# Chapter 2 – Electrostatic Potential And Capacitance Questions and Answers: NCERT Solutions for Class 12 Physics

Class 12 Physics NCERT book solutions for Chapter 2 - Electrostatic Potential And Capacitance Questions and Answers.

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Class 12 Physics NCERT book solutions for Chapter 2 - Electrostatic Potential And Capacitance Questions and Answers.

5×10−8C5×10−8C

and

−3×10−8C−3×10−8C

are located

16 cm16 cm

apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

First charge, q1=5×10−8Cq1=5×10−8C

Second charge,

q2=−3×10−8Cq2=−3×10−8C

Distance between the two given charges, d=16cm=0.16md=16cm=0.16m

Case 1. When point P is inside the system of two charges.

Consider a point named P on the line connecting the two charges.

r is the distance of point P from q1q1.

Potential at point P will be,

V=q14πεor+q24πεo(d−r)V=q14πεor+q24πεo(d−r)

Where, εoεois the Permittivity of free space

But V=0V=0so,

0=q14πεor+q24πεo(d−r)0=q14πεor+q24πεo(d−r)

⇒q14πεor=−q24πεo(d−r)⇒q14πεor=−q24πεo(d−r)

⇒q1r=−q2(d−r)⇒q1r=−q2(d−r)

⇒5×10−8r=−−3×10−8(0.16−r)⇒5×10−8r=−−3×10−8(0.16−r)

⇒0.16r=85⇒0.16r=85

We get,

r=0.1m=10cmr=0.1m=10cm

Therefore, the potential is zero at

10 cm10 cm

distance from the positive charge.

Case 2. When point P is outside the system of two charges.

Potential at point P will be,

V=q14πεos+q24πεo(s−d)V=q14πεos+q24πεo(s−d)

Where, εoεois the Permittivity of free space

But V=0V=0so,

0=q14πεos+q24πεo(s−d)0=q14πεos+q24πεo(s−d)

⇒q14πεos=−q24πεo(s−d)⇒q14πεos=−q24πεo(s−d)

⇒q1s=−q2(s−d)⇒q1s=−q2(s−d)

⇒5×10−8s=−−3×10−8(s−0.16)⇒5×10−8s=−−3×10−8(s−0.16)

⇒0.16s=25⇒0.16s=25

We get,

s=0.4m=40cms=0.4m=40cm

Therefore, the potential is zero at

40 cm40 cm

distance from the positive charge.

10 cm10 cm

has a charge

5μC5μC

at each of its vertices. Calculate the potential at the centre of the hexagon.

Sides of the hexagon, AB=BC=CD=DE=EF=FA=10cmAB=BC=CD=DE=EF=FA=10cm

The distance of O from each vertex, d=10cmd=10cm

Electric potential at point O,

V=6q4πεodV=6q4πεod

Where, εoεois the Permittivity of free space

Value of

14πεo=9×109NC−2m−214πεo=9×109NC−2m−2

⇒V=6×9×109×5×10−60.1⇒V=6×9×109×5×10−60.1

⇒V=2.7×106V⇒V=2.7×106V

Clearly, the potential at the hexagon’s centre is 2.7×106V2.7×106V.

2μC2μC

and

−2μC−2μC

are placed at points A and B,

6 cm6 cm

apart.

a) Identify an equipotential surface of the system.

An 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.

12 cm12 cm

has a charge of 1.6×10−7C1.6×10−7C distributed uniformly on its surface. What is the electric field,

a) inside the sphere?

Spherical conductor’s radius,

r=12cm=0.12mr=12cm=0.12m

The 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.

E=q4πεor2E=q4πεor2

Where, εoεois the Permittivity of free space

Value of

14πεo=9×109NC−2m−214πεo=9×109NC−2m−2

⇒E=1.6×10−7×9×109(0.12)2⇒E=1.6×10−7×9×109(0.12)2

⇒E=105NC−1⇒E=105NC−1

Clearly, the electric field just outside the sphere is 105NC−1105NC−1.

18 cm18 cm

from the centre of the sphere?

18 cm18 cm

from the sphere centre = E1E1

Distance of the given point from the centre, d=18 cm=0.18md=18 cm=0.18m

The formula for electric field is given by,

E1=q4πεod2E1=q4πεod2

Where, εoεois the Permittivity of free space

Value of

14πεo=9×109NC−2m−214πεo=9×109NC−2m−2

⇒E=1.6×10−7×9×109(0.18)2⇒E=1.6×10−7×9×109(0.18)2

⇒E=4.4×104NC−1⇒E=4.4×104NC−1

Therefore, the electric field at a given point

18 cm18 cm

from the sphere centre is 4.4×104NC−14.4×104NC−1.

8pF 8pF

. 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

66

?

Capacitance between the capacitor’s parallel plates,

C=8pFC=8pF

Originally, the distance separating the parallel plates was d, and the air was filled in it.

Dielectric constant of air,

k=1k=1

The formula for Capacitance is given by:

C=kεoAdC=kεoAd

Here,

k=1k=1

, so,

C=εoAdC=εoAd

Where, εoεois the Permittivity of free space

A is the area of each plate.

If the distance separating the plates is decreased to half and the substance has a dielectric constant of

66

filled in between the plates.

Then,

k′=6,d′=d2k′=6,d′=d2

Hence, capacitor’s capacitance becomes,

C′=k′εoAd′C′=k′εoAd′

⇒C′=6εoAd2⇒C′=6εoAd2

⇒C′=12C⇒C′=12C

⇒C′=12×8=96pF⇒C′=12×8=96pF

Therefore, the capacitance when the substance of dielectric constant 66 is filled between the plates is

96 pF96 pF

9 pF9 pF

are connected in series.

a) What is the total capacitance of the combination?

Capacitance of each three capacitors,

C=9pFC=9pF

The formula for equivalent capacitance (C′)(C′) of the capacitors’ series combination is given by

1C′=1C+1C+1C1C′=1C+1C+1C

⇒1C′=19+19+19⇒1C′=19+19+19

⇒1C′=39⇒1C′=39

⇒C′=3pF⇒C′=3pF

Clearly, total capacitance of the combination of the capacitors is

3pF3pF

120 V120 V

supply?

Supply voltage,

V=120 VV=120 V

Potential difference

(V′)(V′)

across each capacitor will be one-third of the supply voltage.

V′=120/3=40VV′=120/3=40V

Clearly, the potential difference across each capacitor is

40 V40 V

2pF, 3 pF and 4pF2pF, 3 pF and 4pF

are connected in parallel.

a) What is the total capacitance of the combination?

Capacitances of the given capacitors are,

C1=2pF ; C2=3pF ; C3=4pFC1=2pF ; C2=3pF ; C3=4pF

The formula for equivalent capacitance (C′)(C′) of the capacitors’ parallel combination is given by

C′=C1+C2+C3C′=C1+C2+C3

⇒C′=2+3+4=9pF⇒C′=2+3+4=9pF

Therefore, total capacitance of the combination is

9pF.9pF.

100 V100 V

supply.

Supply voltage,

V=100 VV=100 V

Charge on a capacitor with capacitance C and potential difference V is given by,

q=CVq=CV……(i)

For C=2pFC=2pF,

Charge =VC=100×2=200pF=VC=100×2=200pF

For C=3pFC=3pF,

Charge =VC=100×3=300pF=VC=100×3=300pF

For C=4pFC=4pF,

Charge =VC=100×4=400pF=VC=100×4=400pF

3 mm3 mm Calculate the capacitance of the capacitor. If this capacitor is connected to a

100 V100 V

supply, what is the charge on each plate of the capacitor?

Area of parallel plate capacitor’s each plate, A=6×10−3m2A=6×10−3m2

Distance separating the plates,

d=3mm=3×10−3md=3mm=3×10−3m

Supply voltage,

V=100 VV=100 V

The formula for parallel plate capacitor’s Capacitance is given by,

C=εoAdC=εoAd

Where, εoεois the Permittivity of free space

εo=8.854×10−12C2N−1m−2εo=8.854×10−12C2N−1m−2

⇒C=8.854×10−12×6×10−33×10−3⇒C=8.854×10−12×6×10−33×10−3

⇒C=17.71×10−12F⇒C=17.71×10−12F

⇒C=17.71pF⇒C=17.71pF

The formula for Potential V is related with charge q and capacitance C is given by,

V=qCV=qC

⇒q=CV=100×17.71×10−12⇒q=CV=100×17.71×10−12

⇒q=1.771×10−9C⇒q=1.771×10−9C

Clearly, the capacitor’s capacitance is

17.71 pF17.71 pF

and charge on each plate is 1.771×10−9C1.771×10−9C.

3 mm3 mm

thick mica sheet (of dielectric constant

=6=6

) were inserted between the plates,

a) while the voltage supply remained connected.

Mica sheet’s Dielectric constant, k = 6

Initial capacitance,

C=17.71×10−12FC=17.71×10−12F

New capacitance,

C′=kC=6×17.71×10−12=106pFC′=kC=6×17.71×10−12=106pF

Supply voltage,

V=100VV=100V

New charge, q′=C′V′=106×100pC=1.06×10−8Cq′=C′V′=106×100pC=1.06×10−8C

Potential across the plates will remain

100 V100 V

Mica sheet’s Dielectric constant, k = 6

Initial capacitance,

C=17.71×10−12FC=17.71×10−12F

New capacitance,

C′=kC=6×17.71×10−12=106pFC′=kC=6×17.71×10−12=106pF

If supply voltage is disconnected, then there will be no influence on the charge amount on the plates.

The formula for potential across the plates is given by,

V′=qC′V′=qC′

V′=1.771×10−9106×10−12=16.7VV′=1.771×10−9106×10−12=16.7V

The potential across the plates when the supply was removed is 16.7V16.7V.

12pF12pF

capacitor is connected to a

50 V50 V

battery. How much electrostatic energy is stored in the capacitor?

Capacitance of the capacitor,

C=12×10−12FC=12×10−12F

Potential difference,

V=50 VV=50 V

The formula for stored electrostatic energy in the capacitor is given by,

E=12CV2E=12CV2

⇒E=12×12×10−12×502⇒E=12×12×10−12×502

⇒E=1.5×10−8J⇒E=1.5×10−8J

Therefore, the stored electrostatic energy in the capacitor is 1.5×10−8J1.5×10−8J.

600 pF600 pF

capacitor is charged by a

200 V200 V

supply. It is then disconnected from the supply and is connected to another uncharged

600 pF600 pF

capacitor. How much electrostatic energy is lost in the process?

Capacitance of the capacitor,

C=600 pFC=600 pF

Potential difference,

V=200 VV=200 V

The formula for stored electrostatic energy in the capacitor is given by,

E=12CV2E=12CV2

⇒E=12×600×10−12×2002⇒E=12×600×10−12×2002

⇒E=1.2×10−5J⇒E=1.2×10−5J

If supply is removed from the capacitor and another capacitor of capacitance

C=600 pFC=600 pF

is joined to it, then equivalent capacitance (C′)(C′) of the series combination is given by

1C′=1C+1C1C′=1C+1C

⇒1C′=1600+1600⇒1C′=1600+1600

⇒1C′=2600⇒1C′=2600

⇒C′=300pF⇒C′=300pF

New electrostatic energy will be,

E′=12C′V2E′=12C′V2

⇒E′=12×300×10−12×2002⇒E′=12×300×10−12×2002

⇒E′=0.6×10−5J⇒E′=0.6×10−5J

Loss in electrostatic energy =E−E′=E−E′

⇒E−E′=1.2×10−5−0.6×10−5=0.6×10−5J⇒E−E′=1.2×10−5−0.6×10−5=0.6×10−5J

⇒E−E′=6×10−6J⇒E−E′=6×10−6J

Clearly, the lost electrostatic energy in the process is 6×10−6J6×10−6J.

8 mC8 mC

is located at the origin. Calculate the work done in taking a small charge of −2×10−9C−2×10−9C from a point

P(0,0,3 cm)P(0,0,3 cm)

to a point

Q(0,4 cm,0)Q(0,4 cm,0)

, via a point

R(0,6 cm,9 cm)R(0,6 cm,9 cm)

A small charge is moved from a point P to point R to point Q, q1=−2×10−9Cq1=−2×10−9C

The figure given below represents all points.

Potential at point P,

V1=q4πεod1V1=q4πεod1

Where d1=3cm=0.03md1=3cm=0.03m

Potential at point Q,

V2=q4πεod2V2=q4πεod2

Where d2=4cm=0.04md2=4cm=0.04m

Work done by the electrostatic force does not depend on the path.

W=q1[V2−V1]W=q1[V2−V1]

⇒W=q1[q4πεod2−q4πεod1]⇒W=q1[q4πεod2−q4πεod1]

⇒W=q1q4πεo[1d2−1d1]⇒W=q1q4πεo[1d2−1d1]

Value of

14πεo=9×109NC−2m−214πεo=9×109NC−2m−2

⇒W=9×109×8×10−3×(−2×10−9)[10.04−10.03]⇒W=9×109×8×10−3×(−2×10−9)[10.04−10.03]

⇒W=1.27J⇒W=1.27J

Clearly, work done during the process is 1.27J1.27J.

bb

has a charge

at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.

Charge at each of cube’s vertices =q=q

This figure shows the cube of side b.

d is the length of the diagonal of face of the cube.

d=b2+b2−−−−−−√=2b2−−−√d=b2+b2=2b2

d=2–√bd=2b

l is the length of the diagonal of the cube.

l=3–√bl=3b

r is the distance between the cube’s centre and one of the eight vertices of the cube.

r=3√b2r=3b2

The electric potential (V) at the cube’s centre is due to the presence of eight charges present at the vertices of the cube is given by,

V=8q4πεorV=8q4πεor

⇒V=8q4πεo(3√b2)⇒V=8q4πεo(3b2)

⇒V=4q3√πεob⇒V=4q3πεob

Clearly, the potential at the cube’s centre is 4q3√πεob4q3πεob.

The 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.

1.5μC1.5μC

and

2.5μC2.5μC

are located

30 cm30 cm

apart. Find the potential and electric field:

a) at the mid-point of the line joining the two charges, and

O is the mid-point of the line connecting the two charges.

(Image will Be Updated Soon)

Magnitude of charge located at A,

q1=1.5μCq1=1.5μC

Magnitude of charge located at B,

q2=2.5μCq2=2.5μC

Distance between the two charges,

d=30cm=0.3md=30cm=0.3m

Let

V1V1

and E1E1 are the electric potential and electric field respectively at O point.

V1V1

is the sum of potential due to charge at A and potential due to charge at B.

V1=q14πεo(d2)+q24πεo(d2)=14πεo(d2)(q1+q2)V1=q14πεo(d2)+q24πεo(d2)=14πεo(d2)(q1+q2)

Where, εoεois the Permittivity of free space

εo=8.854×10−12C2N−1m−2εo=8.854×10−12C2N−1m−2

Value of

14πεo=9×109NC−2m−214πεo=9×109NC−2m−2

⇒V1=9×109×10−6(0.32)(1.5+2.5)=2.4×105V⇒V1=9×109×10−6(0.32)(1.5+2.5)=2.4×105V

And, E1E1is the electric field due to

q2q2

−− electric field due to

q1q1

E1=q24πεo(d2)2−q14πεo(d2)2=14πεo(d2)2(q2−q1)E1=q24πεo(d2)2−q14πεo(d2)2=14πεo(d2)2(q2−q1)

⇒E1=9×109×10−6(0.32)2(2.50−1.5)=4×105Vm−1⇒E1=9×109×10−6(0.32)2(2.50−1.5)=4×105Vm−1

Therefore, the potential at mid-point is

2.4×105V2.4×105V

and the electric field at mid-point is

4×105Vm−14×105Vm−1

. The field is pointed from the greater charge to the smaller charge.

10 m10 cm

from this midpoint in a plane normal to the line and passing through the mid-point.

OZ=10 cm=0.1 mOZ=10 cm=0.1 m

, as shown in figure,

V2V2

and E2E2 are the electric potential and electric field respectively at Z point.

BZ=AZ=0.12+0.152−√=0.18mBZ=AZ=0.12+0.152=0.18m

V2V2

is the sum of potential due to charge at A and potential due to charge at B.

V2=q14πεo(AZ)+q24πεo(BZ)=V2=q14πεo(AZ)+q24πεo(BZ)=

Where, εoεois the Permittivity of free space

εo=8.854×10−12C2N−1m−2εo=8.854×10−12C2N−1m−2

Value of

14πεo=9×109NC−2m−214πεo=9×109NC−2m−2

⇒V2=9×109×10−60.18(1.5+2.5)=2×105V⇒V2=9×109×10−60.18(1.5+2.5)=2×105V

Electric field due to

q1q1

at Z,

EA=q14πεo(AZ)2EA=q14πεo(AZ)2

⇒EA=9×109×1.5×10−6(0.18)2=0.416×106Vm−1⇒EA=9×109×1.5×10−6(0.18)2=0.416×106Vm−1

along AZ

Electric field due to

q2q2

at Z,

EB=q24πεo(BZ)2EB=q24πεo(BZ)2

⇒EB=9×109×2.5×10−6(0.18)2=0.69×106Vm−1⇒EB=9×109×2.5×10−6(0.18)2=0.69×106Vm−1

along BZ.

The resultant field intensity at Z,

E=E2A+E2B+2EAEBcos2θ−√E=EA2+EB2+2EAEBcos2θ

From figure,

cosθ=0.100.18=0.5556cosθ=0.100.18=0.5556

⇒θ=cos−1(0.5556)=56.25∘⇒θ=cos−1(0.5556)=56.25∘

⇒2θ=112.5∘⇒2θ=112.5∘

⇒cos2θ=−0.38⇒cos2θ=−0.38

Now,

E=(0.416×106)2+(0.69×106)2+2×0.416×106×0.416×106×(−0.38)−−√E=(0.416×106)2+(0.69×106)2+2×0.416×106×0.416×106×(−0.38)

E=6.6×105Vm−1E=6.6×105Vm−1

Therefore, the potential at a point

10 cm10 cm

(perpendicular to the mid-point) is

2×105V2×105V

and electric field is

6.6×105Vm−16.6×105Vm−1

a) 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?

Charge located at the centre of the shell is

+q+q

. Hence, a charge of magnitude

−q−q

will be induced to the inner surface of the shell. Therefore, net charge on the shell’s inner surface is

−q−q

Surface charge density at the shell’s inner surface is given by the relation,

σ1=−q4πr21σ1=−q4πr12

A charge of

+q+q

is 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’s outer surface is,

σ2=−q4πr22σ2=−q4πr22

b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.

Ans: 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.

a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by

(E⃗ 2−E⃗ 1)⋅n^=σεo(E→2−E→1)⋅n^=σεo

Where

n^n^

is a unit vector normal to the surface at a point and

σσ

is the surface charge density at that point (The direction of

n^n^

is from side 1 to side 2). Hence show that just outside a conductor, the electric field is

n^σεon^σεo

.

If infinite plane body has a uniform thickness, then electric field due to one surface is given by,

E⃗ 1=−σ2εon^E→1=−σ2εon^

……(1)

Where,

n^n^

is the unit vector normal to the surface at a point

σσ

is the surface charge density at that point

Electric field due to the other surface of the charged body,

E⃗ 2=−σ2εon^E→2=−σ2εon^

……(2)

Electric field at any point due to the two surfaces

(E⃗ 2−E⃗1)=σ2εon^+σ2εon^=σεon^(E→2−E→1)=σ2εon^+σ2εon^=σεon^

⇒(E⃗ 2−E⃗ 1)⋅n^=σεon^⇒(E→2−E→1)⋅n^=σεon^

Inside a closed conductor,

E⃗ 1=0E→1=0

E⃗ 2=E⃗ 1=−σ2εon^E→2=E→1=−σ2εon^

Clearly, the electric field just outside the conductor is

σεon^σεon^

λλ

is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?

There is a long-charged cylinder of length L and radius r and having charge density is

λλ

Another cylinder of equal length encloses the previous cylinder. The radius of this cylinder is R.

Let E be the electric field generated in the space between the two cylinders.

According to Gauss’s theorem, electric flux is given by,

ϕ=E(2πd)Lϕ=E(2πd)L

d is the distance of a point from the cylinder’s common axis.

Let q be the cylinder’s total charge.

So, ϕ=E(2πd)L=qεoϕ=E(2πd)L=qεo

Where, q is the Charge on the outer cylinder’s inner sphere.

εoεois the Permittivity of free space

⇒E(2πd)L=λLεo⇒E(2πd)L=λLεo

⇒E=λ2πdεo⇒E=λ2πdεo

Clearly, the electric field in the space between the two cylinders is λ2πdεoλ2πdεo.

a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton.

The distance separating electron-proton of a hydrogen atom, d=0.53Aod=0.53Ao

Charge on an electron, q1=−1.6×10−19Cq1=−1.6×10−19C

Charge on a proton, q2=1.6×10−19Cq2=1.6×10−19C

The value of potential is zero at infinity.

Potential energy of the system is,

U=0−q1q24πεodU=0−q1q24πεod

⇒U=0−9×109×(1.6×10−19)20.53×10−10⇒U=0−9×109×(1.6×10−19)20.53×10−10

⇒U=−43.7×10−19J⇒U=−43.7×10−19J

⇒V=−27.2eV⇒V=−27.2eV

Clearly, the potential energy of the system is −27.2eV−27.2eV.

Kinetic energy is half of the potential energy by magnitude.

Kinetic energy

=13.6eV=13.6eV

Total energy

=13.6−27.2=− 13.6eV=13.6−27.2=− 13.6eV

Therefore, the minimum work needed to free the electron is

13.6 eV.13.6 eV.

1.06Ao1.06Ao

separation?

Potential energy of the system is,

U=q1q24πεod1−27.2U=q1q24πεod1−27.2

⇒U=9×109×(1.6×10−19)21.06×10−10−27.2⇒U=9×109×(1.6×10−19)21.06×10−10−27.2

⇒U=21.7.3×10−19−27.2⇒U=21.7.3×10−19−27.2

⇒U=−13.6eV⇒U=−13.6eV

Clearly, the potential energy of the system is −13.6eV−13.6eV.

1.5Ao1.5Ao

, and the electron is roughly

1Ao1Ao

from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.

(Image will Be Updated Soon)

Charge on proton 1, q1=1.6×10−19Cq1=1.6×10−19C

Charge on proton 2, q2=1.6×10−19Cq2=1.6×10−19C

Charge on electron, q3=−1.6×10−19Cq3=−1.6×10−19C

Distance between protons 1 and 2, d1=1.5×10−10md1=1.5×10−10m

Distance between proton 1 and electron, d2=1×10−10md2=1×10−10m

Distance between proton 2 and electron, d3=1×10−10md3=1×10−10m

There is zero potential energy at infinity.

The formula for potential energy of the system is given by:

U=q1q24πεod1+q2q34πεod3+q1q34πεod2U=q1q24πεod1+q2q34πεod3+q1q34πεod2

⇒U=9×109×(1.6×109)210−10[−1+11.5−1]⇒U=9×109×(1.6×109)210−10[−1+11.5−1]

⇒U=−30.72J⇒U=−30.72J

⇒U=−19.2eV⇒U=−19.2eV

Clearly, the potential energy of the system is

−19.2eV−19.2eV

QAQA

be the charge on the sphere A, and

CACA

be the capacitance of the sphere A.

Let b be the radius of sphere B,

QBQB

be the charge on the sphere B, and

CBCB

be the capacitance of the sphere B.

The two spheres are joined with a wire, their potential will be equal.

Let EAEA and

EBEB

are the electric field of sphere A and sphere B respectively.

Now,

EAEB=QA4πεoa2×4πεob2QBEAEB=QA4πεoa2×4πεob2QB

⇒EAEB=QAa2×b2QB⇒EAEB=QAa2×b2QB……(1)

And, QAQB=CAVCBVQAQB=CAVCBV, CACB=abCACB=ab

⇒QAQB=ab⇒QAQB=ab……(2)

Now from (1) and (2),

⇒EAEB=aa2×b2b=ba⇒EAEB=aa2×b2b=ba

Hence, at surface, the ratio of the electric fields is baba.

A 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.

−q−q

and

+q+q

are located at points

(0,0,−a)(0,0,−a)

and

(0,0,a)(0,0,a)

, respectively.

a) What is the electrostatic potential at the points

(0,0,z)(0,0,z)

and

(x,y,0)?(x,y,0)?

−q−q

is placed at

(0,0,−a)(0,0,−a)

and charge

+q+q

is placed at

(0,0,a)(0,0,a)

. Hence, they make a dipole. Point

(0,0,z)(0,0,z)

is on the dipole axis and point

(x,y,0)(x,y,0)

is perpendicular to the axis of the dipole. So, electrostatic potential at point

(x,y,0)(x,y,0)

is zero.

Electrostatic potential at point

(0,0,z)(0,0,z)

is given by,

V=14πεo[qz−a−qz+a]V=14πεo[qz−a−qz+a]

⇒V=2qa4πεo(z2−a2)⇒V=2qa4πεo(z2−a2)

⇒V=p4πεo(z2−a2)⇒V=p4πεo(z2−a2)

Where, εoεois the Permittivity of free space

p is the dipole moment of two charges systems.

rr

of a point from the origin when

r/a >>1r/a >>1

rr

is much larger than half of the distance separating the two charges. Hence, the potential (V) at a distance

rr

is inversely proportional to the distance’s square.

V∝1r2V∝1r2

(5, 0, 0)(5, 0, 0)

to

(−7,0,0)(−7,0,0)

along the x-axis? Does the answer change if the path the test charge between the same points is not along the x-axis?

(5,0,0)(5,0,0)

to point

(−7,0,0)(−7,0,0)

along the x-axis.

Electrostatic potential

(V1)(V1)

at point

(5,0,0)(5,0,0)

is given by

V1=14πεo[−q(5−0)2+(−a)2√+q(5−0)2+(a)2√]V1=14πεo[−q(5−0)2+(−a)2+q(5−0)2+(a)2]

⇒V1=14πεo[−q25+a2√+q25+a2√]=0⇒V1=14πεo[−q25+a2+q25+a2]=0

Electrostatic potential

(V2)(V2)

at point

(−7,0,0)(−7,0,0)

is given by

V2=14πεo[−q(−7)2+(−a)2√+q(−7)2+(a)2√]V2=14πεo[−q(−7)2+(−a)2+q(−7)2+(a)2]

⇒V2=14πεo[−q49+a2√+q49+a2√]=0⇒V2=14πεo[−q49+a2+q49+a2]=0

Hence, zero work is done in taking a small test charge from point

(5,0,0)(5,0,0)

to point

(−7,0,0)(−7,0,0)

along 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.

r/a >>1r/a >>1

and contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge).

(Image will Be Updated Soon)

X, Y, Y, and ZX, Y, Y, and Z

respectively, as shown in the below figure,

(Image will Be Updated Soon)

A point P is r distance far away from point Y.

The system of charges makes an electric quadrupole. It can be taken that the electric quadrupole has three charges, that is, charge

+q+q

located at point X, charge

−2q−2q

located at point Y and charge

+q+q

located at point Z.

Now,

XY=YZ=aXY=YZ=a

YP=rYP=r

PX =r+aPX =r+a

PZ =r-aPZ =r-a

The Electrostatic potential created by the three charges system at point P is given by,

V=14πεo[qXP−2qYP+qZP]V=14πεo[qXP−2qYP+qZP]

⇒V=14πεo[qr+a−2qr+qr−a]⇒V=14πεo[qr+a−2qr+qr−a]

⇒V=q4πεo[2a2r(r2−a2)]⇒V=q4πεo[2a2r(r2−a2)]

⇒V=q4πεor3[2a2(1−a2r2)]⇒V=q4πεor3[2a2(1−a2r2)]

Since,

r/a >>1r/a >>1

⇒a/r 1⇒a/r 1

We can take a2r2a2r2 as negligible.

⇒V=q2a24πεor3⇒V=q2a24πεor3

It can be said that V∝1r3V∝1r3.

We know for dipole’s potential, V∝1r2V∝1r2 and for a monopole, V∝1rV∝1r.

2μF2μF

in a circuit across a potential difference of

1 kV.1 kV.

A large number of

1μF1μF

capacitors are available to him each of which can withstand a potential difference of not more than

400 V400 V

. Suggest a possible arrangement that requires the minimum number of capacitors.

Total required capacitance,

C=2μFC=2μF

Potential difference,

V=1kV=1000VV=1kV=1000V

Capacitance of each capacitor,

C1=1μFC1=1μF

Each capacitor can withstand a potential difference,

V1=400VV1=400V

Suppose 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

1000 V1000 V

and potential difference across each capacitor must be

400 V.400 V.

Clearly, the capacitors’ numbers in each row is given by,

1000400=2.51000400=2.5

Hence, there are three capacitors across each row. Capacitance of each row,

11+1+1=3μF11+1+1=3μF

There are n rows and each row has three capacitors, which are joined in parallel. Clearly, circuit’s equivalent capacitance is given by,

13+13+13+13……n terms=n313+13+13+13……n terms=n3

However, circuit’s capacitance is given as

2μF.2μF.

⇒n3=2⇒n3=2

⇒n=6⇒n=6

Clearly, there are 6 rows and each row has three capacitors. A minimum of

6×3=186×3=18

capacitors are needed for the given arrangement.

2 F2 F

parallel plate capacitor, given that the separation between the plates is

0.5 cm0.5 cm

?

Capacitance of a parallel capacitor,

C=2 FC=2 F

Distance separating the two plates, d=0.5 cm=0.5×10−2md=0.5 cm=0.5×10−2m

The formula for parallel plate capacitor’s capacitance is given by,

C=εoAdC=εoAd

Where, εoεois the Permittivity of free space

εo=8.854×10−12C2N−1m−2εo=8.854×10−12C2N−1m−2

We get,

A=CdεoA=Cdεo

A=2×0.5×10−28.854×10−12=1130Km2A=2×0.5×10−28.854×10−12=1130Km2

300 V300 V

supply, determine the charge and voltage across each capacitor.

(Image will Be Updated Soon)

Capacitance of capacitor C1C1 is

100 pF.100 pF.

Capacitance of capacitor C2C2 is

200 pF.200 pF.

Capacitance of capacitor C3C3 is

200 pF.200 pF.

Capacitance of capacitor C4C4 is

100 pF.100 pF.

Supply potential,

V=300 VV=300 V

Capacitors C2C2 and C3C3 are connected in series, therefore, their equivalent capacitance be

C′C′

1C′=1200+12001C′=1200+1200

⇒1C′=2200⇒1C′=2200

⇒C′=100pF⇒C′=100pF

Capacitors C1C1 and

C′C′

are in parallel, therefore their equivalent capacitance be

C′′.C″.

C′′=C1+C′C″=C1+C′

⇒C′′=100+100=200pF⇒C″=100+100=200pF

C′′C″

and C4C4 are connected in series, therefore, their equivalent capacitance be

CC

1C′=1200+11001C′=1200+1100

⇒1C′=3200⇒1C′=3200

⇒C′=2003pF⇒C′=2003pF

Clearly, the equivalent capacitance of the circuit is 2003pF2003pF.

Potential difference across C′′=V′′C″=V″

Potential difference across C4=V4C4=V4

⇒V4+V′′=V=300V⇒V4+V″=V=300V

Charge on C4C4 is given by,

Q4=VCQ4=VC

⇒Q4=300×2003×10−12⇒Q4=300×2003×10−12

⇒Q4=2×10−8C⇒Q4=2×10−8C

⇒V4=Q4C4⇒V4=Q4C4

⇒V4=2×10−8100×10−12=200V⇒V4=2×10−8100×10−12=200V

Voltage across C1C1is given by,

V1=V−V4V1=V−V4

⇒V1=200−100=100V⇒V1=200−100=100V

Hence, the potential difference, V1V1, across C1C1 is

100 V100 V

Charge on C1C1 is given by,

Q1=V1C1Q1=V1C1

⇒Q1=100×100×10−12⇒Q1=100×100×10−12

⇒Q1=10−8C⇒Q1=10−8C

C2C2 and C3C3 having the same capacitances have a

100 V100 V

potential difference together. Since C2C2 and C3C3 are in series, the potential difference across C2C2 and C3C3 is given by,

V2=V3=50VV2=V3=50V

Charge on C2C2 is given by,

Q2=V2C2Q2=V2C2

⇒Q2=50×200×10−12⇒Q2=50×200×10−12

⇒Q2=10−8C⇒Q2=10−8C

Charge on C3C3 is given by,

Q3=V3C3Q3=V3C3

⇒Q3=50×200×10−12⇒Q3=50×200×10−12

⇒Q3=10−8C⇒Q3=10−8C

Clearly, the equivalent capacitance of the given circuit is 2003pF2003pFand,

Q1=10−8CQ1=10−8C

,

Q2=10−8CQ2=10−8C

,

Q3=10−8CQ3=10−8C

,

Q4=2×10−8CQ4=2×10−8C

,

V1=100VV1=100V

, V2=50VV2=50V, V3=50VV3=50V, V4=200VV4=200V.

2.5 mm2.5 mm

. The capacitor is charged by connecting it to a

400 V400 V

supply.

a) How much electrostatic energy is stored by the capacitor?

Area of the parallel capacitor’s plates, A=90cm2=90×10−4m2A=90cm2=90×10−4m2

Distance separating the plates, d=2.5mm=2.5×10−3md=2.5mm=2.5×10−3m

Potential difference across the pates,

V=400VV=400V

The formula for capacitance will be,

C=εoAdC=εoAd

Electrostatic energy stored in capacitor is given by,

E1=12CV2E1=12CV2

⇒E1=12εoAdV2⇒E1=12εoAdV2

Where, εoεois the Permittivity of free space

εo=8.854×10−12C2N−1m−2εo=8.854×10−12C2N−1m−2

⇒E1=128.854×10−12×90×10−42.5×10−34002⇒E1=128.854×10−12×90×10−42.5×10−34002

⇒E1=2.55×106J⇒E1=2.55×106J

The stored electrostatic energy inside the capacitor is, 2.55×106J2.55×106J.

⇒V′=90×10−4×2.5×10−3=2.25×10−5m3⇒V′=90×10−4×2.5×10−3=2.25×10−5m3

Energy density in the capacitor is given by,

u=E1V′u=E1V′

u=2.55×10−62.25×10−5=0.113Jm−3u=2.55×10−62.25×10−5=0.113Jm−3

Also, u=E1V′u=E1V′

⇒u=12εoAdV2Ad=12εo(Vd)2⇒u=12εoAdV2Ad=12εo(Vd)2

Where, Vd=EVd=E, E is electric field.

⇒u=12εoE2⇒u=12εoE2

Hence, derived.

27. A

4μF4μF

capacitor is charged by a

200 V200 V

supply. It is then disconnected from the supply, and is connected to another uncharged

2μF2μF

capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

Ans: It is provided that,

A charged capacitor has capacitance,

C1=4μF=4×10−6FC1=4μF=4×10−6F

Supply voltage,

V1=200VV1=200V

Electrostatic energy stored in C1C1capacitor is given by,

E1=12C1V21E1=12C1V12

⇒E1=12×4×10−6×2002⇒E1=12×4×10−6×2002

⇒E1=8×10−2J⇒E1=8×10−2J

An uncharged capacitor’s capacitance,

C2=2μF=2×10−6FC2=2μF=2×10−6F

When C2C2 is joined to the circuit, the potential attained by it is V2V2.

According to the conservation of charge,

V2(C1+C2)=V1C1V2(C1+C2)=V1C1

⇒V2(4+2)×10−6=200×4×10−6⇒V2(4+2)×10−6=200×4×10−6

⇒V2=4003V⇒V2=4003V

The formula for electrostatic energy for the two capacitors combination is given by,

E2=12V22(C1+C2)E2=12V22(C1+C2)

⇒E2=12(4003)2(4+2)×10−6⇒E2=12(4003)2(4+2)×10−6

⇒E2=5.33×10−2J⇒E2=5.33×10−2J

The amount of lost electrostatic energy by capacitor is,

=E1−E2=0.08−0.0533=0.0267J=E1−E2=0.08−0.0533=0.0267J

Therefore, the lost electrostatic energy is 0.0267J0.0267J.

ΔxΔx

. Hence, work done by the force

=FΔx=FΔx

As a result, the capacitor’s potential energy rises by an amount given as

uAΔxuAΔx

Where, u is the energy density,

A is the area of each plate,

d is the Distance separating the plates,

V is the difference in potential across the plates.

The work done will be equal to the rise in the potential energy i.e.,

FΔx=uAΔxFΔx=uAΔx

⇒F=uA=12εoE2A⇒F=uA=12εoE2A

The formula for electric intensity is given by,

E=VdE=Vd

⇒F=12εoE(Vd)A⇒F=12εoE(Vd)A

Hence, capacitance will be,

C=εoAdC=εoAd

⇒F=12CVE⇒F=12CVE

The formula for charge in the capacitor is,

Q=CVQ=CV

⇒F=12QE⇒F=12QE

The 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.

(Image will Be Updated Soon)

Show that the capacitance of a spherical capacitor is given by

C=4πεor1r2r1−r2C=4πεor1r2r1−r2

Where r1r1 and r2r2 are the radii of outer inner spheres, respectively.

Outer shell’s radius =r1=r1

Inner shell’s radius =r2=r2

The outer shell’s inner surface has charge +Q+Q.

The inner shell’s outer surface has induced charge −Q−Q.

The difference in potential between the two shells is given by,

V=Q4πεor2−Q4πεor1V=Q4πεor2−Q4πεor1

Where, εoεois the Permittivity of free space

⇒V=Q4πεo[1r2−1r1]⇒V=Q4πεo[1r2−1r1]

⇒V=Q(r1−r2)4πεor1r2⇒V=Q(r1−r2)4πεor1r2

The formula for capacitance is given by,

C=QVC=QV

⇒C=4πεor1r2r1−r2⇒C=4πεor1r2r1−r2

This proved.

12 cm12 cm

and an outer sphere of radius

13 cm13 cm

. The outer sphere is earthed and the inner sphere is given a charge of

2.5μC2.5μC

. The space between the concentric spheres is filled with a liquid of dielectric constant

3232

a) Determine the capacitance of the capacitor.

Outer cylinder’s radius,

r1=13cm=0.13mr1=13cm=0.13m

Radius of inner cylinder,

r2=12cm=0.12mr2=12cm=0.12m

Charge on the inner cylinder, q=2.5μC=2.5×10−6Cq=2.5μC=2.5×10−6C

The formula for capacitor’s capacitance is given by,

C=4πεoεrr1r2r1−r2C=4πεoεrr1r2r1−r2

Where, εoεois the Permittivity of free space

εo=8.854×10−12C2N−1m−2εo=8.854×10−12C2N−1m−2

Value of

14πεo=9×109NC−2m−214πεo=9×109NC−2m−2

⇒C=32×0.12×0.139×109(0.13−0.12)⇒C=32×0.12×0.139×109(0.13−0.12)

⇒C≈5.5×10−9F⇒C≈5.5×10−9F

Hence, the capacitor’s capacitance is approximately 5.5×10−9F5.5×10−9F.

V=qCV=qC

⇒V=2.5×10−65.5×10−9=4.5×102V⇒V=2.5×10−65.5×10−9=4.5×102V

Hence, the inner sphere’s potential is 4.5×102V4.5×102V.

c) Compare the capacitance of this capacitor with that of an isolated sphere of radius

12 cm12 cm

. Explain why the latter is much smaller.

The formula for sphere’s capacitance is given by,

C′=4πεorC′=4πεor

⇒C′=4π×8.854×10−12×12×10−2⇒C′=4π×8.854×10−12×12×10−2

⇒C′=1.33×10−11F⇒C′=1.33×10−11F

The 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.

a) 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 Q1Q24πεor2Q1Q24πεor2, where r is the distance between their centres?

(=80)(=80)

than say, mica

(=6)(=6)

.

15 cm15 cm

and radii

1.5 cm1.5 cm

and

1.4 cm1.4 cm

. The outer cylinder is earthed and the inner cylinder is given a charge of

3.5μC3.5μC

. 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).

Co-axial cylinder’s length,

l=15cm=0.15ml=15cm=0.15m

Outer cylinder’s radius,

r1=1.5cm=0.015mr1=1.5cm=0.015m

Radius of inner cylinder,

r2=1.4cm=0.014mr2=1.4cm=0.014m

Charge on the inner cylinder, q=3.5μC=3.5×10−6Cq=3.5μC=3.5×10−6C

The formula for co-axial cylinder’s capacitance of radii r1r1 and r2r2 is given by,

C=2πεolloge(r1r2)C=2πεolloge(r1r2)

Where, εoεois the Permittivity of free space

εo=8.854×10−12C2N−1m−2εo=8.854×10−12C2N−1m−2

⇒C=2π×8.854×10−12×0.152.303log10(0.0150.014)⇒C=2π×8.854×10−12×0.152.303log10(0.0150.014)

⇒C=1.2×10−10F⇒C=1.2×10−10F

The difference in potential of the inner cylinder is given by,

V=qCV=qC

⇒V=3.5×10−61.2×10−10=2.92×104V⇒V=3.5×10−61.2×10−10=2.92×104V

The difference in potential will be 2.92×104V2.92×104V.

1 kV1 kV

, using a material of dielectric constant

33

and dielectric strength about

107 Vm−1107 Vm−1

. (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

1010

of the dielectric strength. What minimum area of the plates is required to have a capacitance of

50 pF50 pF

?

Parallel plate capacitor’s potential,

V=1kV=1000VV=1kV=1000V

Material’s dielectric constant, εr=3εr=3

Dielectric strength

=107 Vm−1=107 Vm−1

For safety, the electric field intensity never exceeds 10% of the dielectric strength.

Electric field intensity, E=0.1×107=106Vm−1E=0.1×107=106Vm−1

Parallel plate capacitor’s capacitance,

C=50 pF=50×10−12FC=50 pF=50×10−12F

The formula for distance separating the plates is given by,

d=VEd=VE

⇒d=1000106=10−3m⇒d=1000106=10−3m

The formula for capacitance is given by,

C=εoεrAdC=εoεrAd

A is the area of each plate,

εoεois the Permittivity of free space

εo=8.854×10−12C2N−1m−2εo=8.854×10−12C2N−1m−2

A=CdεoεrA=Cdεoεr

A=50×10−12×10−38.85×10−12×3≈19cm2A=50×10−12×10−38.85×10−12×3≈19cm2

Clearly, the area of each plate is about 19cm219cm2.

a) A constant electric field in the z-direction,

15×106V15×106V

electrode. The dielectric strength of the gas surrounding the electrode is

5×107 Vm−15×107 Vm−1

. 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).

Potential difference,

V=15×106VV=15×106V

surroundings gas’s dielectric strength=5×107 Vm−1=5×107 Vm−1

Electric field intensity is equal to the dielectric strength, E=5×107 Vm−1E=5×107 Vm−1

The formula for spherical shell’s minimum radius required for the purpose is given by,

r=VEr=VE

⇒r=15×1065×107=0.3m=30cm⇒r=15×1065×107=0.3m=30cm

Clearly, the required minimum radius of the spherical shell is 30cm30cm.

r2r2

and charge

q2q2

. Show that if

q1q1

is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge

q2q2

on the shell is.

q1q1

on a small sphere. Hence, the potential difference between the sphere and the shell does not depend on the charge

q2q2

. For positive charge

q1q1

, the potential difference V is always positive.

a) The top of the atmosphere is at about

400 kV400 kV

with 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

100Vm−1100Vm−1

. 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!)

1m21m2

. Will he get an electric shock if he touches the metal sheet in the morning?

1800 A1800 A

on 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?

1800 A1800 A

, the atmosphere is not discharged entirely. The two reversing currents are in equilibrium, and the atmosphere persists in neutral.

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