Capacitors Question and Answer with Explanation

Q.1. What is the capacitance and what is its unit?

Ans. Capacitance is the property of a capacitor to store electric charge required to raise its potential through unity and is given by

C = \dfrac{Q}{V} and its unit is farad.

Q.2. What is the permittivity of evacuated space?

Ans. Permittivity of evacuated space is  8.854 \times 10^{-12}\dfrac{F}{m}.

Q.3. If air is replaced by a dielectric permittivity ɛr in a condenser of capacity C with air as medium, what will be effect on its capacity?

Ans. Capacity of the condenser will increase and become ɛr C.

Q.4. There are two concentric conducting spherical surface of radii a and b (a<b). the inner spherical surface carries a charge Q and the outer surface is grounded. What is the potential of the inner spherical surface?

Ans. V = \dfrac{Q}{4\pi \epsilon_0 \epsilon_r}(\dfrac{1}{a}-\dfrac{1}{b}) volts

Q.5. What will be the capacitance of the capacitor given in Q.4?

Ans. C = 4 \pi \epsilon_0 \epsilon_e \dfrac{ab}{(a-b)} farads.

Q.6. What is the physical significance of capacitance?

Ans. Capacitance of an arrangement tells us the potential to which the arrangement will be raised when a given charge is placed on it. For a given charge, smaller the potential, larger must be the capacitance and vice-versa.

Q.7. Where is the knowledge of dielectric strength helpful?

Ans. Dielectric strength is the maximum strength of electric field that can be tolerated by the dielectric without electric breakdown. Its knowledge helps us in designing a capacitor by determining the maximum potential that can be applied across the plates of the capacitor.

Q.8. During lightning, you are safer inside a house than under a tree. Why?

Ans. When we stand under a tree, we provide an easy path for the lightning to pass through our body. But when we are in the house, discharge due to lightning may be conducted to the ground through iron pipes/wall etc. Therefore, we are safer.

Q.9. The safest way to protect yourself from lightning is to be inside a car. Comment.

Ans. The body of the car is metallic. It provides electrostatic shielding to the person in the car, because electric field inside the car is zero. The discharging due to lightning passes to the ground through the metallic body of the car.

Q.10. A sensitive instrument is to be shifted from the strong electrostatic field in its environment, Suggest a possible way.

Ans. For this, the instrument must be enclosed fully in a metallic cover. This will provide electrostatic shielding to the instrument.

Q.11. Faraday entered a big metallic cage supported on insulating pillars and then charged the cage by a powerful electric machine. He remained quite safe inside the cage. Do you believe on this happening?

Ans. Yes. As the cage is insulated, the entire charge remained on the outer surface of the cage. The potential at all points inside the cage remained the same. As there was no potential difference between any two points in the cage, Faraday could not get the shock and remained quite safe inside the cage.

Q.12. Can ever the whole charge of a body be transferred to the other? If yes, how and if not, why?

Ans. Yes, the whole charge of a body. A can be transferred to the conducting body B, when A is enclosed by B and is connected to it. This is because charge resides on the outer surface of a conductor.

Q.13. What will you do for your safety, when there is a danger of lightening and you are travelling by car?

Ans. We have to close the windows of our car.

Q.14. A technician has only two capacitors. By using them singly, in series or in parallel, he is able to obtain the capacitance of 4, 5, 20 and 25 micro farad. What are the capacitance of the two capacitors?

Ans. In series grouping, the resultant capacitance is minimum, i.e., C_s = \dfrac{C_1 C_2}{C_1 + C_2} = 25 micro farad.

In parallel grouping, the resultant capacitance is maximum, i.e., CP = C1 + C2 = 25 micro Farad.

It implies that the remaining values 5 micro farad and 20 micro farad represent the individual values of two capacitors.

Q.15. If two isolated conductors each having a definite capacity are far apart and are connected to each other by a fine wire, how do you calculate the capacity of the combination? In joining them with the wire, have you connected them in parallel or in series?

Ans. On connecting, both the spheres acquire a common potential V. If C is total capacity, then total charge, q = CV = q1 + q2, as charge is conserved.

Therefore,  C = \dfrac{q_1}{V} \dfrac{q_2}{V} = C_1 + C_2.

Hence the conductors have been connected in parallel.

Q.16. Two spheres of radii R and 2 R are charged, so that both of these have same surface charge density \sigma. The spheres are located away from each other and are connected by a thin conducting wire. Find the new charge density on the two spheres.

Ans. If q1, q2, are the initial charges on the spheres of radii Rand 2 R respectively, then total charge on the two spheres  q = q_1 + q_2 = 4 \pi R^2 \sigma + 4 \pi (2R)^2 \sigma = 20 \pi R^2 \sigma . When the spheres are connected by a thin wire, they will share charges till their potentials become equal. The charges on the two spheres would then be in the ratio of their capacity. If q11, q21, are the new values of charge on the two spheres, then

 q_{11} = q \times \dfrac{C_1}{C_1 + C_2} and  q_{21} = q \times \dfrac{C_2}{C_1 + C_2}

As capacity of a spherical conductor is directly proportional to the radius of sphere,

Therefore,  q_{11} = q \times \dfrac{R}{R + 2R} = \dfrac{q}{3} = \dfrac{20}{3}\pi R^2 \sigma

q_{21} = q \times \dfrac{2R}{R + 2R} = \dfrac{2q}{3} = \dfrac{40}{3}\pi R^2 \sigma

New charge density of the two spheres would be

\sigma_1 = \dfrac{q_{11}}{4 \pi R^2} = \dfrac{20}{3} \dfrac{\pi R^2 \sigma}{4 \pi R^2} = \dfrac{5}{3} \sigma \sigma_2 = \dfrac{q_{21}}{4 \pi (2R)^2} = \dfrac{40}{3} \dfrac{\pi R^2 \sigma}{16 \pi R^2} = \dfrac{5}{6} \sigma

Q.17. Two conductors identical in shape and size, but one of copper and the other of aluminum (which is less conducting) are both placed in an identical electric field. In which metal, more charge will be induced.

Ans. Maximum induced charge is given by  q_1 = -q(1-\dfrac{1}{K})

For both metals, Cu and Al, K = infinity, Therefore, q1 = -q, i.e., in both the metals, induced charge will be equal. Note that in conduction, current in copper will be more than in aluminum as resistivity of copper is smaller than that of aluminum.

Q.18. A capacitor is charged by using a battery, which is then disconnected. A dielectric slab is then slipped between the plates. Describe qualitatively what happens to the charge, the capacitance, the potential difference, electric field strength and stored energy?

Ans. Charge remains the same; capacitance increases; potential difference decreases; Electric field strength decreases and stored energy decreases.

Q.19. Can there be a potential difference between two conductors of same volume carrying equal positive charges?

Ans. Yes, because two conductors of same volume may have different shapes and hence different capacitances.

Q.20. What is the net charge on a charged capacitor?

Ans. Zero, because one plate has positive charge and the other carries an equal negative charge.

Q.21. Why does the electric conductivity of earth’s atmosphere increase with altitude?

Ans. This is because of ionization caused by highly energetic cosmic ray particles from cosmos, which are hitting the atmosphere of earth.

Q.22. Where does the energy of a capacitor reside?

Ans. The energy resides in the dielectric medium separating the two plates.

Q.23. What is the basic difference between a charged capacitor and an electric cell?

Ans. A capacitor supplies electrical energy stored in it. A cell supplied electric energy by converting chemical energy into electrical energy at constant potential difference.

Q.24. The dielectric constant of a condenser can be taken as infinitely large. Why?

Ans. When a conductor is held inside an electric field, the field inside the conductor becomes zero. The dielectric constant of conductor is the ratio of strength of applied electric field to the reduced value of electric field inside, and it becomes infinity.

Q.25. Difference between series combination of capacitor and parallel combination of capacitor.

S.N.Series Combination of CapacitorS.N.Parallel Combination of Capacitor
1.Negative plate of one condenser connected to positive plate of other and so on.1.Positive plates of all condensers connected to one point and negative plates of all condensers connected to another point.
2.Charge on each condenser is same +/-q2.Potential difference across each condenser is same (V).
3.Net potential difference between A and B is

V = V1 + V2 + V3 + ….

3.Net charge on equivalent condenser is

q = q1 + q2 + q3 + ….

4.Capacity (C) of equivalent condenser is

 \dfrac{1}{C} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dfrac{1}{C_3} i.e. Capacitance decreases

4.Capacity (C) of equivalent condenser is

C = C1 + C2 + C3 + … i.e. Capacity increases.

Q.26. What is A Van de graaf generator?

Ans. A Van  de graaf generator is a powerful machine used for generating high positive potential = 10volt. It is based on the action of sharp points and the property that charge always resides on the outer surface of a hollow conductor. Such high positive potentials are needed for accelerating positive ions.

One Thought to “Capacitors Question and Answer with Explanation”

  1. andrew

    The questions is very nice. If there is other please post to me with answers

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