# Capacitors and dielectric – Question & Answer

In this post, we have presented fundamental and conceptual questions and answers based on parallel plate capacitors and dielectrics between the plates. This Capacitor and dielectric Question set will certainly help to build the concepts.

## Capacitor and dielectric Question & Answer

1 ) A parallel plate capacitor is charged by a battery which is then disconnected. Then, the distance between the plates is decreased. What are the changes in (i) charge on the plates (ii) capacitance (iii) PD between plates (iv) electric field between plates (v) stored energy?

(i) As the capacitor is isolated after charging, hence the charge Q on the plates remains unchanged.

ii) Capacitance is indirectly proportional to the distance d between the plates. Hence, capacitance increases when d is decreased.

iii) As Q = CV, and Q is constant here, hence as C increases Potential Difference V between the plates falls.

iv) The electric field remains unchanged. [E = σ/ε0 = Q/(A ε0)]

v) The stored energy = (1/2) Q2/C. Here, Q = unchanged, but C increases. Hence, the stored energy falls.

2 ) A parallel plate capacitor is charged by a battery which is then disconnected. Then, the distance between the plates is increased. What are the changes in (i) charge on the plates (ii) capacitance (iii) PD between plates (iv) electric field between plates (v) stored energy?

(i) As the capacitor is isolated after charging, hence the charge Q on the plates remains unchanged.

ii) Capacitance is indirectly proportional to the distance d between the plates. Hence, capacitance decreases when d is increased.

iii) As Q = CV, and Q is constant here, hence as C decreases Potential Difference V between the plates rises.

iv) The electric field between the 2 plates remains unchanged. [E = σ/ε0 = Q/(A ε0)]

v) The stored energy = (1/2) Q2/C. Here, Q = unchanged, but C decreases. Hence, the stored energy rises.

3 ) A parallel plate capacitor of plate area A and separation d is charged to a potential difference V. The charging battery is then disconnected and the plates are pulled apart until their separation is 2d.

Write expressions in terms of A, d, and V for

(i) the new PD (ii) the initial stored energy (iii) the final stored energy (iv) the work done in increasing the separation between plates.

[Note: here the answers need to be expressed in terms of A,d, and V.]

As the charging battery is disconnected, that means Q remains unchanged.

i) C0= Aε0/d

V0=V

Q0 = C0V ………. (1)

As plates are pulled apart to 2d.

Cf= Aε0/(2d) = (½) C0

Vf=?

Qf =Cf Vf = (½) C0 Vf …………… (2)

Q constant.

Q0= Qf

C0V = (½) C0 Vf

Vf = 2V

ii) Initial stored energy = Ui = (½) C0V2 = (1/2)(Aε0/d)V2

iii) Final stored energy = Uf = (½) Cf Vf2 = (1/2)(Aε0/(2d))(2V)2 = (Aε0/d)V2

iv) Work done in increasing the separation between plates

= final stored energy – initial stored energy

= (Aε0/d)V2 – (1/2)(Aε0/d)V2 = (1/2)(Aε0/d)V2

4) A capacitor is connected to a battery. If we move its plates further apart, work will be done against the electrostatic attraction between the plates. What will happen to this work?

What will be the effect on the energy of the capacitor?

Answer: [Note: Here the capacitor remains connected to the battery.]

As the separation d increases, capacitance C falls.

Here, V remains constant.

We know, Q = CV

Hence, charge Q on the capacitor plates falls. (By definition: charge storing capacity of the plates falls as capacitance C falls)

The extra charge now flows to the battery from the plates.

The work done as mentioned in the question is utilized to transfer charge from plates to the battery.

As C falls, with V remaining constant, hence the energy of the capacitor (U=CV2/2) also falls. (extra energy is transferred to the battery)

5) Under which of the following two cases more work will be done in moving the plates of a charged capacitor further apart and why?

(i) The charging battery remains connected to the capacitor

(ii) After charging the capacitor, the battery is disconnected.

(i) In this case, V remains constant. (as the charging battery remains connected to the capacitor). Plates are moved further apart, this means d increases, and capacitance C falls.

The stored energy ( following this equation U=CV2/2) falls in this case.

ii) In this case, after charging the battery is disconnected. So, the charge Q remains constant. Plates are moved further apart, this means d increases, and capacitance C falls.

As a result, V increases to keep the Q unchanged. (following this equation Q = CV)

The stored energy [ following this equation U=(1/2)QV or this equation U=(1/2)Q2/C ] rises in this case.

This means, in the 2nd case, more work is done in moving the plates of a charged capacitor further apart.

6) How does the electric field between the plates get affected under the following two cases in moving the plates of a charged capacitor further apart and why?

(i) The charging battery remains connected to the capacitor

(ii) After charging the capacitor, the battery is disconnected.

(i) In this case, V remains constant. (as the charging battery remains connected to the capacitor). Plates are moved further apart, this means d increases, and capacitance C falls.

As Q = CV, the charge on the plates falls in this case.

Electric field = σ/ε0 = Q/(A ε0)

As Q on the plates decreases in the case, the electric field also decreases.

ii) In this case, after charging the battery is disconnected. So, the charge Q remains constant.

[Plates are moved further apart, this means d increases, and capacitance C falls.

As a result, V increases to keep the Q unchanged. (following this equation Q = CV)]

Electric field = σ/ε0 = Q/(A ε0)

As Q remains unchanged, hence electric field also remains unchanged.

7) A capacitor is charged and the charging battery is removed. A slab of dielectric material is then inserted between the plates of the capacitor. How will the electric field between the plates be affected? Why?

Due to polarization in the dielectric, an electric field is generated in the dielectric that opposes the electric field due to the charged plates of the capacitor. Thus the electric field between the plates is weakened as the result of dielectric insertion between the plates.

(2nd approach) Using formula,

E(initial) = σ/ε0 = Q/(A ε0)

and E (after dielectric insertion) = σ/(Kε0) = Q/(KA ε0)

As K>1,

E (after dielectric insertion) is less than E (initial).

Capacitors and dielectric – Question & Answer
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