Question

A parallel plate capacitor is charged by a battery and disconnected. A dielectric slab is then introduced between the plates leaving no gap. Determine the changes that would occur in the values of the charge capacitance and potential difference​

Answer 1

Answer:

Given: Capacitor is charged and then battery is DISCONNECTED

Hence, the only thing that will be constant is the charge stored by the capacitor.

So, Q = constant = Q'

Capacitance will become, C' = K * C (let K is the dielectric constant)

This is because the formula for capacitance is  $C=\frac{Ae}{d}$

and, the new capacitance $C'= \frac{A e.e'}{d}$

where, e = dielectric constant for vacuum

and, e' = dielectric constant for dielectric slab (same as K)

We also know, Q = C * V  .............................( 1 )

So, Q' = C' * V'.................................................( 2 )

Dividing the above 2 equations,

$\frac{Q'}{Q} = \frac{C' V'}{CV}$

But, Q=Q' and C'= KC.

So,

$1=\frac{KC V'}{C V}\\ i.e, 1 = \frac{KV'}{V} \\Hence, V'= \frac{V}{K}$

I used to have a lot of doubt in this question before too, until someone ine brainly helped me. Hope I could help you too  :)