Question

A capacitor plates are charged by a battery with ‘V’ volts. After charging battery is disconnected and a dielectric slab with dielectric constant ‘K’ is inserted between its plates, the potential across the plates of a capacitor will become
a) zero
b) v/2
c) v/k
d) KV​

Answer 1

Answer:

C

Explanation:

As the battery is disconnected, we can say that the capacitor is isolated and the charge present on it cannot go anywhere, neither any charge can come to it.

Hence Q is constant

As we know,

Q = CV

and introducing a dielectric slab make C' = kC

Q = Q'

CV = C' V'

CV = kC.V'

V' = V/k

This answers your question. If you're interested, read further

When I read about this I was curious that where does the energy go as we can see that the capacitor is isolated and the energy stored by it decreases.

It turns out, that the capacitor attracts the dielectric into it but the external agent which actually puts in the dielectric does some negative work on the system!

Answer 2

The correct option to the following question is (c), which is v/k.

Given: Potential difference = V

           Dielectric constant = K

To find The final potential difference of capacitor, when the dielectric is inserted.

Solution:

• A capacitor of capacitance C is connected to a battery, then the capacitor will get charged by Q. The potential difference across the capacitor will be V. The charge will be equal to the product of the capacitor and potential difference.

• When the battery is disconnected, the charge remains constant. Hence Q remains constant.

• As we know Q = CV

When we introduce a slab with a dielectric capacitor will become C' = KC

As the charge will remain constant.

Q = Q'

CV = C'V'

CV = KCV'

V' = V/K

From this, we can interfere that the potential difference across the capacitor will get decrease when a slab gets introduced into the capacitor of dielectric constant.