WBCHSE Class 12 Physics Capacitance And Capacitor Question and Answers

Class 12 Physics Capacitance And Capacitor  Long Questions and Answers

Question 1. What is meant by the 1μF capacitance of a capacitor?
Answer:

By the statement, we mean that, to build up a potential difference of 1 volt between the two plates of the capacitor, 1μC = 10-6 coulomb of charge is to be given to its insulated plate.

Question 2. What is meant by the statement that the dielectric constant of water is 80?
Answer:

By this statement, we mean that the capacitance of a capacitor will become 80 times if water is used as a dielectric instead of air between its plates.

Question 3. Two conductors have equal amounts of the same type
of charge. Can there be a difference of potential between them?
Answer:

Potential of a conductor \(=\frac{\text { charge }}{\text { capacitance }}\)

So, in spite of the fact that the two conductors have equal charge, their potentials may not be equal if they have different capacitances. There may exist a potential difference between them.

Question 4. Two copper spheres have the same radius. One of them is hollow and the other is solid. If they are charged to same potential, which sphere will hold a greater amount of charge?
Answer:

The capacitance of a hollow or of a solid sphere placed in air is proportional to its radius. In this example, the capacitances of both spheres are equal. Again amount of charge = capacitance x potential. So, if the two spheres are charged to the same potential, they will hold the same amount of charge.

WBBSE Class 12 Capacitor Q&A

Question 5. What are the advantages of using a solid insulator as a dielectric of a capacitor?
Answer:

1. The value of the dielectric constant (x) of a solid dielectric is large. So, the capacitance increases many times.

2. The two plates of the capacitor cannot come in contact with each other.

3. The capacitor can be charged to a higher potential.

Question 6. Is it possible to charge a capacitor to any high potential at will?
Answer:

It is not possible to charge a capacitor to any high potential at will. Because, as the potential applied to it becomes high enough, the insulation of the intervening medium breaks down. So, the electric discharge takes place between the capacitor and the intervening medium and also between the two plates.

Question 7. The potential of plate A of a parallel plate capacitor is zero and its other plate B is maintained at a potential +V. How does the potential vary from point to point between these plates? Neglect the end effect.
Answer:

Let the potential of a point at a distance x from the plate A be V’ and the intensity of the electric field be E, then

⇒ \(E=\frac{V^{\prime}}{x} \quad \text { or, } V^{\prime}=E x\)

The intensity of the electric field (E) in the space between the two plates is a constant.

∴ \(V^{\prime} \propto x\)

i.e., the potential V’ increases uniformly with x. So, the graph of variation of potential with distance from A to B is a straight line passing through the origin.

Question 8. The dielectric constant of water is very high. Yet why is water not used as a dielectric in a capacitor?
Answer:

Only pure water is a good insulator. But, impure water acts as an electrolytic conductor. So, water is not suitable as a dielectric in a capacitor. Also, water is a liquid and so it is inconvenient to use it as a dielectric in a capacitor as it can spill.

Question 9. Two protons A and B are placed in between the two plates of a parallel plate capacitor charged to a potential difference V. Are the forces on the two protons equal?

Answer:

The electric field in between the two plates of a parallel plate capacitor charged to a potential difference V is uniform everywhere. So equal forces will act on the two protons.

Short Answer Questions on Capacitance

Question 10. If a soap bubble is electrified, will its shape be changed? If so, how will the potential of the bubble change?
Answer:

A soap bubble is considered a spherical conductor. When it is electrified, the charge distribution on its surface will have a repelling effect. Hence, the radius of the bubble increases. The capacitance of a spherical conductor varies with its radius, but it still remains spherical. So, the capacitance of the charged bubble increases.

The potential, \(V=\frac{\text { charge }}{\text { capacitance }}\); if the charge remains constant, the potential of the bubble will decrease the potential of the bubble will decrease.

Question 11. A capacitor is charged with a battery and then disconnected. A dielectric slab Is then inserted between the plates. How are the

1. Charge
2. Capacitance
3. Potential difference
4. The stored energy related to the capacitor affected?

Answer:

1. Since the capacitor remains disconnected from the battery, t the amount of charge on the capacitor will not change.

2. Due to the insertion of the dielectric slab between the plates, the capacitance will Increase. If K be the dielectric constant of the slab, its capacitance will be K times its previous value.

3. As V = \(\frac{Q}{C}\) and Q = constant,

we have \(V \propto \frac{1}{C}\).

So the potential difference is reduced.

The final potential difference will be \(\frac{1}{k}\) times of its previous value.

4. Energy stored in the capacitor,

⇒ \(U=\frac{1}{2} \frac{Q^2}{C} ; \text { so, } U \propto \frac{1}{C} \text { as } Q=\text { constant. }\)

Due to the insertion of the dielectric slab, the amount of energy stored is reduced. The final energy stored in the capacitor will be \(\frac{1}{k}\) times of its previous value.

Common Questions on Capacitance with Answers

Question 12. A capacitor is charged by a battery and then disconnected. How are the capacitance, potential difference and stored energy related to the capacitor affected if

1. The distance between the plates is decreased or
2. The plates are connected by a metallic wire.

Answer:

The capacitance of a parallel plate capacitor,

⇒ \(C=\frac{\kappa \epsilon_0 \alpha}{d}\)

1. If the distance between the plates is decreased, the capacitance C of the capacitor will increase. Since the charge remains constant, due to the increase of capacitance, the potential difference between the plates ( V = \(\frac{Q}{C}\)) and the energy stored in the capacitor \(\left(U=\frac{1}{2} \frac{Q^2}{C}\right)\) will decrease.

2. If the plates of the charged capacitor are connected by a metallic wire, the capacitor will be discharged immediately. It will not act as a capacitor any more. The potential difference between the plates of the capacitor as well as the energy stored will be zero.

Question 13. A mica slab of thickness equal to the distance between the two plates of a parallel plate air capacitor is Inserted in the space between the plates. Explain the changes in capacitance in the following cases:

1. When the mica slab is Inserted partially;
2. When the space between the plates of the capacitor is totally filled by the mica slab.

Answer:

1. The partial insertion of the mica slab in the space between the plates of the parallel plate air capacitor. Let the area of the plates of the capacitor be a; the area of the mica slab inside the capacitor be < a); the distance between the plates of the capacitor be d and the dielectric constant of mica be K.

Therefore, total capacitance of the capacitor, C = capacitance of parallel plate capacitor of area α₁ with mica as dielectric + capacitance of parallel plate air capacitor of area (α – α₁)

⇒ \(\frac{\epsilon_0 \kappa \alpha_1}{d}+\frac{\epsilon_0\left(\alpha-\alpha_1\right)}{d}=\frac{\epsilon_0 \alpha}{d}+\frac{\epsilon_0 \alpha_1(\kappa-1)}{d}\)

So due to the partial insertion of the mica slab, the capacitance increases by

⇒ \(\frac{\epsilon_0 \alpha_1(k-1)}{d}\)

2. The capacitance will be \(\frac{\epsilon_0 \kappa \alpha}{d}\)

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Question 14. A parallel plate air capacitor is connected to a battery. Its charge, potential difference, electric field and the stored energy between the plates are Q₀, V₀, E₀ and U₀ respectively. Keeping the battery connection unchanged, the capacitor is completely filled with a dielectric material. The charge, potential difference, electric field and energy stored become Q, V, E and U respectively. Which of the following is correct?

1. Q > Q₀
2. V> V₀,
3. E > E₀
4. U > U₀

Answer:

1. When the dielectric material is inserted in the capacitor, its capacitance increases. As the capacitor is still connected to the battery, so its voltage will remain constant.

Since, charge = capacitance x voltage, so Q > Q₀.

2. Since the capacitor remains connected to the battery its voltage remains constant.

Hence, V > V₀ is not correct.

3. Since voltage and distance between the plates remain consistent, so electric field remains constant.

Hence, E > E₀ is not correct.

4. Energy stored in a capacitor

= \(\frac{1}{2}\) x capacitance x (voltage)²

Since the voltage of the capacitor remains constant but its capacitance increases, so the energy storedin the capacitor will increase.

Hence, U > U₀ is correct.

Question 15. If an uncharged capacitor is connected to a battery, then show that half of the energy supplied by the battery to charge the capacitor is dissipated as heat.
Answer:

Let a capacitor of capacitance C be connected to a battery of emf V.

The final charge on the capacitor, Q = CV

The work done by the battery to fully charge the capacitor, W = VQ

So energy supplied by the battery = VQ

The energy storedin the fully charged capacitor = \(\frac{1}{2} C V^2=\frac{1}{2} Q V\)

∴ The remaining amount of energy dissipated as heat

⇒ \(V Q-\frac{1}{2} Q V=\frac{1}{2} Q V\)

Hence half of the energy supplied by the battery is dissipated as heat.

Question 16. What is the force of attraction between the two plates of a parallel plate capacitor? Assume that, the area of each plate of the capacitor is A and one plate is charged with +Q and the other with -Q.
Answer:

The intensity of the electric field at a point near a charged plate having a surface density of charge cr is,

⇒ \(E=\frac{\sigma}{2 \epsilon_0}\)

Now, \(\sigma=\frac{Q}{A}\)

∴ \(E=\frac{Q}{2 A \epsilon_0}\)

The magnitude of the charge on the other plate of the capacitor = Q.

So, the force experienced by the other plate is,

⇒ \(F=Q E=Q \cdot \frac{Q}{2 A \epsilon_0}=\frac{Q^2}{2 A \epsilon_0}\)

Practice Questions on Capacitor Circuits

Question 17. What will be the change in the capacitance of a parallel plate air capacitor if

1. A dielectric slab
2. A conducting slab fills the space between the plates of the capacitor.

Answer:

1. If the dielectric constant of the slab be K, the capacitance will be K times the capacitance of an air capacitor.

2. Due to the insertion of the conducting slab, the capacitor will be totally discharged and it will no more act as a capacitor.

Question 18. Can a single conductor be treated as a capacitor? Which is the second plate in that case?
Answer:

A single conductor can be treated as a capacitor. Earth is taken to be its second plate in that case.

Question 19. In the circuit, the ammeter shows a deflection as soon as the circuit is closed. But after a while, the pointer returns to zero. Explain it.

Answer:

As soon as the circuit is closed, the capacitor begins to accumulate charge. The initial current through the circuit is high as the battery transports charge from one plate to the other of the capacitor. The charging current asymptotically approaches to zero with time.

Finally the sum of potential differences across R and that between the plates of the capacitor becomes equal to the voltage source. Hence current ceases to flow through the ciiÿ cuit and so the pointer of the ammeter returns to zero.

Question 20. Why is the spherical surface of a Van de Graaff generation made very smooth?
Answer:

Electric discharge takes place from rough edges if the spherical surface of V a de Graaff generator is even slightly rough and the potential of the sphere diminishes. To avoid discharge, the outside surface of the sphere is made very smooth.

Question 21. Why is the case of a Van de Graaff generator filled with some gas at a high pressure?
Answer:

At a high potential difference between the two spheres of a Van de Graaff generator, electric discharge may start in the neighboring air medium, because air cannot bear high potential difference under normal pressure. The case of the Van de Graaff generator is thus filled with nitrogen or freon at high pressure. Even under high potential difference, nitrogen or freon molecules hardly ionize. The gas under high pressure, hence, decreases the amount of discharge.

Question 22. Why is It difficult to construct a conductor of capacitance 1F?
Answer:

If the conductor is assumed to be spherical, then its capacitance is C = 4π∈₀r

∴ \(r=\frac{C}{4 \pi \epsilon_0}\)

= \(\frac{1}{\frac{1}{9 \times 10^9}}\)

= \(9 \times 10^9 \mathrm{~m}\)

So the radius of the spherical conductor of capacitance 1 F is very large and even greater than the radius of the earth (6.4 x 106 m). Hence it is difficult to construct a conductor of capacitance 1 F.

Question 23. The graph shown here shows the variation of the total energy (E) stored in a capacitor against the value of the capacitance(C) itself. Which of the two—the charge on the capacitor or the potential used to charge it is kept constant for this graph?
Answer:

Energy stored in a capacitor,

⇒ \(E=\frac{Q^2}{2 C}=\frac{1}{2} C V^2\)

The nature of the graph suggests that it is a plot of the equation,

⇒ \(E=\frac{Q^2}{2 C}\)

Hence in this graph, the charge on the capacitor (Q) is kept consistent.

Important Definitions in Capacitance

Question 24. A, B, C, and D arc four similar metallic parallel plates, equally separated by distance d and connected to cell of emf V.

1. Write the potentials of the plates A, B, C and D.

2.

1. If plates B and C are connected by a wire then what will be the potential of the plates?
2. How will the electric field change in the spacing between the plates?
3. Will the charges on the plates A and D change?

Answer:

1. The potential of plate A is V and the potential of plate D (earthed) is zero. Since the plates are equidistant, the potentials of B and C are \(\frac{2V}{3}\) and \(\frac{V}{3}\) respectively.

2.

1. On connecting plates B and C with a wire, they will come to the same potential. Plates B and C will each have a potential = \(\frac{1}{2}\left(\frac{2 V}{3}+\frac{V}{3}\right)=\frac{V}{2}\)

Plates A and D will still remain at V and zero potential tial respectively.

2. The electric field between plates A and B will increase from \(\frac{2V}{3d}\) to \(\frac{V}{2d}\) and become zero between plates B and C. It will increase from to between plates C and D.

3. Initially the capacitance of the system was \(\frac{\epsilon_0 \alpha}{3 d}\), where α = area of each plate.

Now it has increased to \(\frac{\epsilon_0 \alpha}{2 d}\). Hence charges on plates A and D will increase.

Question 25. The plate A of a parallel plate capacitor is connected to a spring of force constant k and can move, while the plate B is fixed. The arrangement is held between two rigid supports. If a charge +q is placed on plate A and -q on plate B, how much does the spring elongate?

Answer:

The force of attraction between the two plates of the capacitor,

F = -kl (l = elongation of the Spring, k = force constant)…..(1)

If a is the area of the two plates and the distance between the plates is x, then the capacitance of the capacitor,

⇒ \(C=\frac{\epsilon_0 \alpha}{x}\)….(2)

We know energy stored in the capacitor

⇒ \(U=\frac{1}{2} \frac{q^2}{C}\)

From equation (2) we may write,

⇒ \(U=\frac{1}{2} \frac{q^2}{C}\)

As the electrostatic force (F) is a conservative force, it is equal to the negative gradient of the corresponding potential.

∴ \(F=-\frac{d U}{d x}=-\frac{d}{d x}\left(\frac{q^2 x}{2 \epsilon_0 \dot{\alpha}}\right)\)

or, \(F=-\frac{q^2}{2 \epsilon_0 \alpha}\) [negative sign implies that the force is attractive in nature] ….(3)

So from equations (1) and (3) we have,

⇒ \(-k l=-\frac{q^2}{2 \epsilon_0 \alpha}\)

∴ \(l=\frac{q^2}{2 \epsilon_0 \alpha k}\)

Examples of Capacitor Calculation Questions

Question 26. A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants, k₂ and k₃. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then what will be its dielectric constant?

Answer:

Capacitance due to first dielectric,

⇒ \(C_1=\frac{\kappa_1 \epsilon_0\left(\frac{A}{2}\right)}{\left(\frac{d}{2}\right)}=\frac{\kappa_1 \epsilon_0 A}{d}\)

Capacitance due to the second dielectric,

⇒ \(C_2=\frac{\kappa_2 \epsilon_0\left(\frac{A}{2}\right)}{\left(\frac{d}{2}\right)}\)

= \(\frac{\kappa_2 \epsilon_0 A}{d}\)

Capacitance due to the third dielectric,

⇒ \(C_3=\frac{\kappa_3 \epsilon_0 A}{\left(\frac{d}{2}\right)}\)

= \(\frac{2 \kappa_3 \epsilon_0 A}{d}\)

The capacitors C₁ and C₂ are in parallel and their equivalent capacitance is,

⇒ \(C^{\prime}=C_1+C_2=\frac{\epsilon_0 A}{d}\left(\kappa_1+\kappa_2\right)\)

This combination is in series with C₃. Hence the equivalent capacitance is,

⇒ \(\frac{1}{C^{\prime \prime}}=\frac{1}{C^{\prime}}+\frac{1}{C_3}\)

= \(\frac{d}{\epsilon_0 A\left(\kappa_1+\kappa_2\right)}+\frac{d}{2 \kappa_3 \epsilon_0 A}\)

⇒ \(\frac{d}{\epsilon_0 A}\left[\frac{1}{\left(\kappa_1+\kappa_2\right)}+\frac{1}{2 \kappa_3}\right]\)….(1)

When a single dielectric of dielectric constant K is used

⇒ \(\frac{1}{C^{\prime \prime}}=\frac{d}{\epsilon_0 K A}\)….(2)

Comparing equations (1) and (2) we get,

⇒ \(\frac{1}{K}=\frac{1}{\left(K_1+K_2\right)}+\frac{1}{2 \kappa_3}\)

∴ \(\kappa=\frac{2 \kappa_3\left(\kappa_1+\kappa_2\right)}{\kappa_1+\kappa_2+2 \kappa_3}\)

Question 27. Two identical metal plates are positively charged to Q₁ and Q₂ (Q₂ < Q₁). If they are brought near each other to form a parallel plate capacitor of capacitance C, then what will be the potential difference between the plates?
Answer:

Let the area of the metal plates be A and the intensity of the electric fields at any point between the plates due to the first and second metal plates being E₁ and E₂ respectively.

The electric field at any point between the plates due to the first plate,

⇒ \(E_1=\frac{Q_1}{2 A \epsilon_0}\)

The electric field at any point between the plates due to the second plate,

⇒ \(E_2=\frac{Q_2}{2 A \epsilon_0}\)

So-net electric field,

⇒ \(E=E_1-E_2\)

= \(\left(\frac{Q_1-Q_2}{2 A \epsilon_0}\right)\) [where ∈₀ = permittivity of free space]

Again capacitance of parallel plate capacitor

⇒ \(C=\frac{\epsilon_0 A}{d}\) [d = distance between the two plates]

We know, that potential difference = net electric field x distance

∴ \(V=\left(E_1-E_2\right) d\)

= \(\left(\frac{Q_1-Q_2}{2 A \epsilon_0}\right) d\)

= \(\left(\frac{Q_1-Q_2}{2}\right) \frac{d}{A \epsilon_0}\)

Hence potential differences,

⇒ \(V=\frac{Q_1-Q_2}{2 C}\left[∵ C=\frac{\epsilon_0 A}{d}\right]\)