Question

The separation between the plates of a parallel plate capacitor is made three times while it remains connected to a battery (A) the battery absorbs some energy (B) the electric field between plates remains same (c) the charge on the capacitor decrease by 66.7% (D) some work has to be done by an external agent on the plates

Answer 1

As the capacitor is connected to the cell so the potential difference (V) between the plates remains constant. The field between the plate is E=

d

V

and initial charge Q=CV.

when d=2d , the field E

′

=

2d

V

=

2

E

As the capacitnace C∝

d

1

, when d is double the capacitance becomes C/2

now the charge on capacitor is Q

′

=C/2V=CV/2=Q/2

Initial energy U=1/2QV and when separation double the energy U

′

=

2

1

Q

′

V=

4

QV

=

2

U

. Thus the energy of the capacitor is lost.

As energy of capacitor is lost , some energy absorbs by the cell.

As cell is connected to capacitor, the charge will flow to maintain constant potential and some work has to be done by the external agent on the plates.