Capacitors and Dielectrics — Hit the Target

Assuming the earth as an insulated spherical conductor of radius 6400 km. Calculate its capacitance.

Answer

Given,

• Radius of the earth, a = 6400 km = 6.4 × 10⁶ m
• $\dfrac{1}{4\pi\epsilon_0} = 9.0 \times 10^9$ N-m² C^-2

For an isolated spherical conductor,

$\text C = 4\pi\epsilon_0 \text a \\[1em] \text C = \dfrac{\text a}{9.0 \times 10^9}$

Substituting the value of a,

$\text C = \dfrac{6.4 \times 10^6}{9.0 \times 10^9} \\[1em] \text C = 7.11 \times 10^{-4}\ \text F \\[1em] \text C = 711\ \mu\text F$

Hence, the capacitance of the earth is 711 µF.

A parallel plate capacitor of plate area A = 600 cm² and plate separation, d = 2.0 mm is connected to a DC source of 200 V. Calculate in SI unit :

(i) the magnitude of the uniform electric field E between the plates.

(ii) the charge density σ on any plate.

Answer

Given,

• Area of each plate, A = 600 cm² = 6 × 10^-2 m²
• Separation between the plates, d = 2.0 mm = 2.0 × 10^-3 m
• Potential difference, V = 200 V
• ε₀ = 8.85 × 10^-12 C² N^-1 m^-2

(i) The field between the plates is uniform, so

$\text E = \dfrac{\text V}{\text d} \\[1em] = \dfrac{200}{2.0 \times 10^{-3}} \\[1em] \text E = 1 \times 10^5\ \text V\ \text m^{-1}$

(ii) The field between the plates in terms of the surface charge density is E = σ/ε₀, so

$\sigma = \epsilon_0 \text E \\[1em] \sigma = (8.85 \times 10^{-12})(1 \times 10^5) \\[1em] \sigma = 8.85 \times 10^{-7}\ \text C\ \text m^{-2}$

Hence, E = 10⁵ V m^-1 and σ = 8.85 × 10^-7 C m^-2.

A parallel plate capacitor of capacitance C, with air between the plates, is shown in Fig. (a). A sheet of dielectric constant K = 5 is introduced between the plates so as to fill the space, as shown in Fig. (b).

What are the values of the potential differences V₁ and V₂ across the plates in case (a) and (b) respectively? What are the charges Q₁ and Q₂ on the plates? In both cases, a battery of emf E is kept connected to the capacitor.

Answer

Given,

• Capacitance with air, C
• Dielectric constant of the sheet, K = 5
• The battery of emf E remains connected in both cases

Potential differences : In both the cases the battery remains connected across the plates, so the potential difference is fixed by the battery,

$\text V_1 = \text V_2 = \text E$

Charges : In case (a) the capacitance is C, hence

$\text Q_1 = \text{CE}$

In case (b) the dielectric fills the whole space, so the capacitance becomes KC and

$\text Q_2 = (\text{KC})\text E \\[1em] \text Q_2 = 5\thinspace\text{CE}$

Hence, V₁ = V₂ = E, while Q₁ = CE and Q₂ = 5CE.

A 900 pF capacitor is charged by a 100 V battery. How much energy is stored by the capacitor?

Answer

Given,

• Capacitance, C = 900 pF = 900 × 10^-12 F
• Potential difference, V = 100 V

The energy stored in a capacitor is

$\text U = \dfrac{1}{2}\text{CV}^2$

Substituting the values,

$\text U = \dfrac{1}{2}(900 \times 10^{-12})(100)^2 \\[1em] \text U = \dfrac{1}{2}(900 \times 10^{-12})(10^4) \\[1em] \text U = 4.5 \times 10^{-6}\ \text J$

Hence, the energy stored by the capacitor is 4.5 × 10^-6 J.

Derive the expression for the energy stored in a capacitor of capacitance C, charged to a potential V.

Answer

While a capacitor is being charged, work has to be done in carrying charge from one plate to the other against the potential difference that already exists between them. This work is stored as the electric potential energy of the capacitor.

Let q be the charge on the capacitor at some instant, so that the potential difference between the plates is q/C. The work done in transferring a further small charge dq is

$\text{dW} = \left(\dfrac{\text q}{\text C}\right)\text{dq}$

Integrating from q = 0 to the final charge Q,

$\text W = \int_0^{\text Q}\dfrac{\text q}{\text C}\thinspace\text{dq} \\[1em] = \dfrac{1}{\text C}\left[\dfrac{\text q^2}{2}\right]_0^{\text Q} \\[1em] \text W = \dfrac{\text Q^2}{2\text C}$

This work is stored as the energy of the capacitor. Using Q = CV,

$\text U = \dfrac{\text Q^2}{2\text C} \\[1em] = \dfrac{1}{2}\text{CV}^2 \\[1em] \text U = \dfrac{1}{2}\text{QV}$

Derive an expression for the capacitance of a parallel plate capacitor in terms of the areas of the plates and the distance between them. Assume there is a vacuum in between the plates with the electric field between the plates E = σ/ε₀ (where the symbols have their usual meaning).

Answer

Consider a parallel-plate capacitor whose plates each have area A and are held a distance d apart, with vacuum between them. Let the plates carry charges +Q and −Q, so that the surface density of charge is

$\sigma = \dfrac{\text Q}{\text A}$

The electric field in the space between the plates is uniform and is given by

$\text E = \dfrac{\sigma}{\epsilon_0} \\[1em] \text E = \dfrac{\text Q}{\epsilon_0 \text A}$

Since the field is uniform, the potential difference between the plates is

$\text V = \text{Ed} \\[1em] \text V = \dfrac{\text{Qd}}{\epsilon_0 \text A}$

By definition the capacitance is

$\text C = \dfrac{\text Q}{\text V} \\[1em] \text C = \dfrac{\text Q\thinspace\epsilon_0 \text A}{\text{Qd}} \\[1em] \text C = \dfrac{\epsilon_0 \text A}{\text d}$

Thus the capacitance is directly proportional to the area of the plates and inversely proportional to the distance between them.

Find the equivalent capacitance of the circuit between the points A and B, given C₁ = C₂ = C₃ = C₄ = 10 µF and C₅ = 5 µF.

Answer

Given,

• C₁ = C₂ = C₃ = C₄ = 10 µF
• C₅ = 5 µF

The network is a Wheatstone-type bridge in which C₅ forms the bridge arm. The balance condition is

$\dfrac{\text C_1}{\text C_2} = \dfrac{\text C_3}{\text C_4} \\[1em] \dfrac{10}{10} = \dfrac{10}{10}$

The bridge is balanced, so the two ends of C₅ are at the same potential and it carries no charge. It may therefore be removed.

The remaining arms give two series branches,

$\text C' = \dfrac{\text C_1 \text C_2}{\text C_1 + \text C_2} \\[1em] = \dfrac{10 \times 10}{20} \\[1em] \text C' = 5\ \mu\text F \\[1em] \text C'' = \dfrac{\text C_3 \text C_4}{\text C_3 + \text C_4} \\[1em] = \dfrac{10 \times 10}{20} \\[1em] \text C'' = 5\ \mu\text F$

These two branches are in parallel between A and B,

$\text C_\text{AB} = 5 + 5 \\[1em] \text C_\text{AB} = 10\ \mu\text F$

Hence, the equivalent capacitance between A and B is 10 µF.

Note: The answer is printed in the textbook as "10 mF". The correct unit is the microfarad, since all the given capacitances are in microfarads.

Find the electric charge Q₁ on plates of capacitor C₁, shown in figure.

Answer

Given,

• C₁ = 8 µF, C₂ = 16 µF, C₃ = 24 µF, joined in series
• Applied potential difference, V = 110 V

For a series combination,

$\dfrac{1}{\text C_\text{eq}} = \dfrac{1}{8} + \dfrac{1}{16} + \dfrac{1}{24} \\[1em] = \dfrac{6 + 3 + 2}{48} \\[1em] \dfrac{1}{\text C_\text{eq}} = \dfrac{11}{48} \\[1em] \text C_\text{eq} = \dfrac{48}{11}\ \mu\text F$

In a series combination every capacitor carries the same charge, which is the charge drawn from the source,

$\text Q_1 = \text C_\text{eq}\text V \\[1em] \text Q_1 = \left(\dfrac{48}{11} \times 10^{-6}\right)(110) \\[1em] \text Q_1 = 480 \times 10^{-6}\ \text C \\[1em] \text Q_1 = 4.8 \times 10^{-4}\ \text C$

Hence, the charge on the plates of C₁ is 4.8 × 10^-4 C.

An isolated 16 µF parallel plate air capacitor has a potential difference of 100 V (Fig. a). A dielectric slab having relative permittivity (i.e., dielectric constant) = 5 is introduced to fill the space between the two plates completely (Fig. b). Calculate :

(i) the new capacitance of the capacitor,

(ii) the new potential difference between the two plates of capacitor.

Answer

Given,

• Capacitance with air, C₀ = 16 µF
• Initial potential difference, V₀ = 100 V
• Dielectric constant, K = 5

(i) When the dielectric fills the space between the plates completely,

$\text C = \text{KC}_0 \\[1em] = 5 \times 16 \\[1em] \text C = 80\ \mu\text F$

(ii) The capacitor is isolated, so the charge on the plates remains unchanged,

$\text Q = \text C_0 \text V_0 \\[1em] = (16)(100) \\[1em] \text Q = 1600\ \mu\text C$

Hence the new potential difference is

$\text V = \dfrac{\text Q}{\text C} \\[1em] = \dfrac{1600}{80} \\[1em] \text V = 20\ \text V$

Hence, the new capacitance is 80 µF and the new potential difference is 20 V.

Three capacitors each of capacitance C are connected in series. Their equivalent capacitance is C_s. The same three capacitors are now connected in parallel. Their equivalent capacitance becomes C_p. Find the ratio $\dfrac{\text C_p}{\text C_s}$ .

Answer

Given,

• Three capacitors each of capacitance C

For the series combination,

$\dfrac{1}{\text C_s} = \dfrac{1}{\text C} + \dfrac{1}{\text C} + \dfrac{1}{\text C} \\[1em] \dfrac{1}{\text C_s} = \dfrac{3}{\text C} \\[1em] \text C_s = \dfrac{\text C}{3}$

For the parallel combination,

$\text C_p = \text C + \text C + \text C \\[1em] \text C_p = 3\text C$

Therefore

$\dfrac{\text C_p}{\text C_s} = \dfrac{3\text C}{\text C/3} \\[1em] \dfrac{\text C_p}{\text C_s} = 9$

Hence, C_p : C_s = 9 : 1.

Three capacitors C₁ = 6 µF, C₂ = 12 µF and C₃ = 20 µF are connected to a 100 V battery, as shown in Fig.

(i) Charge on each plate of capacitor C₁.

(ii) Electrostatic potential energy stored in capacitor C₃.

Answer

Given,

• C₁ = 6 µF, C₂ = 12 µF, C₃ = 20 µF
• EMF of the battery, V = 100 V

From the figure, C₁ and C₂ are in series and this branch is in parallel with C₃ across the battery.

(i) For the series branch,

$\text C' = \dfrac{\text C_1 \text C_2}{\text C_1 + \text C_2} \\[1em] = \dfrac{6 \times 12}{6 + 12} \\[1em] \text C' = 4\ \mu\text F$

The full 100 V acts across this branch, and in series both capacitors carry the same charge,

$\text Q_1 = \text C'\text V \\[1em] = (4)(100) \\[1em] \text Q_1 = 400\ \mu\text C$

(ii) The capacitor C₃ is directly across the battery, so it has 100 V across it,

$\text U_3 = \dfrac{1}{2}\text C_3 \text V^2 \\[1em] \text U_3 = \dfrac{1}{2}(20 \times 10^{-6})(100)^2 \\[1em] \text U_3 = 0.1\ \text J$

Hence, the charge on each plate of C₁ is 400 µC and the energy stored in C₃ is 0.1 J.

A parallel plate air capacitor has a capacitance of 5 µF. It becomes 50 µF when a dielectric medium occupies the entire space between its two plates. What is the dielectric constant of the medium?

Answer

Given,

• Capacitance with air, C₀ = 5 µF
• Capacitance with the dielectric, C = 50 µF

The dielectric constant is the ratio of the two capacitances,

$\text K = \dfrac{\text C}{\text C_0} \\[1em] \text K = \dfrac{50}{5} \\[1em] \text K = 10$

Hence, the dielectric constant of the medium is 10.

You provided with 8 µF capacitors. Show with the help of a diagram how you will arrange minimum number of them to get a resultant capacitance of 20 µF.

Answer

Given,

• Capacitance of each capacitor = 8 µF
• Required resultant capacitance = 20 µF

Two capacitors of 8 µF joined in series give

$\dfrac{8 \times 8}{8 + 8} = 4\ \mu\text F$

If this series pair is joined in parallel with two more single capacitors of 8 µF each, the resultant capacitance is

$\text C = 8 + 8 + 4 \\[1em] \text C = 20\ \mu\text F$

Hence, a minimum of 4 capacitors is required — two of them joined in series, and this pair connected in parallel with the other two.

Deduce an expression for equivalent capacitance C when three capacitors C₁, C₂ and C₃ are connected in parallel.

Answer

In a parallel combination, one plate of each capacitor is connected to the point A and the other plate to the point B. Hence each capacitor has the same potential difference V across it, while the charges on them are different.

The charges on the three capacitors are

$\text Q_1 = \text C_1 \text V, \qquad \text Q_2 = \text C_2 \text V, \qquad \text Q_3 = \text C_3 \text V$

The total charge drawn from the source is

$\text Q = \text Q_1 + \text Q_2 + \text Q_3 \\[1em] \text Q = (\text C_1 + \text C_2 + \text C_3)\text V$

If a single capacitor of capacitance C between A and B takes the same charge Q at the same potential difference V, then Q = CV. Comparing the two expressions,

$\text C = \text C_1 + \text C_2 + \text C_3$

Thus in a parallel combination the equivalent capacitance is the sum of the individual capacitances.

Three capacitors of capacitance C₁ = 3 µF, C₂ = 6 µF and C₃ = 10 µF are connected to a 10 V battery as shown in Fig. Calculate :

(i) Equivalent capacitance.

(ii) Electrostatic potential energy stored in the system.

Answer

Given,

• C₁ = 3 µF, C₂ = 6 µF, C₃ = 10 µF
• EMF of the battery, V = 10 V

(i) From the figure, C₁ and C₂ are in series and this branch is in parallel with C₃.

$\text C' = \dfrac{\text C_1 \text C_2}{\text C_1 + \text C_2} \\[1em] = \dfrac{3 \times 6}{3 + 6} \\[1em] \text C' = 2\ \mu\text F \\[1em] \text C_\text{eq} = \text C' + \text C_3 \\[1em] = 2 + 10 \\[1em] \text C_\text{eq} = 12\ \mu\text F$

(ii) The energy stored is

$\text U = \dfrac{1}{2}\text C_\text{eq}\text V^2 \\[1em] \text U = \dfrac{1}{2}(12 \times 10^{-6})(10)^2 \\[1em] \text U = 6 \times 10^{-4}\ \text J$

Hence, the equivalent capacitance is 12 µF and the energy stored is 6 × 10^-4 J.

A parallel plate capacitor is charged by a battery, which is then disconnected. A dielectric slab having dielectric constant (relative permittivity) K, is now introduced between its two plates in order to occupy the space completely. State, in terms of K, its effect on the following :

(i) the capacitance of the capacitor,

(ii) the potential difference between its plates,

(iii) the energy stored in the capacitor.

Answer

The battery has been disconnected, so the charge Q on the plates remains unchanged.

(i) The capacitance becomes K times its original value,

$\text C' = \text{KC}$

(ii) Since Q is constant and V = Q/C,

$\text V' = \dfrac{\text Q}{\text{KC}} \\[1em] \text V' = \dfrac{\text V}{\text K}$

(iii) The energy stored is U = Q²/2C, so

$\text U' = \dfrac{\text Q^2}{2(\text{KC})} \\[1em] \text U' = \dfrac{\text U}{\text K}$

Hence, C' = KC, V' = V/K and U' = U/K. As K > 1, the capacitance increases while the potential difference and the stored energy both decrease.

Obtain an expression for equivalent capacitance C, when three capacitors having capacitances C₁, C₂ and C₃ are connected in series.

Answer

In a series combination the capacitors are joined end to end. On giving a charge +Q to the first plate, equal and opposite charges appear on the plates of each capacitor by induction, so every capacitor carries the same charge Q, while the potential differences across them are different.

The potential differences are

$\text V_1 = \dfrac{\text Q}{\text C_1}, \qquad \text V_2 = \dfrac{\text Q}{\text C_2}, \qquad \text V_3 = \dfrac{\text Q}{\text C_3}$

If V is the total potential difference between the end points A and B, then

$\text V = \text V_1 + \text V_2 + \text V_3 \\[1em] \text V = \text Q\left(\dfrac{1}{\text C_1} + \dfrac{1}{\text C_2} + \dfrac{1}{\text C_3}\right)$

For a single equivalent capacitor C carrying the same charge Q at the same potential difference V, we have V = Q/C. Comparing,

$\dfrac{1}{\text C} = \dfrac{1}{\text C_1} + \dfrac{1}{\text C_2} + \dfrac{1}{\text C_3}$

Thus in a series combination the reciprocal of the equivalent capacitance is the sum of the reciprocals of the individual capacitances.

Three capacitors C₁ = 6 µF, C₂ = 3 µF and C₃ = 8 µF are connected to a 50 V battery as shown in figure given below. Calculate :

(a) The equivalent capacitance of the circuit between the points X and Y.

(b) The charge on C₁.

Answer

Given,

• C₁ = 6 µF, C₂ = 3 µF, C₃ = 8 µF
• EMF of the battery, V = 50 V

(a) From the figure, C₁ and C₂ are in series between X and Y, and C₃ is in parallel with this branch.

$\text C' = \dfrac{\text C_1 \text C_2}{\text C_1 + \text C_2} \\[1em] = \dfrac{6 \times 3}{6 + 3} \\[1em] \text C' = 2\ \mu\text F \\[1em] \text C_\text{XY} = \text C' + \text C_3 \\[1em] = 2 + 8 \\[1em] \text C_\text{XY} = 10\ \mu\text F$

(b) The full 50 V acts across the series branch, and both capacitors of that branch carry the same charge,

$\text q = \text C'\text V \\[1em] = (2)(50) \\[1em] \text q = 100\ \mu\text C$

Hence, the equivalent capacitance between X and Y is 10 µF and the charge on C₁ is 100 µC.

Consider 10 identical capacitors each of capacitance 5 µF. The ratio of the minimum and maximum possible values of the capacitance that can be obtained from them, is :

1. 1 : 50
2. 1 : 100
3. 50 : 1
4. 100 : 1

Answer

1 : 100

Explanation:
The minimum capacitance is obtained when all ten are joined in series,

$\text C_\text{min} = \dfrac{5}{10} \\[1em] \text C_\text{min} = 0.5\ \mu\text F$

and the maximum when all ten are joined in parallel,

$\text C_\text{max} = 10 \times 5 \\[1em] \text C_\text{max} = 50\ \mu\text F$

Hence

$\dfrac{\text C_\text{min}}{\text C_\text{max}} = \dfrac{0.5}{50} \\[1em] \dfrac{\text C_\text{min}}{\text C_\text{max}} = \dfrac{1}{100}$

The capacitance of a parallel plate capacitor does not depend on :

1. area of each plate
2. potential difference between the two plates
3. nature of the medium between the two plates
4. distance between the two plates

Answer

potential difference between the two plates

Explanation:
The capacitance of a parallel-plate capacitor is

$\text C = \dfrac{\text K \epsilon_0 \text A}{\text d}$

which depends only on the plate area A, the separation d and the dielectric constant K of the medium. On changing the potential difference the charge changes in the same proportion, so the ratio Q/V, that is C, is unaltered.

Calculate equivalent capacitance of the circuit shown in the figure given below:

Answer

Given,

• C₁ = 50 µF, C₂ = 50 µF, C₃ = 25 µF

From the figure, C₁ and C₂ are joined in parallel, and this combination is in series with C₃.

For the parallel part,

$\text C' = \text C_1 + \text C_2 \\[1em] = 50 + 50 \\[1em] \text C' = 100\ \mu\text F$

For the series combination with C₃,

$\text C_\text{AB} = \dfrac{\text C'\text C_3}{\text C' + \text C_3} \\[1em] = \dfrac{100 \times 25}{100 + 25} \\[1em] \text C_\text{AB} = \dfrac{2500}{125} \\[1em] \text C_\text{AB} = 20\ \mu\text F$

Hence, the equivalent capacitance of the circuit is 20 µF.

The air between the two plates of a parallel plate capacitor having capacitance of 20 µF is replaced with a slab of a dielectric material having dielectric constant (Relative permittivity) = 6.

As a result, capacitance of the capacitor

1. remains 20 µF
2. becomes 120 µF
3. becomes 20/6 µF
4. becomes infinity

Answer

becomes 120 µF

Explanation:
When the dielectric completely fills the space between the plates, the capacitance becomes K times its value with air,

$\text C = \text{KC}_0 \\[1em] = 6 \times 20 \\[1em] \text C = 120\ \mu\text F$

Calculate equivalent capacitance :

Answer

Given,

• C₁ = 8 µF, C₂ = 12 µF, C₃ = 12 µF

From the figure, C₂ and C₃ are joined in parallel, and this combination is in series with C₁.

For the parallel part,

$\text C' = \text C_2 + \text C_3 \\[1em] = 12 + 12 \\[1em] \text C' = 24\ \mu\text F$

For the series combination with C₁,

$\text C_\text{AB} = \dfrac{\text C_1 \text C'}{\text C_1 + \text C'} \\[1em] = \dfrac{8 \times 24}{8 + 24} \\[1em] \text C_\text{AB} = \dfrac{192}{32} \\[1em] \text C_\text{AB} = 6\ \mu\text F$

Hence, the equivalent capacitance between A and B is 6 µF.