Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 5th Lesson Electrostatic Potential and Capacitance Textbook Questions and Answers.

## AP Inter 2nd Year Physics Study Material 5th Lesson Electrostatic Potential and Capacitance

Very Short Answer Questions

Question 1.

Can there be electric potential at a point with zero electric intensity? Give an example.

Answer:

Yes, There can exist potential at a point where the electric intensity is zero.

Ex :

- Between two similar charges intensity of electric field is zero. But potential is not zero.
- Inside the charged spherical conductor electric field intensity is zero but potential is not zero.

Question 2.

Can there be electric intensity at a point with zero electric potential ? Give an example.

Answer:

Yes, electic intensity need not be zero at a point where the potential is zero.

Ex :

1) At mid point between two equal opposite charges potential is zero. But intensity is not zero.

Question 3.

What are meant by equipotential surfaces ?

Answer:

Surface at every point of which the value of potential is the same is defined as equipotential surface

For a point charge, concentric spheres centred at a location of the charge are equipotential surfaces.

Question 4.

Why is the electric field always at right angles to the equipotential surface ? Explain.

Answer:

No work is done in moving a charge from one point on equipotential surface to the other. Therefore, component of electric field intensity along the equipotential surface is zero. Hence, the surface is perpendicular to the field lines.

Question 5.

Three capacitors of capacitances 1μF, 2μF and 3μF, are connected in parallel

(a) What is the ratio of charges ?

(b) What is the ratio of potential difference ?

Answer:

When capacitors are connnected in parallel

(a) q_{1} : q_{2} : q_{3} = V: C_{2} V: C_{3} V = 1μF : 2μF : 3μF

∴ q_{1} : q_{2} ; q_{3} = 1 : 2 : 3

(b) V_{1} : V_{2} : V_{3} = V : V : V = 1 : 1 : 1

Question 6.

Three capacitors of capacitances 1μE, 2μF and 3μF are connected in series

(a) What is the ratio of charges ?

(b) What is the ratio of potential differences ?

Answer:

When capacitors are connected in series

(a) q_{1} : q_{2} : q_{3} = q : q : q = 1 : 1 ; 1

(b) V_{1} : V_{2} : V_{3} = \(\frac{\mathrm{q}}{\mathrm{C}_1}: \frac{\mathrm{q}}{\mathrm{C}_2}: \frac{\mathrm{q}}{\mathrm{C}_3}\) = \(\frac{1}{1}: \frac{1}{2}: \frac{1}{3}\)

∴ V_{1} : V_{2} : V_{3} = 6 : 3 : 2

Question 7.

What happens to the capacitance of a parallel plate capacitor if the area of its plates is doubled ?

Answer:

\(\frac{C_2}{C_1}\) = \(\frac{A_2}{A_1}\) [∵ C_{2} = 2C_{1}]

Given A_{2} = 2A_{1} ; \(\frac{C_2}{C_1}\) = \(\frac{2 \mathrm{~A}_1}{\mathrm{~A}_1}\) [∴ C_{2} = 2C_{1}]

Therefore capacity increases by twice.

Question 8.

The dielectric strength of air is 3 × 10^{6}.Vm^{-1} at certain pressure, A parallel plate capacitor with air in between the plates has a plate separation of 1 cm. Can you charge the capacitor

to 3 × 10^{6}V?

Answer:

Dielectric strength of air E_{0} = 3 × 10^{6} Vm^{-1}

Electric field intensity between the plates, E = \(\frac{E_0}{K}\) = 3 × 10^{6} Vm^{-1} [∵ for air K = 1]

Distance between two plates, d = 1 cm = 102m

Electric potential difference between plates, V = Ed = 3 × 10^{6} × 10^{-2}

∴ V = 3 × 10^{4} Volt.

Hence we cant charge the capacitor upto 3 × 10^{6} Volt.

Short Answer Questions

Question 1.

Derive an expression for the electric potential due to a point charge. (T.S. Mar. ’16)

Answer:

Expression for the electric potential due to a point charge:

- Electric potential at a point is defined as the amount of workdone in moving a unit +ve charge from infinity to that point.

- Consider a point P at a distance r from the point charge having charge + q. The electric field at P = E = \(\frac{q}{4 \pi \varepsilon_0 x^2}\)
- Workdone in taking a unit +ve charge from B to A = dV = -E.dx (-ve sign shows that the workdone is +ve in the direction B to A, Whereas the potential difference is +ve in, the direction A to B.
- Therefore, potential at P = The amount of workdone in taking a unit +ve charge from infinity to P

Question 2.

Derive an expression for the electrostatic potential energy of a system of two point charges and find its relation with electric potential of a charge.

Answer:

Expression for the electrostatic potential energy of a system of two point charges:

- Let two point charges q
_{1}and q_{2}are separated by distance ‘r’ in space. - An electric field will develop around the charge q
_{1}

- To bring a charge q2 from infinity to the point B some work must be done.

workdone = q_{2}V_{B}

But V_{B}= \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1}{\mathrm{r}}\)

W = \(\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r}\) - This amount workdone is stored as electrostatic potential energy (U) of a system of two charged particles. Its unit is joule.

∴ U = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}}\) - If the two charges are similar then ‘U’ is positive. This is in accordance with the fact that two similar charges repel one another and positive work has to be done on the system to bring the charges nearer.
- Conversely if the two charges are of opposite sign, they attract one another and potential energy is negative.

Question 3.

Derive an expression for the potential energy of an electric dipole placed in a uniform electric field.

Answer:

Expression for potential energy of an electric dipole placed in a uniform electric field:

- Consider a electric dipole of length 2a having charges + q and -q.
- The electric dipole is placed in uniform electric field E and its axis makes an angle θ with E.

- Force on charges are equal but opposite sign. They constitute torque on the dipole.

Torque \(\tau\) = one of its force (F) × ⊥^{r}distance (BC)

F = qE and sinθ = \(\frac{\mathrm{BC}}{2 \mathrm{a}}\) ⇒ BC = 2a sinθ

∴ Torque \(\tau\) = qE × 2a sinθ = PE sin θ [∴ p = 2aq] - Suppose the dipole is rotated through an angle dθ, the workdone dw is given by

dw = tdθ = PE sinθ dθ - For rotating the dipole from angle θ
_{1}to θ_{2},

workdone W = \(\int_{\theta_1}^{\theta_2} \mathrm{PE} \sin \theta d \theta\) = PE(cosθ_{1}– cosθ_{2}) - This workdone (W) is then stored as potential energy(U) in the dipole.

∴ U = PE(cosθ_{1}– cosθ_{2}) - If θ
_{1}= 90°and θ_{2}= 0°, U = -PE cosθ.

In vector form U = \(-\vec{P} \cdot \vec{E}\)

Question 4.

Derive an expression for the capacitance of a parallel plate capacitor. (Mar.’16 (AP) Mar ’14)

Answer:

Expression for the capacitance of a parallel plate capacitor:

- P and Q are two parallel plates of a capacitor separated by a distance of d.
- The area of each plate is A. The plate P is charged and Q is earth connected.
- The charge on P is + q and surface charge density of charge = σ

∴ q = Aσ

- The electric intensity at poiñt x, E = \(\frac{|\sigma|}{\varepsilon_0}\)
- Potential difference between the plates P and Q,

- Capacitance of the capacitor Farads (In air)

Note : Capacity of a capacitor with dielectric medium is C = \(\frac{\varepsilon_0 \mathrm{~A}}{\left[\mathrm{~d}-\mathrm{t}+\frac{\mathrm{t}}{\mathbf{k}}\right]}\) Farads.

Question 5.

Explain the behaviour of dielectrics in an external field. (A.P. Mar. ’19)

Answer:

1) When an external field is applied across dielectrics, the centre of positive charge distribution shifts in the direction of electric field and that of the negative charge distribution shifts

opposite to the electric field and induce a net electric field within the medium opposite to the external field. In such situation the molecules are said to be polarised.

2) Now consider a capacitor with a dielectric between the plates. The net field in the dielectric becomes less.

3) If E_{0} the external field strength and E_{1} is the electric field strength induced, then the net field \(\overrightarrow{\mathrm{E}}_{\text {net }}\) = \(\overrightarrow{\mathrm{E}}_0\) + \(\overrightarrow{\mathrm{E}}_1\)

(E_{net}) = E_{0} – E_{i} = \(\frac{E}{K}\) where K is the dielectric constant of the medium.

Long Answer Questions

Question 1.

Define electric potential. Derive and expression for the electric potential due to an electric dipole and hence the electric potential at a point (a) the axial line of electric dipole (b) on the equatorial line of electric dipole.

Answer:

Electric potential (V) : The workdone by a unit positive charge from infinite to a point in an electric field is called electric potential.

Expression for the potential at a point due to a dipole:

- Consider A and B having -q and + q charges separated by a distance 2a.
- The electric dipole moment P = q × 2a along AB
- The electric potential at the point ‘P’ is to be calculated.
- P is at a distance ‘r’ from the point ‘O’. θ is the angle between the line OP and AB.
- BN and AM are perpendicular to OP.
- Potential at P due to charge + q at B,

- Potential at P due to charge -q at A, V
_{2}= \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{-\mathrm{q}}{\mathrm{AP}}\right]\)

∴ V_{2}= \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{-\mathrm{q}}{\mathrm{MP}}\right]\) [∵ BP = NP] - Therefore, Resultant potential at P is V = V
_{1}+ V_{2}

V = \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{q}{N P}-\frac{q}{M P}\right]\) …… (1) - In Δle ONB, ON = OB cosθ = a cosθ; ∴ NP = OP – ON = r – a cosθ ….. (2)
- In Δle AMO, OM = AO cosθ = a cosθ; ∴ MP = MO + OP = r + a cos θ ….. (3)
- Substituting (2) and (3) in (1), we get

- As r > >a, a
^{2}cos^{2}θ can be neglected with comparision of r^{2}.

∴ V = \(\frac{\mathrm{P} \cos \theta}{4 \pi \varepsilon_0 \mathbf{r}^2}\) - Electric potential on the axial line of dipole :

(i) When θ = 0°, point p lies on the side of + q

∴ V = \(\frac{\mathrm{P}}{4 \pi \varepsilon_0 \mathrm{r}^2}\) [∵ cos 0° = 1]

ii) When θ = 180°, point p lies on the side of -q.

∴ V = \(\frac{-\mathrm{P}}{4 \pi \varepsilon_0 \mathrm{r}^2}\) [∵ cos 180° = -1]

b) Electric potential on the equitorial line of the diopole:

when θ = 90°, point P lies on the equitorial line.

∴ V = o [∵ cos 90° = 0]

Question 2.

Explain series and parallel combination of capacitors. Derive the formula for equivalent capacitance in each combination. (AP. & T.S. Mar. ‘15)

Answer:

Series combination : If a number of condensers are connected end to end between the fixed points then such combination is called series.

In this combination

- Charge on each capacitor is equal.
- PD’s across the capacitors is not equal.

Consider three capacitors of capacitanceš C_{1}, C_{2} and C_{3} are connected in series across a battery of P.D ‘V’ as shown in figure.

Let ‘Q’ be the charge on each capacitor.

Let V_{1}, V_{2} and V_{3} be the P.D’s of three

V = V_{1} + V_{2} + V_{3} —— (1)

P.D. across I^{st} condenser V_{1} = \(\frac{\mathrm{Q}}{\mathrm{C}_1}\)

P.D. across II^{nd} condenser V_{2} = \(\frac{\mathrm{Q}}{\mathrm{C}_2}\)

RD across III^{rd} condenser V_{3} = \(\frac{\mathrm{Q}}{\mathrm{C}_3}\)

∴ From the equation (1), V = V_{1} + V_{2} + V_{3}

= \(\frac{\mathrm{Q}}{\mathrm{C}_1}\) + \(\frac{\mathrm{Q}}{\mathrm{C}_2}\) + \(\frac{\mathrm{Q}}{\mathrm{C}_3}\) = Q\(\left[\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}\right]\)

For ‘n’ number of capacitors, the effective capacitance can be written as

Parallel Combination : The first plates of different capacitors are connected at one terminal and all the second plates of the capacitors are connected at another terminal then the two terminals are connected to the two terminals of battery is called parallel combination.

In this combination,

1. The PD’s between each capacitor is equal (or) same.

2. Charge on each capacitor is not equal. Consider three capacitors of capacitancé C_{1}, C_{2} and C_{3} are connected in parallel across a RD ‘V’ as shown in fig.

The charge on Ist capacitor Q_{1} = C_{1} V

The charge on IInd capacitor Q_{2} = C_{2} V

The charge on IIIrd capacitor Q_{2} = C_{3} V

∴ The total charge Q = Q_{1} + Q_{2} + Q_{3}

= C_{1} V + C_{2} V + C_{3} V

Q = V(C_{1} + C_{2} + C_{3}) ⇒ \(\frac{Q}{V}\) = C_{1} + C_{2} + C_{3}

C = C_{1} + C_{2} + C_{3} [ ∵ C = \(\frac{\mathrm{Q}}{\mathrm{V}}\)]

for ‘n’ number of capacitors connected in parallel, the equivalent capacitance can be written as C = C_{1} + C_{2} + C_{3} + …. + C_{n}

Question 3.

Derive an expression for the energy stored in a capacitor. What is the energy stored when the space between the plates is filled with a dielectric

(a) with charging battery disconnected?

(b) with charging battery connected in the circuit?

Answer:

Expression for the energy stored in a capacitor : Consider an uncharged capacitor of capacitance ‘c’ and its initial will be zero. Now it is connected across a battery for charging then the final potential difference across the capacitor be V and final charge on the capacitor be ‘Q’

∴ Average potential difference V_{A} = \(\frac{\mathrm{O}+\mathrm{V}}{2}\) = \(\frac{\mathrm{V}}{2}\)

Hence workdone to move the charge Q = W = V_{A} × Q = \(\frac{\mathrm{VQ}}{2}\)

This is stored as electrostatic potential energy U’

∴ U = \(\frac{\mathrm{QV}}{2}\)

We know Q = CV then ‘U’ can be written as given below.

U = \(\frac{\mathrm{QV}}{2}\) = \(\frac{1}{2}\) CV^{2} = \(\frac{1}{2} \frac{\mathrm{Q}^2}{\mathrm{C}}\)

∴ Energy stored in a capacitor

U = \(\frac{\mathrm{QV}}{2}\) = \(\frac{1}{2}\) CV^{2} = \(\frac{1}{2} \frac{\mathrm{Q}^2}{\mathrm{C}}\)

Effect of Dielectric on energy stored :

Case (a) : When the charging battery is disconnected from the circuit:

Let the capacitor is charged by a battery and the disconnected from the circuit. Now the space between the plates is filled with a dielectric of dielectric constant ‘K’ then potential decreases by \(\frac{1}{K}\) times and charge remains constant.

Capacity increases by ‘K times

∴ Energy stored decreases by \(\frac{1}{\mathrm{~K}}\) times.

Case (b): When the charging battery is connected in the circuit:

Let the charging battery is continue the supply of charge. When the dielectric is introduced then potential decreases by \(\frac{1}{\mathrm{~K}}\) times and charge on the plates increases until the potential difference attains the original value = V ,

New charge on the plates Q’ = KQ

Hence new capacity C’ = \(\frac{Q^{\prime}}{V}=\frac{K Q}{V}\) = KC

Energy stored in the capacitor U’ = \(\frac{1}{2}\)Q’V = \(\frac{1}{2}\)(KQ) V = KU

U’ = KU

∴ Energy stored in the capacitor increases by ‘K times.

Problems

Question 1.

An elementary particle of mass ‘m’ and charge +e initially at a very large distance is projected with velocity ‘v’ at a much more massive particle of charge + Ze at rest. The closest possible distance of approach of the incident particle is

Solution:

For an elementary particle, mass = m; charge = +e; velocity = v.

For much more massive particle, charge = + Ze

From law of conservation of energy, we have

K.E of elementary particles = Electrostatic potential energy of elementary particle at a closest distance (d)

∴ The closest possible distance of approach of the incident particle,

d = \(\frac{Z e^2}{2 \pi \varepsilon_0 m v^2}\)

Question 2.

In a hydrogen atom the electron and proton are at a distance of 0.5 A. The dipole moment of the system is

Answer:

In a hydrogen atom the charge of an electron = -1.6 × 10^{-19}C

In a hydrogen atom the charge of proton,

q_{p} = +1.6 × 10^{-19}C

The distance between the proton and an electron

2a = 0.5A = 0.5 × 10^{-10} m

The dipole moment of the system,

P = 2a × q_{p} = 0.5 × 10^{-10} × 1.6 × 10

∴ P = 8 × 10^{-30}cm

Question 3.

There is a uniform electric field in the XOY plane represented by (40\(\hat{i}\) + 30\(\hat{j}\)) Vm^{-1}. If the electric potential at the origin is 200 V, the electric potential at the point with coordinates (2m, 1m) is

Answer:

Given, uniform Electric field intensity,

\(\overrightarrow{\mathrm{E}}\) = (40\(\hat{i}\) + 30\(\hat{j}\)) Vm^{-1}

Electric potential at the origin = 200V

Position vector d\(\begin{aligned}

&\rightarrow \\

&\mathrm{r}

\end{aligned}\) = (2\(\hat{i}\) + 1\(\hat{j}\)) m

We know that,

V_{p} – V_{o} = -(80 + 30) = -110Volt

V_{p} = V_{o} – 110 = (200 – 110) Volt = 90 Volt

∴ potential at point P, V_{p} = 90Volt.

Question 4.

An equilateral triangle has a side length L. A charge +q is kept at the centroid of the triangle. P is a point on the perimeter of the triangle. The ratio of the minimum and maximum possible electric potentials for the point P is

Answer:

Charge at the centroid of an equilateral triangle = +q

The charge + q divides the line segment in ratió 2 : 1.

That means r_{max} = 2 and r_{min} = 1

V_{min} = \(\frac{1}{4 \pi \varepsilon_0} \frac{q}{r_{\max }}\) and V_{max} = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{r}_{\min }}\)

Question 5.

ABC is an equilateral triangle of side 2m. There is a uniform electric field of intensity 100V/m in the plane of the triangle and parallel to BC as shown. If the electric potential at A is 200 V, then the electric potentials at B and C.

Answer:

Given length of side of an equilateral triangle a = 2m

E = 100V/m; V_{A} = 200V

Let D be the mid point between B and C Potential at D = V_{D} = 200V

From fig V_{B} – V_{D} = Ed

⇒ V_{B} – 200 = 100 × 1

∴ Potential at B, V_{B} = 300 V And V_{D} – V_{C} = Ed

200 – V_{C} = 100 × 1

∴ Potential at C, V_{C} = 100V.

Question 6.

An electric dipole of moment p is placed in a uniform electric field E, with p parallel to E. It is then rotated by an angle q. The work done is

Solution:

Let AB be a electric dipole having charges -q and + q

Electric dipole moment of AB = p

Electric field = E The workdone by a dipole, when it is rotated through an angle q from E,

W = \(\int_0^q p E \sin \theta d \theta\)

⇒ W = pE \([\cos \theta]_0^q\) = pE (cos0° – cosq)

∴ W = pE(1 – cosq)

Question 7.

Three identical metal plates each of area ‘A’ are arranged parallel to each other, ’d’ is the distance between the plates as shown. A battery of V volts is connected; as shown. The charge stored in the system of plates is

Answer:

Area of each plate = A

distance between two plates = d capacity of each, parallel plate capacitor,

C = \(\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\)

Two capacitors are connected in parallel as shown in figure

Equivalent capacity of two capacitors connected in parallel, C_{p} = 2C = \(\frac{2 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\)

Charge stored in the system of plates,

q = C_{p}V = \(\frac{2 \varepsilon_0 A}{d} V\)

∴ q = \(\frac{2 A \varepsilon_0 \mathrm{~V}}{\mathrm{~d}}\)

Question 8.

Four identical metal plates each of area A are separated mutually by a distance d and are connected as shown. Find the capacity of the system between the terminals A and B.

Solution:

Area of each plate of a capacitor = A

Distance between two parallel plate capacitors = d

Capacity of each parallel plate capacitor,

C = \(\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\)

Given fig is

The equivalent circuit of the above fig is

Question 9.

In the circuit shown the battery of ‘V’ volts has no internal resistance. All three condensers are equal in capacity. Find the condenser that carries more charge ?

Answer:

The equivalent circuit to the given circuit is as shown

In series combination of capacitors, charge q flows through each capacitor. Then q_{1} = q = C_{1}V_{1}; q_{2} – q = C_{2}V_{2}; q_{3} = q = C_{3}V_{3}

∴ q_{1} = q_{2} = q_{3}

Hence three capacitors C_{1}, C_{2} and C_{3} carry the same charge.

Question 10.

Two capacitors A and B of capacities C and 2C are connected in parallel and the combination is connected to a battery of V volts. After the charging is over, the battery is removed. Now a dielectric slab of K = 2 is inserted between the plates of A so as to fill the completely. The energy lost by the system during the sharing of charges is

Solution:

i) With battery of parallel combination:

C_{1} = C;C_{2} = 2C; V = V

C_{p} = C_{1} + C_{2} = 3C; q = 3Cv

ii) Without battery of parallel combination : .

Question 11.

A condenser of certain capacity is charged to a potential V and stores some energy. A second condenser of twice the capacity is to store half the energy of the first, find to what potential one must be charged?

Answer:

For first capacitor, C_{1} = C; V_{1} = V

And U_{1} = \(\frac{1}{2}\) C_{1}V\(V_1^2\) = \(\frac{1}{2}\) CV^{2} …….. (1)

For second capacitor, C_{2} = 2C_{1} = 2C;

U_{2} = \(\frac{\mathrm{U}_1}{2}\) = \(\frac{1}{4} \mathrm{CV}^2\); Let potential difference across the capacitor = V_{2}

Textual Exercises

Question 1.

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:

Here q1 = 5 × 10^{-8}C, q_{2} = -3 × 10^{-8}C

Let the potential be zero at a distance x cm from the charge q_{1} = 5 × 10^{-8}C.

∴ r_{1} = x × 10^{-2}m

r_{2} = (16 – x) × 10^{-2}m

Now V = \(\frac{\mathrm{q}_1}{4 \pi \varepsilon_0 \mathrm{r}_1}\) + \(\frac{\mathrm{q}_2}{4 \pi \varepsilon_0 \mathrm{r}_2}\)

= \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{q_1}{r_1}+\frac{q_2}{r_2}\right]\)

∴ \(\frac{\mathrm{q}_1}{\mathrm{r}_1}\) = \(\frac{-\mathrm{q}_2}{\mathrm{r}_2}\)

= \(\frac{5 \times 10^{-8}}{x \times 10^{-2}}\) = \(\frac{-\left(-3 \times 10^{-8}\right)}{(16-x) 10^{-2}}\) or \(\frac{5}{x}\) = \(\frac{3}{16-x}\)

3x = 80 – 5x

8x = 80, x = 10cm

Question 2.

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

Answer:

In fig. O is centre of hexagon ABCDEFA of each side 10 cm, As it clear from the figure, OAB, OBC etc. are equilateral triangles.

Therefore

OA = OB = OC = OD = OE = OF = r = 10 cm = 10^{-1}m

As potential is scalar, there for C potential at O is

Question 3.

Two charges 2 μC and -2 μC 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) The plane normal to AB and passing through its middle point has zero potential everywhere.

b) Normal to the plane in the direction AB.

Question 4.

A spherical conductor of radius 12 cm has a charge of 1.6 × 10^{-7}C distributed uniformly on its surface. What is the electric field.

a) inside the sphere

b) just outside the sphere

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

Solution:

a) Here r = 12 cm = 12 × 10^{-2}m,

q = 1.6 × 10^{-7}C. Inside the sphere, E = 0

b) Just coincide the sphere (say on the surface of the sphere)

E = \(\frac{\mathrm{q}}{4 \pi \varepsilon_0 \mathrm{r}^2}\)

= 9 × 10^{9} × \(\frac{1.6 \times 10^{-7}}{\left(12 \times 10^{-2}\right)^2}\) = 10^{5} N/c

c) At r = irm = 18 × 10^{-2}

E = \(\frac{\mathrm{q}}{4 \pi \varepsilon_0 \mathrm{r}^2}\) = \(\frac{9 \times 10^9 \times 1.6 \times 10^{-7}}{\left(18 \times 10^{-2}\right)^2}\)

= 4.4 × 10^{4} N/C

Question 5.

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:

C_{1} = \(\frac{\varepsilon_0 A}{d}\) = 8pF

C_{2} = k\(\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d} / 2}\) = \(\frac{6 \times 2 \varepsilon_0 A}{d}\) = 12 × 8 = 96pF.

Question 6.

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) \(\frac{1}{C_S}\) = \(\frac{1}{9}\) + \(\frac{1}{9}\) + \(\frac{1}{9}\) = \(\frac{3}{9}\) = \(\frac{1}{3}\) ; C_{s} = 3pF

\(\frac{\mathrm{V}}{3}\) = \(\frac{\mathrm{120}}{3}\) = 40V

b) P.d across each capacitor =

\(\frac{\mathrm{V}}{3}\) = \(\frac{\mathrm{120}}{3}\) = 40V

Question 7.

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) C_{p} = 2 + 3 + 4 = 9pF

b) For each capacitor, V is same = 100 Volt

q_{1} = C_{1}V_{1} = 2 × 100 = 200pC

q_{2} = C_{2}V = 3 × 100 = 300pC

q_{3} = C_{3}V = 4 × 100 = 400pC

Question 8.

In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10^{-3} m^{2} 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:

Here A = 6 × 10^{-3}m^{2}, d = 3mm = 3 × 10^{-3}m, C = ?

V = 100V, q = ?

C_{0} = \(\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}=\frac{\left(8.85 \times 10^{-12}\right)\left(6 \times 10^{-3}\right)}{3 \times 10^{-3}}\)

= 1.77 × 10^{-11}F

q = C_{0}V= 1.77 × 10^{-11} × 100

= 1.77 × 10^{-9}C

Question 9.

Explain what would happen if in the capacitor given in Exercise 8, a 3 mm 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:

a) Capacity increases to C = KC_{0}

= 6 × 1.77 × 10^{-11}F

charge increases to

q^{1} = C^{1}V = 6 × 1.77 × 10^{-11} × 10^{2}C

b) After the supply was disconnected new capacity C = KC_{0} = 6 × 1.77 × 10^{-11}F

New voltage V^{1} = \(\frac{q}{C^1}\) = \(\frac{1.77 \times 10^{-9}}{6 \times 1.77 \times 10^{-11}}\)

= 16.67V

Question 10.

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

Solution:

Here C = 12pF = 12 × 10^{-12}F, V = 50Volt, E = ?

E = \(\frac{1}{2} \mathrm{CV}^2\) = \(\frac{1}{2}\left(12 \times 10^{-12}\right)(50)^2\)

= 1.5 × 10^{-8}J

Question 11.

A 600pF 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:

Here C_{1} = C_{2} = 600pF = 600 × 10^{-12}

F = 6 × 10^{-10}F,

V_{1} = 200 V, V_{2} = o

Additional Exercises

Question 1.

A charge of 8mC is located at the origin. Calculate the workdone in taking a small charge of’ -2 × 10^{-9} C from a point P(0, 0, 3 cm) to a point Q(0,4 cm, 0), Via a point R(0,6 cm, 9 cm).

Solution:

From fig. a charge q = 8mc = 8 × 10^{-3}C is located at the origin O. Charge to be carried is

q_{0} = -2 × 10^{-9}C from P to Q

Where OP = r_{p} = 3 cm

= 3 × 10^{-2}m and OQ = r_{Q} = 4cm = 4 × 10^{-2}m

As electrostatic forces are conservative forces, workdone is independent of the path. Therefore there is no relevance of point R.

Question 2.

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.

Answer:

We know that the length of diagonal of thè cube of each side b is \(\sqrt{3 b^2}\) = \(\mathrm{b} \sqrt{3}\)

Distance between centre of the cube and each vertex r = \(\frac{b \sqrt{3}}{2}\)

V = \(\frac{1}{4 \pi \varepsilon_0} \frac{q}{r}\)

and 8 charges each of valué q are present at the eight vertices, of the cube therefore

∴ V = \(\frac{1}{4 \pi \varepsilon_0} \frac{8 q}{b \sqrt{3} / 2}\) or V = \(\frac{4 \mathrm{q}}{\sqrt{3} \pi \varepsilon_0 \mathrm{~b}}\)

Further electric field intensity at the centre due to all the eight charges is zero because the fields due to individual charges cancel in pairs.

Question 3.

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:

Here q_{1} = 1.5,C = 1.5 × 10^{-6} C,

q_{1} = 2.5μC = 2.5 × 10^{-6}C

Distance between the two spheres = 30cm

from figure

b) Let P be the point in a plane normal to the line passing through the mid point,

where OP 10cm = 0. 1m

From figure,

Now PA = PB

Resultant field intensity at P is

E = \(\sqrt{\mathrm{E}_1^2+\mathrm{E}_2^2+2 \mathrm{E}_1 \mathrm{E}_2 \cos \theta}\)

E = 6.58 × 10^{5} Vm^{-1}

Let θ be the angle which resultant intensity \(\overrightarrow{\mathrm{E}}\) makes with \(\overrightarrow{\mathrm{E}_1}\).

Question 4.

A spherical conducting shell of inner radius r_{1} and outer radius r_{2} has a charge Q.

a) A charge q is placed at the centre of the shell. What ¡s the surface charge density on the inner 4nd outer surfaces of the shell?

Answer:

a) The charge of + Q resides on the Outer surface of the shell. The charge q placed at the centre of the shell induces charge -q on the inner surface and charge + q on the outer surface of the shell, from figure.

∴ Total charge on inner surface of the shell is -q and total charge on the outer surface of the shell is (Q + q)

σ_{1} = \(\frac{\mathrm{q}}{4 \pi \mathrm{r}_1^2}\) and σ_{1} = \(\frac{\mathrm{Q}_1+\mathrm{q}}{4 \pi \mathrm{r}_2^2}\)

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.

Solution:

Electric field intensity inside a cavity with no charge is zero, eveñ when the shell has

any irregular shape. If we were to take a closed loop part of which is inside the cavity along a field line and the rest outside it then network done by the field in carrying a rest charge over the closed loop will not be zero. This is impossible for an electrostatic field. Hence electric field intensity inside a cavity with no charge is always zero.

Question 5.

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

another given by (E_{2} – E_{1}). \(\hat{\mathbf{n}}\) = \(\frac{\sigma}{\varepsilon_0}\)

Where \(\hat{\mathbf{n}}\) is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of \(\hat{\mathbf{n}}\) is from side 1 to side 2.)

Hence show that just out side a conductor, the electric field is σ\(\hat{\mathrm{n}} / \varepsilon_0\)

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

(Hint: for (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.)

Answer:

a) Normal component of electric field intensity due to a thin infinite plane sheet of charge on left side.

\(\overrightarrow{\mathrm{E}}_1\) = –\(\frac{\sigma}{2 \varepsilon_0} \hat{\mathbf{n}}\)

and on right side 2 = \(\overrightarrow{\mathrm{E}_2}\) = \(\frac{\sigma}{2 \varepsilon_0} \hat{\mathrm{n}}\)

Discontinuity in the normal component from one side to the other is

b) To show that the tangential component of electostatic field is continous from one side of a charged surface to another, we use the fact that workdone by electrostatic field on a closed loop is zero.

Question 6.

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

Solution:

From figure A is a long charged cylinder of linear charge density λ, lengh l and radius a. A hollow co-axial conducting cylinder B of length L and radius b surrounds A.

The charge q = λL spreads uniformly on the outer surface of λ. It induces – q charge on the cylinder B. which spreads on the inner surface of B. An electric field \(\overrightarrow{\mathrm{E}}\) is produced in the space between the two cylinders which is directed radically outwards. Let us consider a co-axial cylindrical Gaussian surface of radius r. The electric flux through the cylindrical Gaussian surface is

The Electric flux through the end faces of the cylindrical Gaussian surface is zero as \(\overrightarrow{\mathrm{E}}\) is parallel to them. According to Gauss’s theorem

Question 7.

In a hydrogen atom, the electron and proton are bound at a distance of about 0.53Å:

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.

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)?

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

Solution:

a) Here q_{1} = -1.6 × 10^{-19}C; q_{2} = + 1.6 × 10^{-19}C.

r = 0.53λ = 0.53 × 10^{-10}m

Potential energy = P.E at ∞ -P. E at r

= 0 – \(\frac{q_1 q_2}{4 \pi \varepsilon_0 r}\) = \(\frac{-9 \times 10^9\left(1.6 \times 10^{-19}\right)^2}{0.53 \times 10^{-10}}\)

= -43.47 × 10^{-19} Joule

= \(\frac{-43.47 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}\) = -27.16 eV

b) K.E in the Orbit = \(\frac{1}{2}\) (27.16) eV

Total energy = K.E + P.E

= 13.58 – 27.16 = – 13.58 eV

Work required to free the electron = 13.58eV

c) Potential energy at a seperation of r_{1} (= 1.06A) is

= \(\frac{\mathrm{q}_1 \mathrm{q}_2}{4 \pi \varepsilon_0 \mathrm{r}_1}\) = \(\frac{9 \times 10^9\left(1.6 \times 10^{-19}\right)}{1.06 \times 10^{-10}}\)

= 21.73 × 10^{-19}J = 13.58eV

Potential energy of the system, when zero of P.E is taken at r_{1} = 1.06A is

= RE at r_{1} -P.E at r = 13.58 – 27.16 = -13.58eV.

By shifting the zero of potential energy work required to free the electron is not affected. It continues to be the same, being equal to + 13.58 eV

Question 8.

If one of the two electrons of a H_{2} molecule is removed, we get a hydrogen molecular ion \(\mathbf{H}_2^{+}\). In the ground state of an \(\mathbf{H}_2^{+}\), the two protons are separated by roughly 1.5A. and the electron is roughly 1 A from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.

Solution:

Here q_{1} = charge on electron (= -1.6 × 10^{-19}C)

q_{2}, q_{3} = charge on two protons, each = 1.6 × 10^{-19}C

r_{12} = distance between

q_{1} and q_{2} = 1A = 10^{-10}m

r_{23} = distance between

q_{2} and q_{3} = 1.5A = 1.5 × 10^{-10}m

r_{31} = distance between

q_{3} and q_{1} = 1A = 10^{-10}m.

Taking zero of potential energy at infinity, we have

Question 9.

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.

Solution:

The charge flows from the sphere at higher potential to the other at lower potential till their potentials become equal. After sharing the charges on two spheres would be.

\(\frac{\mathrm{Q}_1}{\mathrm{Q}_2}\) = \(\frac{C_1 V}{C_2 V}\) where C_{1}. C_{2} are the capacities of two spheres.

But \(\frac{\mathrm{C}_1}{\mathrm{C}_2}\) = \(\frac{\mathrm{a}}{\mathrm{b}}\) ∴ \(\frac{\mathrm{Q}_1}{\mathrm{Q}_2}\) = \(\frac{a}{b}\)

Ratio of surface density of charge on the two spheres

Hence ratio of electric fields at the surface of two spheres

\(\frac{\mathrm{E}_1}{\mathrm{E}_2}\) = \(\frac{\sigma_1}{\sigma_2}\) = \(\frac{\mathrm{b}}{\mathrm{a}}\)

A sharp and pointed end can be treated as a sphere of very small radius and a flat portion behaves as a sphere of much larger radius. Therefore, charge density on sharp and pointed ends of conductor is much higher than on its flatter portions.

Question 10.

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 >> I.

(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?

Here -q is at (0, 0, -a) and +q is at (0, 0, a)

a) Potential at (0, 0, z) would be

Potential at (x, y, 0) i.e at a point 1 to z-axis where charges are located is zero

b) we have proved that

V = \(\frac{P \cos \theta}{4 \pi \varepsilon_0\left(r^2+a^2 \cos ^2 \theta\right)}\)

If \(\frac{r}{a}\) >> 1 then a << r ∴ V = \(\frac{p \cos \theta}{4 \pi \varepsilon_0 r^2}\)

∴ V = \(\frac{1}{\mathrm{r}^2}\)

i.e potential is inversely proportional to square of the distance

c) Potential at (5, 0, 0) is

As work done = charge (V_{2} – V_{1})

W = zero

As work done by electrostatic field is independent of the path connecting the two points therefore work done will

continue to be zero along every path.

Question 11.

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).

Answer:

As is clear from figure an electric quadrupole may be regarded as a system of three charges +q, -2q and + q at A, B and C respectively.

Let AC = 2a we have to calculate electric potential at any point P where BP = r, using superposition principle. Potential at p is given by

∴ \(\frac{\mathrm{a}^2}{\mathrm{r}^2}\) is negligibly small V = \(\frac{\mathrm{q} \cdot 2 \mathrm{a}^2}{4 \pi \varepsilon_0 \mathrm{r}^3}\)

clearly V ∝ \(\frac{1}{\mathrm{r}}\)

In case of an electric dipole V ∝ \(\frac{1}{\mathrm{r}^2}\) and in case of an electric monopole

(i.e a single charge), V ∝ \(\frac{1}{\mathrm{r}}\)

Question 12.

An electrical technician requires a capacitance of 2 μF in a circuit across a potential difference of 1kV. A large number of 1μF capacitors are available to him each of which can withstand a potential difference of not more than 400V. Suggest a possible arrangement that requires the minimum number of capacitors.

Answer:

Here total capacitance, C = 2μF

Potential difference v = 1KV = 1000 volt

Capacity of each capacitor C_{1} = 1μF

Maximum potential difference across each V = 400 volt

Let n capacitors of 1μF each be connected in series in a row and m such rows be connected in parallel as shown in the figure. As potential difference in each row

= 1000 Volt

∴ Potential difference across each capacitor = \(\frac{1000}{\mathrm{n}}\) = 400

∴ n = \(\frac{1000}{400}\) = 2.5

As n has to be a whole number (not less than 2.5) therefore n = 3

capacitance of each row of 3 condensors of 1μF

Each is series = 1/3

Total capacitance of m such rows in parallel = \(\frac{\mathrm{m}}{3}\)

∴ \(\frac{\mathrm{m}}{3}\) = 2(μf) or m = 6

∴ Total number of capacitors =

n × m = 3 × 6 = 18.

Hence 1μF capacitors should be connected in six parallel rows, each row containing three capacitors in series.

Question 13.

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

Answer:

Here A = ? C = 2F,

d = 0.5 cm = 5 × 10^{-3}m

As C = \(\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\)

∴ A = \(\frac{c d}{\varepsilon_0}=\frac{2 \times 5 \times 10^{-3}}{8.85 \times 10^{-12}}\)

= 1.13 × 10^{9}m^{2}

Which is too large.

That is why ordinary capacitors are in the range of μF or less. However in electrolytic capacitors d is too small. Therefore their capacitance is much larger (=0.1F)

Question 14.

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

Answer:

Here C_{2} and C_{3} are in series

A = \(\frac{c d}{\varepsilon_0}=\frac{2 \times 5 \times 10^{-3}}{8.85 \times 10^{-12}}\)

∴ \(\frac{1}{C_1}\) = \(\frac{1}{200}\) + \(\frac{1}{200}\) = \(\frac{2}{200}\) = \(\frac{1}{100}\)

C_{S} = 100pF

Now C_{S} and C_{1} are in parallel

∴ C_{p} = C_{s} + C_{1} = 100 + 100 = 200pF

Again C_{p} and C_{4} are in series

∴ \(\frac{1}{\mathrm{C}_{\mathrm{s}}}\) = \(\frac{1}{\mathrm{C}_{\mathrm{p}}}\) + \(\frac{1}{\mathrm{C}_4}\) = \(\frac{1}{200}\) + \(\frac{1}{100}\) = \(\frac{3}{200}\)

∴ C = \(\frac{200}{3}\)pF = 66.7 × 10^{-12}F

As C_{p} and C_{4} are in series

∴ V_{p} + V_{4} = 300

Charge on C_{4} is q_{4}

= CV = \(\frac{200}{3}\) × 10^{-12} × 300 = 2 × 10^{-8}C

Potential difference across

C_{4} is V_{4} = \(\frac{\mathrm{q}_4}{\mathrm{C}_4}\) = \(\frac{2 \times 10^{-8}}{100 \times 10^{-12}}\) = 200V

from (i) V_{p} = 300 – V_{4} = 300 – 200 = 100

Potential difference across

C_{1} is V_{1} = V_{p} = 100V

Charge on C_{1} = q_{1} = C_{1}V_{1}

= 100 × 10^{-12} × 100 = 10^{-8}C.

Potential diff across

C_{2} and C_{3} in series = 100V

charge on C_{2};

q_{2} = C_{2} V_{2} = 200 × 10^{-12} × 50 = 10^{-8}C

Charge on C_{3};

q_{3} = C_{3}V_{3} = 200 × 10^{-12} × 50 = 10^{-8}C

Question 15.

The plates of parallel plate capacitor have an area of 90 cm^{2} 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.

Answer:

a) Here A = 90 cm^{2} = 90 × 1o^{-4}m^{2}

= 9 × 10^{-3}m^{2}

d = 2.5mm = 2.5 × 10^{-3}m

Question 16.

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 ?

Answer:

Here C_{1} = 4µF = 4 × 10^{-6}F, V_{1} = 200volt. Initial elctrostatic energy stored in C_{1} is E_{1}

= \(\frac{1}{2} C_1 V_1^2\) = \(\frac{1}{2}\) × 4 × 10^{-6} × 200 × 200

E_{1} = 8 × 10^{-2} Joule

When 4µF capacitor is connected to uncharged capacitor of 2µF charge flows and both acquire a common potential.

∴ final electrostatic energy of both capacitors

E_{2} = \(\frac{1}{2}\)(C_{1} + C_{2})V^{2}

= \(\frac{1}{2}\) × 6 × 10^{-6} × \(\frac{800}{6}\) × \(\frac{800}{6}\)

E_{2} = 5.33 × 10^{-2}Joule.

Energy dissipated in the form of heat and electro magnetic radiation.

E_{1} – E_{2} = 8 × 10^{-2} – 5.33 × 10^{-2}

= 2.67 × 10^{-2} Joule.

Question 17.

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 1/2.

Answer:

If F is the force on each plates of parallel plate capacitor, then work done in increasing the seperation between the plates by Δx = fΔx

This must be the increase in potential energy of the capacitor Now the increase the volume of capacitor is = A Δx

If U = energy density = energy stored/ volume then the increase in potential energy = U.AΔx

∴ fΔx = U. AΔx

The origin of factor 1/2 in force can be explained by the fact that inside the conductor field is zero and outside the conductor, the field is. E. Therefore the average value of the field (i.e E/2) contributes to the force.

Question 18.

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 = \(\frac{4 \pi \varepsilon_0 r_1 r_2}{r_1-r_2}\)

where r_{1} and r_{2} are the radii of outer and inner spheres, respectively.

Solution:

As is clear from the figure +Q charge spreads uniformly on inner surface of outer sphere of radius r_{1}. The induced charge – Q spreads uniformly on the outer surface of inner sphere of radius r_{2}. The outer surface of outer sphere is earthed. Due to electrostatic shielding E = 0 for r < r_{2} and E = 0 for r < r_{2} and E = 0 for r > r_{1}

In the space between the two spheres electric intensity E exists as shown. Potential difference between the two spheres.

Question 19.

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 2.5 μC. 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:

Here r_{a} = 12cm = 12 × 10^{-2}m

r_{b} = 13cm = 13 × 10^{-2}m

q = 2.5μC = 2.5 × 10^{-6}C E_{r} = 32

(a) C = ?

(b)

V = ?

V = \(\frac{q}{c}\) = \(\frac{2.5 \times 10^{-6}}{5.5 \times 10^{-9}}\) = 4.5 × 10^{2}Volt

(c) Capacity of an isolated sphere of radius R

R = 12 × 10^{-2}m is

C^{1} = \(4 \pi \varepsilon_0 R\) = \(\frac{1}{9 \times 10^9}\) × 12 × 10^{-12}

= 1.33 × 10^{-11} Farad.

The capacity of an isolated sphere is much smaller because in a capacitor outer sphere is earthed potential difference decreases and capacitance increases.

Question 20.

Answer carefully:

(a) Two large conducting sphers carrying charges Q_{1} and Q_{2} are brought close to each other. Is the magnitude of electrostatic force between them exactly given by Q_{1} Q_{2}/\(4 \pi \varepsilon_0 \mathbf{r}^2\). where r is the distance between their centres ?

(b) If Coulomb’s law involved 1/r^{3} dependence (instead of 1/r^{2}), would Gauss’s 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).

Answer:

a) When the charged spheres are brought close together the charge distributions on them become non-uniform. Therefore, coloumb’s law is not valid hence the magnitude of force is not given exactly by this formula.

b) No Gauss’s law will not be true if coloumb’s law involved 1/r^{3} dependence instead of 1/r^{2} dependence.

c) The line of force gives the direction of accelaration of charge. If the electric line of force is linear the test charge will move along the line if the line of force is not linear the charge will not go along the line.

d) As force due to the field is discreted towards the nucleus and the electron does not move in the direction of this force, therefore work done is zero when the orbit is circular. This is true even when orbit is elliptical as electric forces are conservative forces.

e) No electric potential is continuous.

f) The capacity of a single conductor implies that the second conductor is infinity.

g) This is because a molecule of water in its normal state has an unsymmetrical shape and therefore it has a permanent dipole moment.

Question 21.

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 effets (i.e., bending of field lines at the ends).

Answer:

Here L = 15cms = 15 × 10^{-2}m

r_{a} = 1.4cm = 1.4 × 10^{-2}m,

r_{b} = 1.5cm = 1.5 × 10^{-2m}

q = 3.5 μC = 3.5 × 10^{-6} coloumb, C = ? V = ?

Question 22.

A parallel plate capacitor is to be designed with a voltage rating 1kV, using a material of dielectric constant 3 and dielectric strength about 10^{7} 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 ?

Answer:

Here V = 1kV= 1000Volt; K = ε_{r} = 3

Dielectric strength = 10^{7}V/m

As electric field at the most should be 10% of dielectric strength due to reasons of safety.

E = 10% of 10^{7}V/m = 10^{6}V/m A = ?

C = 50pF = 50 × 10^{-12}F

Question 23.

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.

Answer:

By definition an equipotential surfaces is that every point of which potential is the same. In the four cases given above:

a) Equipotential surfaces are planes parallel to x – y plane. These are equidistant.

b) Equipotential surfaces are planes parallel to x – y plane. As the field increases uniformly distance between the planes decreases.

c) Equipotential surfaces concentric spheres with origin at the centre.

d) Equipotential surfaces have the shape which changes periodically at far off distances from the grid.

Question 24.

In a Van de Graff type generator a spherical metal shell is to be a 15 × 10^{6} V electrode. The dielectric strength of the gas surrounding the electrode is 5 × 10^{7} Vm^{-1}. What is the minimum radius of the spherical shell required? (You will learn from this exercise why one cannot build and electrostatic generator using a very small shell which requires a small charge to acquire a high potential.)

Answer:

Here V = 15 × 10^{6} volt .

Dielectric strength = 5 × 10^{-7} Vm^{-1}

minimum rodius, r = ?

max. Electric field E = 10% (dielectric strength)

E = \(\frac{10}{100}\) × 5 × 10^{7} = 5 × 10^{6}VM^{-1}

As E = \(\frac{\mathrm{V}}{\mathrm{r}}\) ∴ r = \(\frac{\mathrm{V}}{\mathrm{E}}\) = \(\frac{15 \times 10^6}{5 \times 10^6}\) = 3m

obviously we cannot build an electrostatic generator, using a very small shell.

Question 25.

A small sphere of radius r_{1} and charge q_{1} is enclosed by a spherical shell of radium r_{2} and charge q_{2}. Show that if q_{1} 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 q_{2} on the shell is.

Answer:

As the charge resides always on the outer surface of the shell therefore, when the sphere and shell are connected by a wire, charge will flow essentially from the sphere to the shell, whatever be the magnitude and sign of charge q_{2}.

Question 26.

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 100 Vm^{-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 on evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1m^{2}. 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 it self 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 lighting?

(Hint : The earth has and 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 + 1800C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.)

Answer:

a) since our body and the surface of earth both are conducting therefore our body and the ground form an equipotential surface. As we step out into the open from our house the original equipotential surfaces of open air change, keeping out body and the ground at the same potential that is why we do not get an electric shock.

b) Yes, the man will get a shock This is because the steady discharging current of the atmosphere charges up the aluminium sheet gradually and raises its voltage to an extent depending on the capacitance of the condenser formed by the aluminium sheet and the ground and the insulating slab.

c) The atmosphere is being discharged continuously by understorms and lightning all over the globe. It is also discharging due to the small conductivity of air. The two opposing processes, on an overage, are in equilibrium. Therefore the atmosphere. Keeps charged.

d) During lightning the electric energy of the atmosphere is dissipated in the form of light, heat and sound.