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#420014

Topic: Electrostatics of Conductors

(a) A conductor A with a cavity as shown in Fig. (a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor.

(b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q.

(c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.

Solution

(a) Let us consider a Gaussian surface that is lying wholly within a conductor and enclosing the cavity. The electric field intensity E inside the charged conductor is zero.

Let q is the charge inside the conductor and ∈0 is the permittivity of free space.

According to Gauss's law,

Flux, ϕ =¯E,

¯ds =

q

∈0

Here, E = 0

q

∈0= 0

∵ ∈0 ≠ 0

∴ q = 0

Therefore, charge inside the conductor is zero.

The entire charge Q appears on the outer surface of the conductor.

(b) The outer surface of conductor A has a charge of amount Q. Another conductor B having charge +q is kept inside conductor A and it is insulated from A. Hence, a charge of

amount - q will be induced in the inner surface of conductor A and +q is induced on the outer surface of conductor A. Therefore, total charge on the outer surface of conductor A

is Q + q.

(c) A sensitive instrument can be shielded from the strong electrostatic field in its environment by enclosing it fully inside a metallic surface. A closed metallic body acts as an

electrostatic shield.

#420038


A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (σ /2ϵ0)n where n is the unit vector in the outward normal direction, and

σ is the surface charge density near the hole.

Solution



Let us consider a conductor with a cavity or a hole. Electric field inside the cavity is zero.

Let E is the electric field just outside the conductor, q is the electric charge, σ is the charge density, and ∈0 is the permittivity of free space.

charge | q | =¯σ ×

¯ds

According to Gauss's law,

Flux, ϕ =¯E.

¯ds =

| q |

∈0

Eds =

¯σ ×

¯ds

∈0

∴ E =σ∈0

n

Therefore, the electric field just outside the conductors is σ∈0

n. This field is a superposition of field due to the cavity (E ′ ) and the field due to the rest of the charged conductor

(E ′ ). These fields are equal and opposite inside the conductor, and equal in magnitude and direction outside the conductor.

∴ E ′ + E ′ = E

E ′ =E

2

E ′ =σ

2 ∈0n

Therefore, the field due to the rest of the conductor is σ∈0

n. Hence, proved.

#420100

Topic: Electrostatic Potential

Two charges 5 × 10 − 8 C and −3 × 10 − 8 C are located 16 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.

Solution

Let p be the point of zero potential be p at distance r from the charge 1..

d = 16 cm

Electric potential, v = q1 /4πϵor + q2 /4πϵo(d − r)

for v = 0,

q1

r = −q2

d − r

⇒5 × 10 − 8

r =−3 × 10 − 8

0.16 − r

r = 10 cm

If, point p is outside,

v = q1 /4πϵor + q2 /4πϵo(r − d)

so here, r = 40 cm

#420102


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

Solution

q = 5μC

length of the hexagon = 10 cm

distance of center from the vertex,d = 10 cm

V = 6q/4πϵod

V = 2.7 × 106V



#420106


Two charges 2μC and −2μC are placed at points A and B 6 cm apart.

(a) Identify an equipotential surface of the system.

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

Solution

(a) An equipotential surface is the plane on which total potential is zero.

It is at the midpoint of the line AB and perpendicular to it.

(b) The direction of the electric field at every point on this surface is normal to the plane in direction of AB.

#420117

Topic: Dielectrics in capacitors

A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF =10 − 12 F). What will be the capacitance if the distance between the plates is reduced by half,

and the space between them is filled with a substance of dielectric constant 6?

Solution

Let the distance between the plates initially be d.

Thus initial capacitance C1 = ϵoA/d = 8 pF

Substance of dielectric constant K = 6 is inserted between the plates.

Also, now the distance between the plates is d ′ = 0.5d

Thus new capacitance C2 = KϵoA/d ′

∴ C2 = 6ϵoA/0.5d = 12C1 = 96 pF

#420120

Topic: Equivalent Capacitance in series-parallel

Three capacitors each of capacitance 9 pF are connected in series.

(a) What is the total capacitance of the combination?

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

Solution

(a) In series combination, the equivalent Capacitance is given by

1

Ceq=

1

C1+

1

C2+

1

C3

∴1

Ceq=

1

9+

1

9+

1

9

∴ Ceq = 3pF

(b) When capacitors of equal capacity are connected in series, the potential difference across each capacitor is V

n, where V is total potential and n is number of capacitors. Here,

the p.d across each capacitor will be 120

3= 40V

#420125


Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel.

(a) What is the total capacitance of the combination?

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

Solution



(a) For parallel combination, Ceq = C1 + C2 + C3 = (2 + 3 + 4)pF = 9pF

(b) q = V × C

for parallel array, V remains the same across each capacitor.

for C = 2 pF

q1 = VC = 100 × 2 × 10 − 12 = 2 × 10 − 10C

for C = 3 pF

q2 = VC = 100 × 3 × 10 − 12 = 3 × 10 − 10C

for C = 4 pF

q1 = VC = 100 × 4 × 10 − 12 = 4 × 10 − 10C

#420130

Topic: Types of Capacitors

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

capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?

Solution

A = 6 × 10 − 3 m2, d = 3 × 10 − 3 m

C = ϵoA/d

C = 17.71 pF ≈ 18 pF

Voltage V=100 Volts

So charge on the capacitor is given by,

q = CV = 1.8 × 10 − 9C

#420137


Exercise:

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

capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?

Explain what would happen if in the capacitor given in Exercise a 3mm thick mica sheet (of dielectric constant = 6) were inserted between the plates,

(a) While the voltage supply remained connected.

(b) After the supply was disconnected.

Solution



C =ϵoA

d

According to the values given C = 1.8 × 10 − 11F

Now Q = CV

Hence charge on the positive plate will be 1.8 × 10 − 9C

(a) Dielectric constant, k = 6

C = 1.8 × 10 − 11 F

new capacitance, C ′ = kC = 108 pF

V = 100volts

q ′ = V × C ′ = 1.08 × 10 − 8 C

V remains the same.

(b) C ′ = kC = 6 × 1.8 × 10 − 11 = 108pF

q remains the same.

q = 1.8 × 10 − 9 C

V ′ = q/C ′ = 16.67 V

#420140

Topic: Energy in a Capacitor

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

Solution

The total energy of capacitor is E =1

2CV2 =

1

2× 12 × 10 − 12 × 502 = 1.5 × 10 − 8J

#420144


A 600 pF capacitor is charged by a 200V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic

energy is lost in the process?

Solution

Initial energy, E =1

2CV2

E = 1.2 × 10 − 5 J

when another capacitor is connected,

1

C1=

1

C+

1

C

C1 = 300 pF

E1 =1

2C1V2 = 0.6 × 10 − 5

ΔE = E − E1 = 6 × 10 − 6 J

#420160


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

Solution

Diagonal of the cube face, d = √(b2 + b2) = b√2

Diagonal of cube l = √(d2 + b2) = b√3

Distance between center of cube and vertices, r = l /2 = b√3/2

Potential, v = 8q/4πϵor = 4q/√3πϵob (8 due to presence of 8 charges at the vertices)

by symmetry of charge electric field, E=0.



#420172


Two tiny spheres carrying charges 1.5μC and 2.5μC are located 30 cm apart. Find the potential and electric field:

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

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

Solution

Two charges placed at points A and B are represented in the given figure. O is the mid-point of the line joining the two charges.

Magnitude of charge located at A, q1 = 1.5μC

Magnitude of charge located at B, q2 = 2.5μC

Distance between the two charges, d = 30cm = 0.3m

(a)

Let V1 and E1 are the electric potential and electric field respectively at O.

V1 = Potential due to charge at A + Potential due to charge at B

V1

q1

4π ∈0

d

2 +

q2

4π ∈0

d

2 =

1

4π ∈0

d

2 (q1 + q2)

Where,

∈0 = Permittivity of free space

1

4π ∈0= 9 × 109NC2m − 2

∴ V1 =

9 × 109 × 10 − 6

0.30

2

E1 = Electric field due to q2− Electric field due to q1

=

q2

4π ∈0

d

22

−

q1

4π ∈0

d

22

=

9 × 109

0.30

22

× 106 × (2.5 − 1.5)

= 4 × 105Vm − 1

Therefore, the potential at mid-point is 2.4 × 105V and the electric field at mid-point is 4 × 105Vm − 1. The field is directed from the larger charge to the smaller charge.

(b) Consider a point Z such that normal distance OZ = 10cm = 0.1m, as shown in the following figure.

V2 and E2 are the electric potential and electric field respectively at Z.

It can be observed from the figure that distance,

BZ = AZ = √(0.1)2 + (0.15)2 = 0.18am

V2 = Electric potential due to A + Electric Potential due to B

=

q1

4π ∈0(AZ) +q2

4π ∈0(BZ)

=

9 × 109 × 10 − 6

0.18 (1.5 + 2.5)

= 2 × 105V

Electric field due to q at Z,

( ) ( ) ( )

( )

( ) ( )

( )



EA =q1

4π ∈0(AZ)2

=

9 × 109 × 1.5 × 10 − 6

(0.18)2

= 0.416 × 106V/m

Electric field due to q2 at Z,

EB =q2

4π ∈0(BZ)2

=9 × 109 × 2.5 × 10 − 6

(0.18)2

= 0.69 × 106Vm − 1

The resultant field intensity at Z,

E = E2A

+ E2B

+ 2EAEB cos 2θ

Where, 2θis the angle, ∠AZB

From the figure, we obtain

cos θ =0.10

0.18 =5

9 = 0.5556

θ = cos − 100.5556 = 56.25

∴ 2θ = 112.50

cos 2θ = − 0.38

E = √(0.416 × 106)2 × (0.69 × 106)2 + 2 × 0.416 × 0.69 × 1012 × ( − 0.38)

= 6.6 × 105Vm − 1

Therefore, the potential at a point 10 cm (perpendicular to the mid-point) is 2.0 × 1.5V and electric field is 6.6 × 105Vm − 1.

#420284


If one of the two electrons of a H2 molecule is removed, we get a hydrogen molecular ion H +2

. In the ground state of an H +2

, the two protons are separated by roughly 1.5A , and

the electron is roughly 1A from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.

Solution

√



The system of two protons and one electron is represented in the given figure.

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

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

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

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

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

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

The potential energy at infinity is zero.

Potential energy of the system,

V =q1q2

4π ∈0d1+

q2q3

4π ∈0d3+

q3q1

4π ∈0d2

Substituting 1

4π ∈0= 9 × 109Nm2 C − 2, we obtain

V =

9 × 109 × 10 − 19 × 10 − 19

10 − 19 − (16)2 +(1.6)2

1.5 + − (1.6)2

= − 30.7 × 10 − 19J

= − 19.2 eV

Therefore, the potential energy of the system is −19.2 eV.

#420293


Two charges -q and +q are located at points (0, 0, -a) and (0, 0, a), respectively.

(a) What is the electrostatic potential at the points (0, 0, z) and (x, y, 0) ?

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

(c) How much work is done in moving a small test charge from the point (5,0,0) to (-7,0,0) along the x-axis? Does the answer change if the path of the test charge between the

same points is not along the x-axis?

Solution

[ ]



(a) Zero at both the points

Charge − q is located at (0, 0, − a) and charge + q is located at (0, 0, a). Hence, they form a dipole. Point (0, 0, z) is on the axis of this dipole and point (x, y, 0) is normal to the axis

of the dipole. Hence, electrostatic potential at point (x, y, 0) is zero. Electrostatic potential at point (0, 0, z) is given by,

V =1

4π ∈0

q

z − a +1

4π ∈0−

q

z + a

=q(z + a − z + a)

4π ∈0(z2 − a2)

=2qa

4π ∈0(z2 − a2) =p

4π ∈0(z2 − a2)

Where,

∈0 = Permittivity of free space

p = Dipole moment of the system of two charges = 2qa

(b) Distance r is much greater than half of the distance between the two charges. Hence, the potential (V) at a distance r is inversely proportional to square of the distance i.e.,

V ∝1

r2

(c) Zero. 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) to point (−7, 0, 0) along the x-axis. Electrostatic potential (V1) at point (5, 0, 0) is given by,

Electrostatic potential, V2, at point (− 7, 0, 0) is given by,

V2 =−q

4π ∈0

1

√( − 7)2 + ( − a)2 +q

4π ∈0

1

√( − 7)2 + (a)2

=−q

4π ∈0√49 + a2 +q

4π ∈0

1

√49 + a2

= 0

Hence, no work is done in moving a small test charge from point (5, 0, 0) to point (−7, 0, 0) along the x-axis.

The answer does not change because work done by the electrostatic field in moving a test charge between the two points is independent of the path connecting the two points.

#420306

Topic: Electric Dipole

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 r for r/a >> 1, and contrast your

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

Solution

( ) ( )



Four charges of same magnitude are placed at points X, Y, Y, and Z respectively, as shown in the following figure.

A point is located at P, which is r distance away from point Y.

The system of charges forms an electric quadrupole.

It can be considered that the system of the electric quadrupole has three charges.

Charge +q placed at point X

Charge −2q placed at point Y

Charge +q placed at point Z

XY = YZ = a

YP = r

PX = r + a

PZ = r − a

Electrostatic potential caused by the system of three charges at point P is given by,

V =1

4π ∈0

q

XP −2q

YP +q

ZP

=1

4π ∈0

1

r + a −2q

r +q

r − a

=q

4π ∈0

r(r − a) − 2(r + a)(r − a) + r(r + a)

r(r + a)(r − a)

=q

4π ∈0

r2 − ra − 2r2 + 2a2 + r2 + ra

r(r2 − a2)

=q

4π ∈0

2a2

r(r2 − a2)

=

2qa2

4π ∈0r3 1 −a2

r2

Since r

a >> 1,

∴a

r << 1

a2

r2 is taken as negligible.

∴ V =

2qa2

4π ∈0r3

It can be inferred that potential, V ∝1

r3

However, it is known that for a dipole, V ∝1

r2

And, for a monopole, V ∝

1

r

[ ]

[ ]

[ ]

[ ]

[ ]

( )



#420316


What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realise from your answer why ordinary capacitors

are in the range of F or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]

Solution

Capacitance of a parallel capacitor, V = 2F

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

Capacitance of a parallel plate capacitor is given by the relation,

C =∈0A

d

A =Cd

∈0

Where,

∈0 = Permittivity of free space = 8.85 × 10 − 12C2N − 1m − 2

∴ A =

2 × 0.5 × 10 − 2

8.85 × 10 − 12 = 1130km2

Hence, the area of the plates is too large. To avoid this situation, the capacitance is taken in the range of μF.

#420324


Obtain the equivalent capacitance of the network shown in the figure. For a 300 V supply, determine the charge and voltage across each capacitor.

Solution

Capacitance of capacitor C1 is 100 pF.




Supply potential, V = 300V

Capacitors C2 and C3 are connected in series. Let their equivalent capacitance be C ′ .

∴1

C ′ =1

200 +1

200 =2

200

C ′ = 100pF

Capacitors C1 and C ′ are in parallel. Let their equivalent capacitance be C ″ .

∴ C ″ = C ′ + C1

= 100 + 100 = 200pF

C ″ and C4 are connected in series. Let their equivalent capacitance be C.



∴1

C =1

C ″ +1

C4

=1

200 +1

100 =2 + 1

200

C =200

3 pF

Hence, the equivalent capacitance of the circuit is 200

3 pF.

Potential difference across C ″ = V ″

Potential difference across C4 = V4

∴ V ″ + V4 = V = 300V

Charge on C4 is given by,

Q4 = CV

=200

3 × 10 − 12 × 300

= 2 × 10 − 8C

V4 =Q4

C4

=

2 × 10 − 8

100 × 10 − 12 = 200V

∴ Voltage across C1 is given by,

V1 = V − V4

= 300 − 200 = 100V

Hence, potential difference, V1, across C1 is 100 V.

Charge on C1 is given by,

Q1 = C1V1

= 100 × 10 − 12 × 100

= 10 − 8C

C2 and C3 having same capacitances have a potential difference of 100 V together. Since C2 and C3 are in series, the potential difference across C2 and C3 is given by,

V2 = V3 = 50V

Therefore, charge on C2 is given by,

Q2 = C2V2

= 200 × 10 − 12 × 50

= 10 − 8C

And charge on C3 is given by,

Q3 = C3V3

= 200 × 10 − 12 × 50

= 10 − 8C

Hence, the equivalent capacitance of the given circuit is 200

3pF with,

Q1 = 10 − 8C, V1 = 100V

Q2 = 10 − 8C, V2 = 50V

Q3 = 10 − 8C, V3 = 50V

Q4 = 2 × 10 − 8C, V4 = 200V



#420330


The plates of a parallel plate capacitor have an area of 90 cm2 each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply.

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

(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.

Solution

(a) Area of the plates of a parallel plate capacitor, A = 90cm2 = 90 × 10 − 4m2

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

Potential difference across the plates, V = 400V

Capacitance of the capacitor is given by the relation,

C =∈0A

d

Electrostatic energy stored in the capacitor is given by the relation,

E1 =1

2 CV2

= 1

2

∈0A

d V2

Where,

∈0 = Permittivity of free space = 8.85 × 10 − 12C2N − 1m − 2

∴ E1 =1 × 8.85 × 106−12 × 90 × 10 − 4 × (400)2

2 × 2.5 × 10 − 3 = 2.55 × 10 − 6J

Hence, the electrostatic energy stored by the capacitor is 2.55 × 10 − 6J.

(b) Volume of the given capacitor,

V ′ = A × d

= 90 × 10 − 4 × 25 × 10 − 3

= 2.25 × 10 − 4m3

Energy stored in the capacitor per unit volume is given by,

u =E1

V ′

=2.55 × 10 − 6

2.25 × 10 − 4 = 0.113Jm − 3

Again, u =E1

V ′

=

1

2 CV2

Ad =

∈0A

2d V2

Ad =1

2 ∈0

V

d2

Where,

V

d = Electric intensity = E

∴ u =1

2 ∈0E2

( )



#420342


A 4μF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2μF capacitor. How much electrostatic energy of

the first capacitor is lost in the form of heat and electromagnetic radiation?

Solution

Capacitance of a charged capacitor, C1 = 4μF = 4 × 10 − 6F

Supply voltage, V1 = 200V

Electrostatic energy stored in C1 is given by,

E1 =1

2 C1V21

=1

2 × 4 × 10 − 6 × (200)2

= 8 × 10 − 2J

Capacitance of an uncharged capacitor, C2 = 2μF = 2 × 10 − 6F

When C2 is connected to the circuit, the potential acquired by it is V2.

According to the conservation of charge, initial charge on capacitor C1 is equal to the final charge on capacitors, C1 and C2

∴ V2(C1 + C2) = C1V1

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

V2 =400

3 V

Electrostatic energy for the combination of two capacitors is given by,

E2 =1

2 (C1 + C2)V22

=1

2 (2 + 4) × 10 − 6 ×400

32

= 5.33 × 10 − 2J

Hence, amount of electrostatic energy lost by capacitor C1

= E1 − E2

= 0.08 − 0.0533 = 0.0267

= 2.67 × 10 − 2J

#420347


Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (1

2 ) QE, 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 .

Solution

( )



Let F be the force applied to separate the plates of a parallel plate capacitor by a distance of △x. Hence, work done by the force to do so = F△x

As a result, the potential energy of the capacitor increases by an amount given as uA△x

Where,

u = Energy density

A = Area of each plate

d = Distance between the plates

V = Potential difference across the plates

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

F△x = uA△x

F = uA =1

2 ∈0E2A

Electric intensity is given by,

E =V

d

∴ F =1

2 ∈0

V

d EA =1

2∈0A

V

d E

However, capacitance, C =∈0A

d

∴ F =1

2 (CV)E

Charge on the capacitor is given by,

Q = CV

∴ F =1

2 QE

The physical origin of the factor, 1

2, in the force formula lies in the fact that just outside the conductor, field is E and inside it is zero. Hence, it is the average value, E

2, of the field

that contributes to the force.

#420355

Topic: Capacitance

A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Fig.). Show that the capacitance of a spherical capacitor is

given by

c =4πϵ0 r1r2

r1 − r2

where r1 and r2 are the radii of outer and inner spheres, respectively.

Solution

( )

( ) ( )



Radius of the outer shell = r1

Radius of the inner shell = r2

The inner surface of the outer shell has charge +Q.

The outer surface of the inner shell has induced charge −Q.

Potential difference between the two shells is given by,

V =

Q

4πεor2−

Q

4πεor1

Where,

εo= Permittivity of free space

V =

Q

4πεor2−

Q

4πεor1

V =

Q

4πεo

1

r2−

1

r1

V =

Q(r1 − r2)

4πεor1r2

Capacitance of the given system is given by,

C =Q

V

C =4πεor1r2

r1 − r2

Hence, proved.

#420361

Topic: Capacitance

A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5μC. The

space between the concentric spheres is filled with a liquid of dielectric constant 32.

(a) Determine the capacitance of the capacitor.

(b) What is the potential of the inner sphere?

(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.

Solution

[ ]



Radius of the inner sphere, r2= 12 cm = 0.12 m

Radius of the outer sphere, r1= 13 cm = 0.13 m

Charge on the inner sphere, q = 2.5 × 10 − 6 C

Dielectric constant of a liquid, εr = 32

(a)

Capacitance, C =4πεoεrr1r2

r1 − r2

where,

εo= Permittivity of free space = 8.85 × 10 − 12 C2N − 1m2

∴ C ≈ 5.5 × 10 − 9 F

(b)

Potential of the inner sphere is given by,

V = q/C

= 4.5 × 102V

(c)

Radius of an isolated sphere, r= 12 cm

Capacitance of the sphere is given by the relation,

C ′ = 4πεor

= 1.33 × 10 − 11 F

The capacitance of the isolated sphere is less in comparison to the concentric spheres. This is because the outer sphere of the concentric spheres is earthed. Hence, the potential

difference is less and the capacitance is more than the isolated sphere.

#420371

Topic: Dielectrics and Polarisation

Answer carefully:

(a) Two large conducting spheres carrying charges Q1 and Q2 are brought close to each other. Is the magnitude of electrostatic force between them exactly given by

Q1Q2 /4πϵ0r2, where r is the distance between their centres?

(b) If Coulomb's law involved 1/r3 dependence (instead of 1/r2),would Gausss law be still true ?

(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?

(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?

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

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

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

Solution



(a) No, because there is non-uniform charge distribution on the sphere.

(b) No. Gauss Law would not be valid as the flux then would be dependent on the distance and shape of surface.

(c) Not necessarily. (True only if the field line is a straight line.) The field line gives the direction of acceleration, not that of velocity, in general.

(d) Since the electron returns to original point after completing the orbit, the displacement is zero, and hence the work done is zero. This is independent of the shape of orbit.

(e) Electric field is discontinuous across the surface of a charged conductor. However the electric potential is continuous.

(f) A single conductor is a capacitor with one of the plates at infinity.

(g) Water molecules have a permanent dipole moment, hence has a greater dielectric constant. On the other hand, mica doesn't have any dipole moment,so the dielectric

constant is low.

#420384

Topic: Capacitance

A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5μC.

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).

Solution

Length of a co-axial cylinder, l = 15cm = 0.15m

Radius of outer cylinder, r1 = 1.5cm = 0.015m

Radius of inner cylinder, r2 = 1.4cm = 0.014m

Charge on the inner cylinder, q= 3.5 µC

Capacitance of coaxial cylinder,

C =2πεol

ln(r1 / r2)

= 1.2 × 10 − 10 F

Potential of the inner cylinder is given by,

V =q

C

= 2.92 × 104 V

#420387


A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 107 Vm − 1. (Dielectric strength is the

maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never

to exceed, say 10% of the dielectric strength.What minimum area of the plates is required to have a capacitance of 50 pF ?

Solution



Potential rating of a parallel plate capacitor, V = 1 kV = 1000 V

Dielectric constant of a material, εr = 3

Dielectric strength = 107 V/m

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

Hence, electric field intensity, E = 10% of 107 = 106 V/m

Capacitance of the parallel plate capacitor, C = 50 pF

Distance between the plates is given by,

d =V

E

=1000

106 = 10 − 3m

Capacitance, C =εoεrA

d

∴ A =Cd

εoεr=

50 × 10 − 12 × 10 − 3

8.85 × 10 − 12 × 3≈ 19 cm2

#420389


Describe schematically the equipotential surfaces corresponding to

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

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

(c) a single positive charge at the origin, and

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

Solution

(a) Equidistant planes parallel to the x-y plane are the equipotential surfaces.

(b) Planes parallel to the x-y plane are equipotential surfaces with exception that when the planes get closer, the field increases.

(c) Concentric spheres centered at the origin are equipotential surfaces.

(d) A periodically varying shape near the grid which gradually reaches the shape of planes parallel to the grid at far distances.

#420398


In a Van de Graaff type generator a spherical metal shell is to be a 15 × 106V electrode. The dielectric strength of the gas surrounding the electrode is 5 × 107Vm − 1.What is the

minimum radius of thespherical 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.)

Solution



Potential difference, V = 15 × 106 V

Dielectric strength of the surrounding gas = 5 × 107 V/m

Electric field intensity, E= Dielectric strength = 5 × 107 V/m

Minimum radius of the spherical shell required for the purpose is given by,

r =V

E

=15 × 106

5 × 107= 0.3m

Hence, the minimum radius of the spherical shell required is 30 cm.

#420404


Answer the following:

(a) The top of the atmosphere is at about 400 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 − 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!)

(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 1m2. Will he get an electric shock if he

touches the metal sheet next morning?

(c) The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 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?

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

(Hint: The earth has an electric field of about 100 Vm − 1 at its surface in the downward direction, corresponding to a surface charge density =−10 − 9 Cm − 2. Due to the slight

conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about + 1800 C is pumped every second into the earth as a whole. The earth,

however, does not get discharged since thunderstorms and lightning occurring continually allover the globe pump an equal amount of negative charge on the earth.)

Solution

(a) We do not get an electric shock as we step out of our house because the original equipotential surfaces of open air changes, keeping our body and the ground at the

same potential.

(b) Yes. The steady discharging current in the atmosphere charges up the aluminium sheet gradually and raises its voltage to an extent depending on the capacitance of the

capacitor (formed by the sheet, slab and the ground).

(c) The atmosphere is continually being charged by thunderstorms and lightning all over the globe and discharged through regions of ordinary weather. The two opposing

currents are, on an average, in equilibrium.

(d) Light energy involved in lightning; heat and sound energy in the accompanying thunder.

#420505




Figure shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the

figure). The charging current is constant and equal to 0.15A.

(a) Calculate the capacitance and the rate of change of potential difference between the plates.

(b) Obtain the displacement current across the plates.

(c) Is Kirchhoffs first rule (junction rule) valid at each plate of the capacitor Explain.

Solution

(a)

C = ε0A/d = 80.1 pF

dQ

dt = CdV

dt

dV

dt =0.15

80.1 × 10 − 12 = 1.87 × 109 V s − 1

(b)

The displacement current across the plate is same as the conduction current.

Hence, the displacement current , Id = 0.15 A

(c)

Yes, Kirchhoff's first rule is valid at each plate of the capacitor provided that we take the sum of conduction and displacement for the current.

7/4/2018 ... · the electric ﬁeld intensity e inside the charged conductor is zero. let q is the...

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