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

(i) Zero

(ii) V/2

(iii) V/K

(iv) KV​

Answer 1

The potential across the plates of a  capacitor will become (iii) $\frac{V}{K}$

Explanation:

In this scenario, $Q=$ Charge remains constant

Where $E = \frac{Q}{Eo}$

Capacitor plates are charged by a battery with ‘V’ volts, then

$V= E.d$

Apply $E$ value in $V$

$V=\frac{Q.d}{A.Eo}$

After charging  battery is disconnected,

Now consider

$E^{1}=\frac{q}{AEo} \\E^{1}=\frac{q}{KAEo}$

Known value is $V^{1}=E^{1} .d$

Apply $E^{1}$ value

$V^{1}=E^{1}.d\\V^{1}=\frac{Q.d}{KAE_{o} } \\V^{1}=(\frac{Q.d}{AE_{o} } )(\frac{1}{K})$ { a dielectric slab with dielectric constant ‘K’  is inserted between its plates}

We got $V=\frac{Q.d}{A.Eo}$

Apply in $V^{1}=(\frac{Q.d}{AE_{o} } )(\frac{1}{K})$, then potential across the plates of a  capacitor will become

$V^{1} = \frac{V}{K}$

*Attaching picture for clear understanding