A capacitor of capacitance C is charged fully using a battery of e.m.f, E . It is then
disconnected form the battery . If the separation between the plates of the capacitor is
now doubled. What will happen to : Charge stored by the capacitor,P.D across it, field
strength between the plates and energy stored between the plates of the capacitor.. Answer the question again, if the battery is kept connected

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Solution

First always remember 1.when battery is remove after charging capacitor ,the charge remain constant while doing operation of capacitor (eg. Inserting dielectric)

2. When battery remain connected and then we do operation then potential on capacitor is constant.

A) when plates move the charge remain constant that is equal to =cv

B)as plate distance double the capacitance decreases therefore potential increase asQ=cv Q constant as c decreases v increase.

C)electeic feild remaon constant asE(initial)= v/d =q/(cd) (c=èa/d) therfore E =q/èaE(final)=Vnew/2d =q/(Cnew×2d) (Cnew=èa/2d) therfore E (final)=q/èa

D) energy =1/2q^2/c as c decreases energy increases

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Similar questions

Q. (a) Distinguish with the help of a suitable diagram, the difference in the behaviour of a conductor and a dielectric placed in an external electric field. How does polarised dielectric modify the original external field?
(b) A capacitor of capacitance C is charged fully by connecting it to a battery of emf E. It is then disconnected from the battery. If the separation between the plates of the capacitor is now doubled, how will the following change?
(i) Charge stored by the capacitor.
(ii) Field strength between the plates.
(iii) Energy stored by the capacitor.
Justify your answer in each case.

Explain briefly the process of charging a parallel plate capacitor when it is connected across a d.c. battery.

A capacitor of capacitance ‘C’ is charged to ‘V’ volts by a battery. After some time the battery is disconnected and the distance between the plates is doubled. Now a slab of dielectric constant, 1 < k < 2, is introduced to fill the space between the plates. How will the following be affected?

(a) The electric field between the plates of the capacitor

(b) The energy stored in thecapacitor

Justify your answer by writing the necessary expressions

A parallel plate capacitor of capacitance C is charged and disconnected from the battery. The energy stored in it is E. If a dielectric slab of dielectric constant 6 is inserted between the plates of the capacitor then energy and capacitance will become :

Statement 1 :

A parallel plate capacitor is charged by a battery of voltage

V

. The battery is then disconnected. If the space between the plates is filled with a dielectric, the energy stored in the capacitor will decrease.

Statement 2 :

The capacitance of a capacitor increases due to the introduction of a dielectric between the plates.

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