Question

an airfilled parallel plate capacitor is connected across a battery . after it is fully charged,the battery is disconnected now a dielectric slab is inserted between the plates of the capacitor to fill the space completely . then the a) capacitance will decrease b) electric field between the plates will increase c)pd between the plates will increase d) charge on plates will remain the same.

Answer 1

Answer:

The capacitance without dielectric is C=

d

Aϵ

0

When dielectric slab is inserted, the capacitance becomes, C

′

=

d

AKϵ

0

=KC where K be the dielectric constant.

i)Thus, the capacitance will increase K times of the initial.

ii) As the battery is disconnected so the charge on capacitor remains constant.

Since, Q=CV so potential V will decrease and also E=V/d so the field E will also decrease.

iii) Stored energy, U=

2C

Q

2

. As charge Q is constant and C is increasing so energy will decrease