Question

A capacitor with plate separation d is charged to v volts. The battery is disconnected and a dielectric slab of thickness d2 and dielectric constant '2' is inserted between the plates. The potential difference across its terminals become

Answer 1

Given:

A capacitor with plate separation d is charged to V volts. The battery is disconnected and a dielectric slab of thickness (d/2) and dielectric constant '2' is inserted between the plates.

To find:

Potential difference across the terminals.

Calculation:

The basic concept behind this question is : Whenever the capacitors are disconnected from the battery the charge on the plates become constant.

Before insertion of dielectric slab :

$\rm{ \therefore \: q = CV}$

$\rm{ = > \: q = \dfrac{\epsilon_{0}A}{d} \times V}$

$\rm{ = > \: q = \dfrac{\epsilon_{0}AV}{d}}$

After insertion of dielectric slab:

Net capacitance is constituted by series combination of air filled capacitor and dielectric slab.

$\rm{ \: C_{air} = \dfrac{\epsilon_{0}A}{ (\frac{d}{2} )}}$

$\rm{ = > \: C_{air} = \dfrac{2\epsilon_{0}A}{ d}} = 2C$

$\rm{ \: C_{slab} = \dfrac{2\epsilon_{0}A}{ (\frac{d}{2} )}}$

$\rm{ = > \: C_{slab} = \dfrac{4\epsilon_{0}A}{ d}} = 4C$

Since they are in series , the net capacitance will be :

$\rm{ = > \: C_{net}= \dfrac{4C \times 2C}{4C + 2C} = \dfrac{4C}{3} }$

Charge in both cases will be same:

$\rm{\therefore \: q1 = q2}$

$\rm{ = > \: \dfrac{\epsilon_{0}AV}{d}} = \dfrac{4\epsilon_{0}A(V_{new}) }{3d}$

$\rm{ = > V_{new} = \dfrac{3V}{4} }$

So. final answer is :

$\boxed{ \rm{ V_{new} = \dfrac{3V}{4} }}$