32\u03b5on^E\u21922=\u2212\u03c32\u03b5on^
\n\u2026\u2026(2)
\nElectric field at any point due to the two surfaces
\n(E\u20d7 2\u2212E\u20d71)=\u03c32\u03b5on^+\u03c32\u03b5on^=\u03c3\u03b5on^(E\u21922\u2212E\u21921)=\u03c32\u03b5on^+\u03c32\u03b5on^=\u03c3\u03b5on^
\n\u21d2(E\u20d7 2\u2212E\u20d7 1)\u22c5n^=\u03c3\u03b5on^\u21d2(E\u21922\u2212E\u21921)\u22c5n^=\u03c3\u03b5on^
\nInside a closed conductor,
\nE\u20d7 1=0E\u21921=0
\nE\u20d7 2=E\u20d7 1=\u2212\u03c32\u03b5on^E\u21922=E\u21921=\u2212\u03c32\u03b5on^
\nClearly, the electric field just outside the conductor is
\n\u03c3\u03b5on^\u03c3\u03b5on^<\/h3>\n

b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.<\/h2>\n

Ans: When a charged particle is transferred from one location to the other on a closed-loop, the work performed by the electrostatic field is zero. Hence, the tangential segment of the electrostatic field is continuous from the charged surface’s one side.<\/h3>\n

17. A long charged cylinder of linear charged density
\n\u03bb\u03bb
\nis surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?<\/h2>\n

Ans: Provided that,
\nThere is a long-charged cylinder of length L and radius r and having charge density is
\n\u03bb\u03bb
\nAnother cylinder of equal length encloses the previous cylinder. The radius of this cylinder is R.
\nLet E be the electric field generated in the space between the two cylinders.
\nAccording to Gauss\u2019s theorem, electric flux is given by,
\n\u03d5=E(2\u03c0d)L\u03d5=E(2\u03c0d)L
\nd is the distance of a point from the cylinder\u2019s common axis.
\nLet q be the cylinder\u2019s total charge.
\nSo, \u03d5=E(2\u03c0d)L=q\u03b5o\u03d5=E(2\u03c0d)L=q\u03b5o
\nWhere, q is the Charge on the outer cylinder\u2019s inner sphere.
\n\u03b5o\u03b5ois the Permittivity of free space
\n\u21d2E(2\u03c0d)L=\u03bbL\u03b5o\u21d2E(2\u03c0d)L=\u03bbL\u03b5o
\n\u21d2E=\u03bb2\u03c0d\u03b5o\u21d2E=\u03bb2\u03c0d\u03b5o
\nClearly, the electric field in the space between the two cylinders is \u03bb2\u03c0d\u03b5o\u03bb2\u03c0d\u03b5o.<\/h3>\n

18. In a hydrogen atom, the electron and proton are bound at a distance of about d=0.53Aod=0.53Ao:
\na) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton.<\/h2>\n

Ans: Provided that,
\nThe distance separating electron-proton of a hydrogen atom, d=0.53Aod=0.53Ao
\nCharge on an electron, q1=\u22121.6\u00d710\u221219Cq1=\u22121.6\u00d710\u221219C
\nCharge on a proton, q2=1.6\u00d710\u221219Cq2=1.6\u00d710\u221219C
\nThe value of potential is zero at infinity.
\nPotential energy of the system is,
\nU=0\u2212q1q24\u03c0\u03b5odU=0\u2212q1q24\u03c0\u03b5od
\n\u21d2U=0\u22129\u00d7109\u00d7(1.6\u00d710\u221219)20.53\u00d710\u221210\u21d2U=0\u22129\u00d7109\u00d7(1.6\u00d710\u221219)20.53\u00d710\u221210
\n\u21d2U=\u221243.7\u00d710\u221219J\u21d2U=\u221243.7\u00d710\u221219J
\n\u21d2V=\u221227.2eV\u21d2V=\u221227.2eV
\nClearly, the potential energy of the system is \u221227.2eV\u221227.2eV.<\/h3>\n

b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?<\/h2>\n

Ans: It is mentioned in the question that,
\nKinetic energy is half of the potential energy by magnitude.
\nKinetic energy
\n=13.6eV=13.6eV
\nTotal energy
\n=13.6\u221227.2=\u2212 13.6eV=13.6\u221227.2=\u2212 13.6eV
\nTherefore, the minimum work needed to free the electron is
\n13.6 eV.13.6 eV.<\/h3>\n

c) What are the answers to (a) and (b) above if the zero of potential energy is taken at
\n1.06Ao1.06Ao
\nseparation?<\/h2>\n

Ans: When potential energy is taken as zero, d1=1.06Aod1=1.06Ao
\nPotential energy of the system is,
\nU=q1q24\u03c0\u03b5od1\u221227.2U=q1q24\u03c0\u03b5od1\u221227.2
\n\u21d2U=9\u00d7109\u00d7(1.6\u00d710\u221219)21.06\u00d710\u221210\u221227.2\u21d2U=9\u00d7109\u00d7(1.6\u00d710\u221219)21.06\u00d710\u221210\u221227.2
\n\u21d2U=21.7.3\u00d710\u221219\u221227.2\u21d2U=21.7.3\u00d710\u221219\u221227.2
\n\u21d2U=\u221213.6eV\u21d2U=\u221213.6eV
\nClearly, the potential energy of the system is \u221213.6eV\u221213.6eV.<\/h3>\n

19. If one of the two electrons of a H2H2 molecule is removed, we get a hydrogen molecular ion H2+H2+. In the ground state of an H2+H2+, the two protons are separated by roughly
\n1.5Ao1.5Ao
\n, and the electron is roughly
\n1Ao1Ao
\nfrom each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.<\/h2>\n

Ans: The figure shows two protons and one electron.
\n(Image will Be Updated Soon)
\nCharge on proton 1, q1=1.6\u00d710\u221219Cq1=1.6\u00d710\u221219C
\nCharge on proton 2, q2=1.6\u00d710\u221219Cq2=1.6\u00d710\u221219C
\nCharge on electron, q3=\u22121.6\u00d710\u221219Cq3=\u22121.6\u00d710\u221219C
\nDistance between protons 1 and 2, d1=1.5\u00d710\u221210md1=1.5\u00d710\u221210m
\nDistance between proton 1 and electron, d2=1\u00d710\u221210md2=1\u00d710\u221210m
\nDistance between proton 2 and electron, d3=1\u00d710\u221210md3=1\u00d710\u221210m
\nThere is zero potential energy at infinity.
\nThe formula for potential energy of the system is given by:
\nU=q1q24\u03c0\u03b5od1+q2q34\u03c0\u03b5od3+q1q34\u03c0\u03b5od2U=q1q24\u03c0\u03b5od1+q2q34\u03c0\u03b5od3+q1q34\u03c0\u03b5od2
\n\u21d2U=9\u00d7109\u00d7(1.6\u00d7109)210\u221210[\u22121+11.5\u22121]\u21d2U=9\u00d7109\u00d7(1.6\u00d7109)210\u221210[\u22121+11.5\u22121]
\n\u21d2U=\u221230.72J\u21d2U=\u221230.72J
\n\u21d2U=\u221219.2eV\u21d2U=\u221219.2eV
\nClearly, the potential energy of the system is
\n\u221219.2eV\u221219.2eV<\/h3>\n

20. Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.<\/h2>\n

Ans: Let a be the radius of a sphere A,
\nQAQA
\nbe the charge on the sphere A, and
\nCACA
\nbe the capacitance of the sphere A.
\nLet b be the radius of sphere B,
\nQBQB
\nbe the charge on the sphere B, and
\nCBCB
\nbe the capacitance of the sphere B.
\nThe two spheres are joined with a wire, their potential will be equal.
\nLet EAEA and
\nEBEB
\nare the electric field of sphere A and sphere B respectively.
\nNow,
\nEAEB=QA4\u03c0\u03b5oa2\u00d74\u03c0\u03b5ob2QBEAEB=QA4\u03c0\u03b5oa2\u00d74\u03c0\u03b5ob2QB
\n\u21d2EAEB=QAa2\u00d7b2QB\u21d2EAEB=QAa2\u00d7b2QB\u2026\u2026(1)
\nAnd, QAQB=CAVCBVQAQB=CAVCBV, CACB=abCACB=ab
\n\u21d2QAQB=ab\u21d2QAQB=ab\u2026\u2026(2)
\nNow from (1) and (2),
\n\u21d2EAEB=aa2\u00d7b2b=ba\u21d2EAEB=aa2\u00d7b2b=ba
\nHence, at surface, the ratio of the electric fields is baba.
\nA pointed and sharp end can be arranged as a sphere of a minimum radius, and a flat portion acts as a sphere of a much greater radius. Therefore, the charge density on pointed and sharp ends of a conductor is much higher than on its flatter portions.<\/h3>\n

21. Two charges
\n\u2212q\u2212q
\nand
\n+q+q
\nare located at points
\n(0,0,\u2212a)(0,0,\u2212a)
\nand
\n(0,0,a)(0,0,a)
\n, respectively.
\na) What is the electrostatic potential at the points
\n(0,0,z)(0,0,z)
\nand
\n(x,y,0)?(x,y,0)?<\/h2>\n

Ans: Charge
\n\u2212q\u2212q
\nis placed at
\n(0,0,\u2212a)(0,0,\u2212a)
\nand charge
\n+q+q
\nis placed at
\n(0,0,a)(0,0,a)
\n. Hence, they make a dipole. Point
\n(0,0,z)(0,0,z)
\nis on the dipole axis and point
\n(x,y,0)(x,y,0)
\nis perpendicular to the axis of the dipole. So, electrostatic potential at point
\n(x,y,0)(x,y,0)
\nis zero.
\nElectrostatic potential at point
\n(0,0,z)(0,0,z)
\nis given by,
\nV=14\u03c0\u03b5o[qz\u2212a\u2212qz+a]V=14\u03c0\u03b5o[qz\u2212a\u2212qz+a]
\n\u21d2V=2qa4\u03c0\u03b5o(z2\u2212a2)\u21d2V=2qa4\u03c0\u03b5o(z2\u2212a2)
\n\u21d2V=p4\u03c0\u03b5o(z2\u2212a2)\u21d2V=p4\u03c0\u03b5o(z2\u2212a2)
\nWhere, \u03b5o\u03b5ois the Permittivity of free space
\np is the dipole moment of two charges systems.<\/h3>\n

b) Obtain the dependence of potential on the distance
\nrr
\nof a point from the origin when
\nr\/a >>1r\/a >>1<\/h2>\n

Ans: Distance
\nrr
\nis much larger than half of the distance separating the two charges. Hence, the potential (V) at a distance
\nrr
\nis inversely proportional to the distance\u2019s square.
\nV\u221d1r2V\u221d1r2<\/h3>\n

c) How much work is done in moving a small test charge from the point
\n(5, 0, 0)(5, 0, 0)
\nto
\n(\u22127,0,0)(\u22127,0,0)
\nalong the x-axis? Does the answer change if the path the test charge between the same points is not along the x-axis?<\/h2>\n

Ans: The answer does not change if the path of the test is not along the x-axis. A test charge is moved from point
\n(5,0,0)(5,0,0)
\nto point
\n(\u22127,0,0)(\u22127,0,0)
\nalong the x-axis.
\nElectrostatic potential
\n(V1)(V1)
\nat point
\n(5,0,0)(5,0,0)
\nis given by
\nV1=14\u03c0\u03b5o[\u2212q(5\u22120)2+(\u2212a)2\u221a+q(5\u22120)2+(a)2\u221a]V1=14\u03c0\u03b5o[\u2212q(5\u22120)2+(\u2212a)2+q(5\u22120)2+(a)2]
\n\u21d2V1=14\u03c0\u03b5o[\u2212q25+a2\u221a+q25+a2\u221a]=0\u21d2V1=14\u03c0\u03b5o[\u2212q25+a2+q25+a2]=0
\nElectrostatic potential
\n(V2)(V2)
\nat point
\n(\u22127,0,0)(\u22127,0,0)
\nis given by
\nV2=14\u03c0\u03b5o[\u2212q(\u22127)2+(\u2212a)2\u221a+q(\u22127)2+(a)2\u221a]V2=14\u03c0\u03b5o[\u2212q(\u22127)2+(\u2212a)2+q(\u22127)2+(a)2]
\n\u21d2V2=14\u03c0\u03b5o[\u2212q49+a2\u221a+q49+a2\u221a]=0\u21d2V2=14\u03c0\u03b5o[\u2212q49+a2+q49+a2]=0
\nHence, zero work is done in taking a small test charge from point
\n(5,0,0)(5,0,0)
\nto point
\n(\u22127,0,0)(\u22127,0,0)
\nalong the x-axis because work performed by the electrostatic field in moving a test charge between the two locations is path independent connecting the two points.<\/h3>\n

22. Figure shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on rr for
\nr\/a >>1r\/a >>1
\nand contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge).
\n(Image will Be Updated Soon)<\/h2>\n

Ans: Four charges of equal magnitude are located at points
\nX, Y, Y, and ZX, Y, Y, and Z
\nrespectively, as shown in the below figure,
\n(Image will Be Updated Soon)
\nA point P is r distance far away from point Y.
\nThe system of charges makes an electric quadrupole. It can be taken that the electric quadrupole has three charges, that is, charge
\n+q+q
\nlocated at point X, charge
\n\u22122q\u22122q
\nlocated at point Y and charge
\n+q+q
\nlocated at point Z.
\nNow,
\nXY=YZ=aXY=YZ=a
\nYP=rYP=r
\nPX =r+aPX =r+a
\nPZ =r-aPZ =r-a
\nThe Electrostatic potential created by the three charges system at point P is given by,
\nV=14\u03c0\u03b5o[qXP\u22122qYP+qZP]V=14\u03c0\u03b5o[qXP\u22122qYP+qZP]
\n\u21d2V=14\u03c0\u03b5o[qr+a\u22122qr+qr\u2212a]\u21d2V=14\u03c0\u03b5o[qr+a\u22122qr+qr\u2212a]
\n\u21d2V=q4\u03c0\u03b5o[2a2r(r2\u2212a2)]\u21d2V=q4\u03c0\u03b5o[2a2r(r2\u2212a2)]
\n\u21d2V=q4\u03c0\u03b5or3[2a2(1\u2212a2r2)]\u21d2V=q4\u03c0\u03b5or3[2a2(1\u2212a2r2)]
\nSince,
\nr\/a >>1r\/a >>1
\n\u21d2a\/r 1\u21d2a\/r 1
\nWe can take a2r2a2r2 as negligible.
\n\u21d2V=q2a24\u03c0\u03b5or3\u21d2V=q2a24\u03c0\u03b5or3
\nIt can be said that V\u221d1r3V\u221d1r3.
\nWe know for dipole\u2019s potential, V\u221d1r2V\u221d1r2 and for a monopole, V\u221d1rV\u221d1r.<\/h3>\n

23. An electrical technician requires a capacitance of
\n2\u03bcF2\u03bcF
\nin a circuit across a potential difference of
\n1 kV.1 kV.
\nA large number of
\n1\u03bcF1\u03bcF
\ncapacitors are available to him each of which can withstand a potential difference of not more than
\n400 V400 V
\n. Suggest a possible arrangement that requires the minimum number of capacitors.<\/h2>\n

Ans: It is provided that,
\nTotal required capacitance,
\nC=2\u03bcFC=2\u03bcF
\nPotential difference,
\nV=1kV=1000VV=1kV=1000V
\nCapacitance of each capacitor,
\nC1=1\u03bcFC1=1\u03bcF
\nEach capacitor can withstand a potential difference,
\nV1=400VV1=400V
\nSuppose a number of capacitors are joined in series and these series circuits are joined in parallel to each other. The potential difference in each row must be
\n1000 V1000 V
\nand potential difference across each capacitor must be
\n400 V.400 V.
\nClearly, the capacitors\u2019 numbers in each row is given by,
\n1000400=2.51000400=2.5
\nHence, there are three capacitors across each row. Capacitance of each row,
\n11+1+1=3\u03bcF11+1+1=3\u03bcF
\nThere are n rows and each row has three capacitors, which are joined in parallel. Clearly, circuit\u2019s equivalent capacitance is given by,
\n13+13+13+13……n terms=n313+13+13+13……n terms=n3
\nHowever, circuit\u2019s capacitance is given as
\n2\u03bcF.2\u03bcF.
\n\u21d2n3=2\u21d2n3=2
\n\u21d2n=6\u21d2n=6
\nClearly, there are 6 rows and each row has three capacitors. A minimum of
\n6\u00d73=186\u00d73=18
\ncapacitors are needed for the given arrangement.<\/h3>\n

24. What is the area of the plates of a
\n2 F2 F
\nparallel plate capacitor, given that the separation between the plates is
\n0.5 cm0.5 cm
\n?<\/h2>\n

Ans: It is provided that,
\nCapacitance of a parallel capacitor,
\nC=2 FC=2 F
\nDistance separating the two plates, d=0.5\u00a0cm=0.5\u00d710\u22122md=0.5\u00a0cm=0.5\u00d710\u22122m
\nThe formula for parallel plate capacitor\u2019s capacitance is given by,
\nC=\u03b5oAdC=\u03b5oAd
\nWhere, \u03b5o\u03b5ois the Permittivity of free space
\n\u03b5o=8.854\u00d710\u221212C2N\u22121m\u22122\u03b5o=8.854\u00d710\u221212C2N\u22121m\u22122
\nWe get,
\nA=Cd\u03b5oA=Cd\u03b5o
\nA=2\u00d70.5\u00d710\u221228.854\u00d710\u221212=1130Km2A=2\u00d70.5\u00d710\u221228.854\u00d710\u221212=1130Km2<\/h3>\n

25. Obtain the equivalent capacitance of the network in Figure. For a
\n300 V300 V
\nsupply, determine the charge and voltage across each capacitor.
\n(Image will Be Updated Soon)<\/h2>\n

Ans: It is provided that,
\nCapacitance of capacitor C1C1 is
\n100 pF.100 pF.
\nCapacitance of capacitor C2C2\u00a0 is
\n200 pF.200 pF.
\nCapacitance of capacitor C3C3 is
\n200 pF.200 pF.
\nCapacitance of capacitor C4C4 is
\n100 pF.100 pF.
\nSupply potential,
\nV=300 VV=300 V
\nCapacitors C2C2 and C3C3 are connected in series, therefore, their equivalent capacitance be
\nC\u2032C\u2032
\n1C\u2032=1200+12001C\u2032=1200+1200
\n\u21d21C\u2032=2200\u21d21C\u2032=2200
\n\u21d2C\u2032=100pF\u21d2C\u2032=100pF
\nCapacitors C1C1 and
\nC\u2032C\u2032
\nare in parallel, therefore their equivalent capacitance be
\nC\u2032\u2032.C\u2033.
\nC\u2032\u2032=C1+C\u2032C\u2033=C1+C\u2032
\n\u21d2C\u2032\u2032=100+100=200pF\u21d2C\u2033=100+100=200pF
\nC\u2032\u2032C\u2033
\nand C4C4 are connected in series, therefore, their equivalent capacitance be
\nCC
\n1C\u2032=1200+11001C\u2032=1200+1100
\n\u21d21C\u2032=3200\u21d21C\u2032=3200
\n\u21d2C\u2032=2003pF\u21d2C\u2032=2003pF
\nClearly, the equivalent capacitance of the circuit is 2003pF2003pF.
\nPotential difference across C\u2032\u2032=V\u2032\u2032C\u2033=V\u2033
\nPotential difference across C4=V4C4=V4
\n\u21d2V4+V\u2032\u2032=V=300V\u21d2V4+V\u2033=V=300V
\nCharge on C4C4 is given by,
\nQ4=VCQ4=VC
\n\u21d2Q4=300\u00d72003\u00d710\u221212\u21d2Q4=300\u00d72003\u00d710\u221212
\n\u21d2Q4=2\u00d710\u22128C\u21d2Q4=2\u00d710\u22128C
\n\u21d2V4=Q4C4\u21d2V4=Q4C4
\n\u21d2V4=2\u00d710\u22128100\u00d710\u221212=200V\u21d2V4=2\u00d710\u22128100\u00d710\u221212=200V
\nVoltage across C1C1is given by,
\nV1=V\u2212V4V1=V\u2212V4
\n\u21d2V1=200\u2212100=100V\u21d2V1=200\u2212100=100V
\nHence, the potential difference, V1V1, across C1C1 is
\n100 V100 V
\nCharge on C1C1 is given by,
\nQ1=V1C1Q1=V1C1
\n\u21d2Q1=100\u00d7100\u00d710\u221212\u21d2Q1=100\u00d7100\u00d710\u221212
\n\u21d2Q1=10\u22128C\u21d2Q1=10\u22128C
\nC2C2 and C3C3 having the same capacitances have a
\n100 V100 V
\npotential difference together. Since C2C2 and C3C3 are in series, the potential difference across C2C2 and C3C3 is given by,
\nV2=V3=50VV2=V3=50V
\nCharge on C2C2 is given by,
\nQ2=V2C2Q2=V2C2
\n\u21d2Q2=50\u00d7200\u00d710\u221212\u21d2Q2=50\u00d7200\u00d710\u221212
\n\u21d2Q2=10\u22128C\u21d2Q2=10\u22128C
\nCharge on C3C3 is given by,
\nQ3=V3C3Q3=V3C3
\n\u21d2Q3=50\u00d7200\u00d710\u221212\u21d2Q3=50\u00d7200\u00d710\u221212
\n\u21d2Q3=10\u22128C\u21d2Q3=10\u22128C
\nClearly, the equivalent capacitance of the given circuit is 2003pF2003pFand,
\nQ1=10\u22128CQ1=10\u22128C
\n,
\nQ2=10\u22128CQ2=10\u22128C
\n,
\nQ3=10\u22128CQ3=10\u22128C
\n,
\nQ4=2\u00d710\u22128CQ4=2\u00d710\u22128C
\n,
\nV1=100VV1=100V
\n, V2=50VV2=50V, \u00a0V3=50VV3=50V,\u00a0 V4=200VV4=200V.<\/h3>\n

26. The plates of a parallel plate capacitor have an area of 90cm290cm2 each and are separated by
\n2.5 mm2.5 mm
\n. The capacitor is charged by connecting it to a
\n400 V400 V
\nsupply.
\na) How much electrostatic energy is stored by the capacitor?<\/h2>\n

Ans: It is provided that,
\nArea of the parallel capacitor\u2019s plates, A=90cm2=90\u00d710\u22124m2A=90cm2=90\u00d710\u22124m2
\nDistance separating the plates, d=2.5mm=2.5\u00d710\u22123md=2.5mm=2.5\u00d710\u22123m
\nPotential difference across the pates,
\nV=400VV=400V
\nThe formula for capacitance will be,
\nC=\u03b5oAdC=\u03b5oAd
\nElectrostatic energy stored in capacitor is given by,
\nE1=12CV2E1=12CV2
\n\u21d2E1=12\u03b5oAdV2\u21d2E1=12\u03b5oAdV2
\nWhere, \u03b5o\u03b5ois the Permittivity of free space
\n\u03b5o=8.854\u00d710\u221212C2N\u22121m\u22122\u03b5o=8.854\u00d710\u221212C2N\u22121m\u22122
\n\u21d2E1=128.854\u00d710\u221212\u00d790\u00d710\u221242.5\u00d710\u221234002\u21d2E1=128.854\u00d710\u221212\u00d790\u00d710\u221242.5\u00d710\u221234002
\n\u21d2E1=2.55\u00d7106J\u21d2E1=2.55\u00d7106J
\nThe stored electrostatic energy inside the capacitor is, 2.55\u00d7106J2.55\u00d7106J.<\/h3>\n

b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u. Hence arrive at a relation between u and the magnitude of electric field E between the plates.<\/h2>\n

Ans: Volume of the given capacitor, V\u2032=A\u00d7dV\u2032=A\u00d7d
\n\u21d2V\u2032=90\u00d710\u22124\u00d72.5\u00d710\u22123=2.25\u00d710\u22125m3\u21d2V\u2032=90\u00d710\u22124\u00d72.5\u00d710\u22123=2.25\u00d710\u22125m3
\nEnergy density in the capacitor is given by,
\nu=E1V\u2032u=E1V\u2032
\nu=2.55\u00d710\u221262.25\u00d710\u22125=0.113Jm\u22123u=2.55\u00d710\u221262.25\u00d710\u22125=0.113Jm\u22123
\nAlso, u=E1V\u2032u=E1V\u2032
\n\u21d2u=12\u03b5oAdV2Ad=12\u03b5o(Vd)2\u21d2u=12\u03b5oAdV2Ad=12\u03b5o(Vd)2
\nWhere, Vd=EVd=E, E is electric field.
\n\u21d2u=12\u03b5oE2\u21d2u=12\u03b5oE2
\nHence, derived.
\n27. A
\n4\u03bcF4\u03bcF
\ncapacitor is charged by a
\n200 V200 V
\nsupply. It is then disconnected from the supply, and is connected to another uncharged
\n2\u03bcF2\u03bcF
\ncapacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?
\nAns: It is provided that,
\nA charged capacitor has capacitance,
\nC1=4\u03bcF=4\u00d710\u22126FC1=4\u03bcF=4\u00d710\u22126F
\nSupply voltage,
\nV1=200VV1=200V
\nElectrostatic energy stored in C1C1capacitor is given by,
\nE1=12C1V21E1=12C1V12
\n\u21d2E1=12\u00d74\u00d710\u22126\u00d72002\u21d2E1=12\u00d74\u00d710\u22126\u00d72002
\n\u21d2E1=8\u00d710\u22122J\u21d2E1=8\u00d710\u22122J
\nAn uncharged capacitor\u2019s capacitance,
\nC2=2\u03bcF=2\u00d710\u22126FC2=2\u03bcF=2\u00d710\u22126F
\nWhen C2C2 is joined to the circuit, the potential attained by it is V2V2.
\nAccording to the conservation of charge,
\nV2(C1+C2)=V1C1V2(C1+C2)=V1C1
\n\u21d2V2(4+2)\u00d710\u22126=200\u00d74\u00d710\u22126\u21d2V2(4+2)\u00d710\u22126=200\u00d74\u00d710\u22126
\n\u21d2V2=4003V\u21d2V2=4003V
\nThe formula for electrostatic energy for the two capacitors combination is given by,
\nE2=12V22(C1+C2)E2=12V22(C1+C2)
\n\u21d2E2=12(4003)2(4+2)\u00d710\u22126\u21d2E2=12(4003)2(4+2)\u00d710\u22126
\n\u21d2E2=5.33\u00d710\u22122J\u21d2E2=5.33\u00d710\u22122J
\nThe amount of lost electrostatic energy by capacitor is,
\n=E1\u2212E2=0.08\u22120.0533=0.0267J=E1\u2212E2=0.08\u22120.0533=0.0267J
\nTherefore, the lost electrostatic energy is 0.0267J0.0267J.<\/h3>\n

28. Show that the force on each plate of a parallel plate capacitor has a magnitude equal to 12QE12QE where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor 1212.<\/h2>\n

Ans: Let F be the applied force to separate the parallel plates of a capacitor by a distance of
\n\u0394x\u0394x
\n. Hence, work done by the force
\n=F\u0394x=F\u0394x
\nAs a result, the capacitor\u2019s potential energy rises by an amount given as
\nuA\u0394xuA\u0394x
\nWhere, u is the energy density,
\nA is the area of each plate,
\nd is the Distance separating the plates,
\nV is the difference in potential across the plates.
\nThe work done will be equal to the rise in the potential energy i.e.,
\nF\u0394x=uA\u0394xF\u0394x=uA\u0394x
\n\u21d2F=uA=12\u03b5oE2A\u21d2F=uA=12\u03b5oE2A
\nThe formula for electric intensity is given by,
\nE=VdE=Vd
\n\u21d2F=12\u03b5oE(Vd)A\u21d2F=12\u03b5oE(Vd)A
\nHence, capacitance will be,
\nC=\u03b5oAdC=\u03b5oAd
\n\u21d2F=12CVE\u21d2F=12CVE
\nThe formula for charge in the capacitor is,
\nQ=CVQ=CV
\n\u21d2F=12QE\u21d2F=12QE
\nThe actual origin of the force formula’s half factor is that just outside the conductor, the field is E, and it is zero inside it. Henceforth, it is the average amount of the field that contributes to the force.<\/h3>\n

29. A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Figure).
\n(Image will Be Updated Soon)
\nShow that the capacitance of a spherical capacitor is given by
\nC=4\u03c0\u03b5or1r2r1\u2212r2C=4\u03c0\u03b5or1r2r1\u2212r2
\nWhere r1r1 and r2r2 are the radii of outer inner spheres, respectively.<\/h2>\n

Ans: It is provided that,
\nOuter shell\u2019s radius =r1=r1
\nInner shell\u2019s radius =r2=r2
\nThe outer shell\u2019s inner surface has charge +Q+Q.
\nThe inner shell\u2019s outer surface has induced charge \u2212Q\u2212Q.
\nThe difference in potential between the two shells is given by,
\nV=Q4\u03c0\u03b5or2\u2212Q4\u03c0\u03b5or1V=Q4\u03c0\u03b5or2\u2212Q4\u03c0\u03b5or1
\nWhere, \u03b5o\u03b5ois the Permittivity of free space
\n\u21d2V=Q4\u03c0\u03b5o[1r2\u22121r1]\u21d2V=Q4\u03c0\u03b5o[1r2\u22121r1]
\n\u21d2V=Q(r1\u2212r2)4\u03c0\u03b5or1r2\u21d2V=Q(r1\u2212r2)4\u03c0\u03b5or1r2
\nThe formula for capacitance is given by,
\nC=QVC=QV
\n\u21d2C=4\u03c0\u03b5or1r2r1\u2212r2\u21d2C=4\u03c0\u03b5or1r2r1\u2212r2
\nThis proved.<\/h3>\n

30. A spherical capacitor has an inner sphere of radius
\n12 cm12 cm
\nand an outer sphere of radius
\n13 cm13 cm
\n. The outer sphere is earthed and the inner sphere is given a charge of
\n2.5\u03bcC2.5\u03bcC
\n. The space between the concentric spheres is filled with a liquid of dielectric constant
\n3232
\na) Determine the capacitance of the capacitor.<\/h2>\n

Ans: It is provided that,
\nOuter cylinder\u2019s radius,
\nr1=13cm=0.13mr1=13cm=0.13m
\nRadius of inner cylinder,
\nr2=12cm=0.12mr2=12cm=0.12m
\nCharge on the inner cylinder, q=2.5\u03bcC=2.5\u00d710\u22126Cq=2.5\u03bcC=2.5\u00d710\u22126C
\nThe formula for capacitor\u2019s capacitance is given by,
\nC=4\u03c0\u03b5o\u03b5rr1r2r1\u2212r2C=4\u03c0\u03b5o\u03b5rr1r2r1\u2212r2
\nWhere, \u03b5o\u03b5ois the Permittivity of free space
\n\u03b5o=8.854\u00d710\u221212C2N\u22121m\u22122\u03b5o=8.854\u00d710\u221212C2N\u22121m\u22122
\nValue of
\n14\u03c0\u03b5o=9\u00d7109NC\u22122m\u2212214\u03c0\u03b5o=9\u00d7109NC\u22122m\u22122
\n\u21d2C=32\u00d70.12\u00d70.139\u00d7109(0.13\u22120.12)\u21d2C=32\u00d70.12\u00d70.139\u00d7109(0.13\u22120.12)
\n\u21d2C\u22485.5\u00d710\u22129F\u21d2C\u22485.5\u00d710\u22129F
\nHence, the capacitor\u2019s capacitance is approximately 5.5\u00d710\u22129F5.5\u00d710\u22129F.<\/h3>\n

b) What is the potential of the inner sphere?<\/h2>\n

Ans: The inner sphere\u2019s potential is given by,
\nV=qCV=qC
\n\u21d2V=2.5\u00d710\u221265.5\u00d710\u22129=4.5\u00d7102V\u21d2V=2.5\u00d710\u221265.5\u00d710\u22129=4.5\u00d7102V
\nHence, the inner sphere\u2019s potential is 4.5\u00d7102V4.5\u00d7102V.
\nc) Compare the capacitance of this capacitor with that of an isolated sphere of radius
\n12 cm12 cm
\n. Explain why the latter is much smaller.<\/h2>\n

Ans: Isolated sphere\u2019s radius, r=12\u00d710\u22122mr=12\u00d710\u22122m
\nThe formula for sphere\u2019s capacitance is given by,
\nC\u2032=4\u03c0\u03b5orC\u2032=4\u03c0\u03b5or
\n\u21d2C\u2032=4\u03c0\u00d78.854\u00d710\u221212\u00d712\u00d710\u22122\u21d2C\u2032=4\u03c0\u00d78.854\u00d710\u221212\u00d712\u00d710\u22122
\n\u21d2C\u2032=1.33\u00d710\u221211F\u21d2C\u2032=1.33\u00d710\u221211F
\nThe isolated sphere’s capacitance is less in contrast to the concentric spheres because the concentric spheres’ outer sphere is earthed. Hence, the difference in potential is minor, and the capacitance is higher than the isolated sphere.<\/h3>\n

31. Answer carefully:
\na) Two large conducting spheres carrying charges Q1Q1 and Q2Q2 are brought close to each other. Is the magnitude of electrostatic force between them exactly given by Q1Q24\u03c0\u03b5or2Q1Q24\u03c0\u03b5or2, where r is the distance between their centres?<\/h2>\n

Ans: The expression does not precisely give the force between two conducting spheres, Q1Q24\u03c0\u03b5or2Q1Q24\u03c0\u03b5or2 Because there is an irregular charge distribution on the spheres, hereafter, Coulomb\u2019s law is not valid.<\/h3>\n

b) If Coulomb\u2019s law involved 1r31r3 dependence (instead of 1r21r2), would Gauss\u2019s law be still true?<\/h2>\n

Ans: Gauss\u2019s law would not be valid if Coulomb\u2019s law included dependency on 1r31r3 instead of 1r21r2.<\/h3>\n

c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?<\/h2>\n

Ans: Yes, suppose a small test charge is discharged at rest at an electrostatic field configuration location. In that case, it will move along the field lines crossing through the point, mainly if the field lines are straight because the field lines provide the acceleration’s direction and not of velocity.<\/h3>\n

d) What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical?<\/h2>\n

Ans: Whenever the electron transits in a circular orbit, the work performed by the force of the nucleus is zero, making an orbit either circular or elliptical; the work done by the field of a nucleus is zero.<\/h3>\n

e) We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?<\/h2>\n

Ans: No, the electric field is not continuous across the covering of a charged conductor. However, an electric potential is continuous.<\/h3>\n

f) What meaning would you give to the capacitance of a single conductor?<\/h2>\n

Ans: A single conductor’s capacitance is considered a parallel plate capacitor with one of the two plates at infinity.<\/h3>\n

f) Guess a possible reason why water has a much greater dielectric constant
\n(=80)(=80)
\nthan say, mica
\n(=6)(=6)
\n.<\/h2>\n

Ans: Water has an irregular space as contrasted to mica. Since it has a persistent di pole moment, it has a higher dielectric constant than mica.<\/h3>\n

32. A cylindrical capacitor has two co-axial cylinders of length
\n15 cm15 cm
\nand radii
\n1.5 cm1.5 cm
\nand
\n1.4 cm1.4 cm
\n. The outer cylinder is earthed and the inner cylinder is given a charge of
\n3.5\u03bcC3.5\u03bcC
\n. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e., bending of field lines at the ends).<\/h2>\n

Ans: It is provided that,
\nCo-axial cylinder\u2019s length,
\nl=15cm=0.15ml=15cm=0.15m
\nOuter cylinder\u2019s radius,
\nr1=1.5cm=0.015mr1=1.5cm=0.015m
\nRadius of inner cylinder,
\nr2=1.4cm=0.014mr2=1.4cm=0.014m
\nCharge on the inner cylinder, q=3.5\u03bcC=3.5\u00d710\u22126Cq=3.5\u03bcC=3.5\u00d710\u22126C
\nThe formula for co-axial cylinder\u2019s capacitance of radii r1r1 and r2r2 is given by,
\nC=2\u03c0\u03b5olloge(r1r2)C=2\u03c0\u03b5olloge(r1r2)
\nWhere, \u03b5o\u03b5ois the Permittivity of free space
\n\u03b5o=8.854\u00d710\u221212C2N\u22121m\u22122\u03b5o=8.854\u00d710\u221212C2N\u22121m\u22122
\n\u21d2C=2\u03c0\u00d78.854\u00d710\u221212\u00d70.152.303log10(0.0150.014)\u21d2C=2\u03c0\u00d78.854\u00d710\u221212\u00d70.152.303log10(0.0150.014)
\n\u21d2C=1.2\u00d710\u221210F\u21d2C=1.2\u00d710\u221210F
\nThe difference in potential of the inner cylinder is given by,
\nV=qCV=qC
\n\u21d2V=3.5\u00d710\u221261.2\u00d710\u221210=2.92\u00d7104V\u21d2V=3.5\u00d710\u221261.2\u00d710\u221210=2.92\u00d7104V
\nThe difference in potential will be 2.92\u00d7104V2.92\u00d7104V.<\/h3>\n

33. A parallel plate capacitor is to be designed with a voltage rating
\n1 kV1 kV
\n, using a material of dielectric constant
\n33
\nand dielectric strength about
\n107 Vm\u22121107 Vm\u22121
\n. (Dielectric strength is the maximum electric filed a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionization.) For safety, we should like the field never to exceed, say
\n1010
\nof the dielectric strength. What minimum area of the plates is required to have a capacitance of
\n50 pF50 pF
\n?<\/h2>\n

Ans: It is provided that,
\nParallel plate capacitor\u2019s potential,
\nV=1kV=1000VV=1kV=1000V
\nMaterial\u2019s dielectric constant, \u03b5r=3\u03b5r=3
\nDielectric strength
\n=107 Vm\u22121=107 Vm\u22121
\nFor safety, the electric field intensity never exceeds 10% of the dielectric strength.
\nElectric field intensity, E=0.1\u00d7107=106Vm\u22121E=0.1\u00d7107=106Vm\u22121
\nParallel plate capacitor\u2019s capacitance,
\nC=50 pF=50\u00d710\u221212FC=50 pF=50\u00d710\u221212F
\nThe formula for distance separating the plates is given by,
\nd=VEd=VE
\n\u21d2d=1000106=10\u22123m\u21d2d=1000106=10\u22123m
\nThe formula for capacitance is given by,
\nC=\u03b5o\u03b5rAdC=\u03b5o\u03b5rAd
\nA is the area of each plate,
\n\u03b5o\u03b5ois the Permittivity of free space
\n\u03b5o=8.854\u00d710\u221212C2N\u22121m\u22122\u03b5o=8.854\u00d710\u221212C2N\u22121m\u22122
\nA=Cd\u03b5o\u03b5rA=Cd\u03b5o\u03b5r
\nA=50\u00d710\u221212\u00d710\u221238.85\u00d710\u221212\u00d73\u224819cm2A=50\u00d710\u221212\u00d710\u221238.85\u00d710\u221212\u00d73\u224819cm2
\nClearly, the area of each plate is about 19cm219cm2.<\/h3>\n

34. Describe schematically the equipotential surface corresponding to
\na) A constant electric field in the z-direction,<\/h2>\n

Ans: Equipotential surfaces are the equidistant planes parallel to the x-y plane for the constant electric field in the z-direction.<\/h3>\n

b) A field that uniformly increases in magnitude but remains in a constant (say, z) direction,<\/h2>\n

Ans: Equipotential surfaces are the parallel planes to the x-y plane, except that the field increases when the planes get closer.<\/h3>\n

c) A single positive charge at the origin, and<\/h2>\n

Ans: Equipotential surfaces are the concentric spheres centered at the origin.<\/h3>\n

d) A uniform grid consisting of long equally spaced parallel charged wires in a plane.<\/h2>\n

Ans: A periodically changing shape near the provided grid is the equipotential surface. This shape slowly reaches the shape of planes that are parallel to the grid at a greater distance.<\/h3>\n

35. In a Van de Graaff type generator a spherical metal shell is to be a
\n15\u00d7106V15\u00d7106V
\nelectrode. The dielectric strength of the gas surrounding the electrode is
\n5\u00d7107 Vm\u221215\u00d7107 Vm\u22121
\n. What is the minimum radius of the spherical shell required? (You will learn from this exercise why one cannot build an electrostatic generator using a very small shell which requires a small charge to acquire a high potential).<\/h2>\n

Ans: It is provided that,
\nPotential difference,
\nV=15\u00d7106VV=15\u00d7106V
\nsurroundings gas\u2019s dielectric strength=5\u00d7107\u00a0Vm\u22121=5\u00d7107\u00a0Vm\u22121
\nElectric field intensity is equal to the dielectric strength,\u00a0 E=5\u00d7107\u00a0Vm\u22121E=5\u00d7107\u00a0Vm\u22121
\nThe formula for spherical shell\u2019s minimum radius required for the purpose is given by,
\nr=VEr=VE
\n\u21d2r=15\u00d71065\u00d7107=0.3m=30cm\u21d2r=15\u00d71065\u00d7107=0.3m=30cm
\nClearly, the required minimum radius of the spherical shell is 30cm30cm.<\/h3>\n

36. A small sphere of radius r1r1 and charge q1q1 is enclosed by a spherical shell of radius
\nr2r2
\nand charge
\nq2q2
\n. Show that if
\nq1q1
\nis positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge
\nq2q2
\non the shell is.<\/h2>\n

Ans: According to Gauss’s law, the electric field between a sphere and a shell is concluded by the charge
\nq1q1
\non a small sphere. Hence, the potential difference between the sphere and the shell does not depend on the charge
\nq2q2
\n. For positive charge
\nq1q1
\n, the potential difference V is always positive.<\/h3>\n

37. Answer the following:
\na) The top of the atmosphere is at about
\n400 kV400 kV
\nwith respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about
\n100Vm\u22121100Vm\u22121
\n. Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside!)<\/h2>\n

Ans: When we step out of our house, we do not get an electric shock because the primary equipotential surfaces of open-air change, putting our body and the ground at the same potential.<\/h3>\n

b) A man fixes outside his house one evening a two-metre-high insulating slab carrying on its top a large aluminium sheet of area
\n1m21m2
\n. Will he get an electric shock if he touches the metal sheet in the morning?<\/h2>\n

Ans: Yes, the man gets an electric shock if he touches the metal slab the following day. The constant discharging current in the atmosphere charges up the aluminium sheet. As an outcome, its voltage increases gradually. The increment in the voltage depends on the capacitance produced by the aluminium slab and the ground.<\/h3>\n

c) The discharging current in the atmosphere due to the small conductivity of air is known to be
\n1800 A1800 A
\non an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged?<\/h2>\n

Ans: The existence of thunderstorms and lightning energizes the atmosphere continuously. Therefore, indeed with a discharging current of
\n1800 A1800 A
\n, the atmosphere is not discharged entirely. The two reversing currents are in equilibrium, and the atmosphere persists in neutral.<\/h3>\n

d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during lightning?<\/h2>\n

Ans: Light energy, sound energy, and heat energy are wasted in the atmosphere during lightning and thunderstorms.<\/h3>\n","protected":false},"excerpt":{"rendered":"

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