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

1. Two charges
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.

Ans: It is provided that,
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.

2. A regular hexagon of side
10 cm10 cm
has a charge
5μC5μC
at each of its vertices. Calculate the potential at the centre of the hexagon.

Ans: The given figure represents six equal charges, q=5×10−6Cq=5×10−6C, at the hexagon’s vertices.
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.

3. Two charges
2μC2μC
and
−2μC−2μC
are placed at points A and B,
6 cm6 cm
apart.
a) Identify an equipotential surface of the system.

Ans: The given figure represents two charges.
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.

b) What is the direction of the electric field at every point on this surface?

Ans: The electric field’s direction is perpendicular to the plane in the line AB direction at every location on this surface.

4. A spherical conductor of radius
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?

Ans: It is provided that,
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.

b) just outside the sphere?

Ans: Just outside the conductor, Electric field E is given by
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.

c) at a point
18 cm18 cm
from the centre of the sphere?

Ans: Let electric field at a given point which is
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.

5. A parallel plate capacitor with air between the plates has a capacitance of
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
?

Ans: It is provided that,
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

6. Three capacitors each of capacitance
9 pF9 pF
are connected in series.
a) What is the total capacitance of the combination?

Ans: It is provided that,
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

b) What is the potential difference across each capacitor if the combination is connected to a
120 V120 V
supply?

Ans: Provided that,
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

7. Three capacitors of capacitance
2pF, 3 pF and 4pF2pF, 3 pF and 4pF
are connected in parallel.
a) What is the total capacitance of the combination?

Ans: Provided that,
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.

b) Determine the charge on each capacitor if the combination is connected to a
100 V100 V
supply.

Ans: We have,
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

8. In a parallel plate capacitor with air between the plates, each plate has an area of 6×10−3m26×10−3m2 and the distance between the plates is
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?

Ans: It is provided that,
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.

9. Explain what would happen if in the capacitor given in Exercise 8, a
3 mm3 mm
thick mica sheet (of dielectric constant
=6=6
) were inserted between the plates,
a) while the voltage supply remained connected.

Ans: It is provided that,
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

c) after the supply was disconnected.

Ans: It is provided that,
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.

10. A
12pF12pF
capacitor is connected to a
50 V50 V
battery. How much electrostatic energy is stored in the capacitor?

Ans: It is provided that,
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.

11. A
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?

Ans: It is provided that,
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.

12. A charge of
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)

Ans: Charge located at the origin, q=8 mC=8×10−3Cq=8 mC=8×10−3C
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.

13. A cube of side
bb
has a charge
qq
at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.

Ans: Length of each cube’s side =l=l
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.

14. Two tiny spheres carrying charges
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

Ans: Two charges placed at points A and B are represented in the given figure.
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.

b) at a point
10 m10 cm
from this midpoint in a plane normal to the line and passing through the mid-point.

Ans: Consider a point Z such that
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+2EAEBcos⁡2θ
From figure,
cosθ=0.100.18=0.5556cos⁡θ=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⇒cos⁡2θ=−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

15. A spherical conducting shell of inner radius r1r1 and outer radius r2r2 has a charge Q.
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?

Ans: Provided that,
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.

16.
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
.

Ans: Electric field on the charged body’s one side is E1E1 and the electric field on the same   body’s other side is E2E2.
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^

b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.

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.

17. A long charged cylinder of linear charged density
λλ
is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?

Ans: Provided that,
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.

18. In a hydrogen atom, the electron and proton are bound at a distance of about d=0.53Aod=0.53Ao:
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.

Ans: Provided that,
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.

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)?

Ans: It is mentioned in the question that,
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.

c) What are the answers to (a) and (b) above if the zero of potential energy is taken at
1.06Ao1.06Ao
separation?

Ans: When potential energy is taken as zero, d1=1.06Aod1=1.06Ao
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.

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
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.

Ans: The figure shows two protons and one electron.
(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

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.

Ans: Let a be the radius of a sphere A,
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.

21. Two charges
−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)?

Ans: Charge
−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.

b) Obtain the dependence of potential on the distance
rr
of a point from the origin when
r/a >>1r/a >>1

Ans: Distance
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

c) How much work is done in moving a small test charge from the point
(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?

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
(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.

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
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)

Ans: Four charges of equal magnitude are located at points
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.

23. An electrical technician requires a capacitance of
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.

Ans: It is provided that,
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.

24. What is the area of the plates of a
2 F2 F
parallel plate capacitor, given that the separation between the plates is
0.5 cm0.5 cm
?

Ans: It is provided that,
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

25. Obtain the equivalent capacitance of the network in Figure. For a
300 V300 V
supply, determine the charge and voltage across each capacitor.
(Image will Be Updated Soon)

Ans: It is provided that,
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.

26. The plates of a parallel plate capacitor have an area of 90cm290cm2 each and are separated by
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?

Ans: It is provided that,
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.

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.

Ans: Volume of the given capacitor, V′=A×dV′=A×d
⇒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.

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.

Ans: Let F be the applied force to separate the parallel plates of a capacitor by a distance of
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.

29. A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Figure).
(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.

Ans: It is provided that,
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.

30. A spherical capacitor has an inner sphere of radius
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.

Ans: It is provided that,
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.

b) What is the potential of the inner sphere?

Ans: The inner sphere’s potential is given by,
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.

Ans: Isolated sphere’s radius, r=12×10−2mr=12×10−2m
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.

31. Answer carefully:
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?

Ans: The expression does not precisely give the force between two conducting spheres, Q1Q24πεor2Q1Q24πεor2 Because there is an irregular charge distribution on the spheres, hereafter, Coulomb’s law is not valid.

b) If Coulomb’s law involved 1r31r3 dependence (instead of 1r21r2), would Gauss’s law be still true?

Ans: Gauss’s law would not be valid if Coulomb’s law included dependency on 1r31r3 instead of 1r21r2.

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?

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.

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?

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.

e) We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?

Ans: No, the electric field is not continuous across the covering of a charged conductor. However, an electric potential is continuous.

f) What meaning would you give to the capacitance of a single conductor?

Ans: A single conductor’s capacitance is considered a parallel plate capacitor with one of the two plates at infinity.

f) Guess a possible reason why water has a much greater dielectric constant
(=80)(=80)
than say, mica
(=6)(=6)
.

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.

32. A cylindrical capacitor has two co-axial cylinders of length
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).

Ans: It is provided that,
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.

33. A parallel plate capacitor is to be designed with a voltage rating
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
?

Ans: It is provided that,
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.

34. Describe schematically the equipotential surface corresponding to
a) A constant electric field in the z-direction,

Ans: Equipotential surfaces are the equidistant planes parallel to the x-y plane for the constant electric field in the z-direction.

b) A field that uniformly increases in magnitude but remains in a constant (say, z) direction,

Ans: Equipotential surfaces are the parallel planes to the x-y plane, except that the field increases when the planes get closer.

c) A single positive charge at the origin, and

Ans: Equipotential surfaces are the concentric spheres centered at the origin.

d) A uniform grid consisting of long equally spaced parallel charged wires in a plane.

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.

35. In a Van de Graaff type generator a spherical metal shell is to be a
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).

Ans: It is provided that,
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.

36. A small sphere of radius r1r1 and charge q1q1 is enclosed by a spherical shell of radius
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.

Ans: According to Gauss’s law, the electric field between a sphere and a shell is concluded by the charge
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.

37. Answer the following:
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!)

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.

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
1m21m2
. Will he get an electric shock if he touches the metal sheet in the morning?

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.

c) The discharging current in the atmosphere due to the small conductivity of air is known to be
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?

Ans: The existence of thunderstorms and lightning energizes the atmosphere continuously. Therefore, indeed with a discharging current of
1800 A1800 A
, the atmosphere is not discharged entirely. The two reversing currents are in equilibrium, and the atmosphere persists in neutral.

d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during lightning?

Ans: Light energy, sound energy, and heat energy are wasted in the atmosphere during lightning and thunderstorms.