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. The potential difference across its terminals becomes

Question

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. The potential difference across its terminals becomes (A) V (B) 2 V (C) (4 V/3) (D) (3 V/4)

Tardigrade Admin · Accepted Answer

q=C V C'=(A ε0/d-(d)2(1-(1)2) =(4 A ε0/3 d)=(4 C/3) q=(4 C V'/3) C V=(4 C V'/3) V'=(3 V/4)

Q. A capacitor with plate separation $d$ is charged to $V$ volts. The battery is disconnected and a dielectric slab of thickness $\frac{d}{2}$ and dielectric constant ' $2$ ' is inserted between the plates. The potential difference across its terminals becomes

$V$

$2 V$

$\frac{4 V}{3}$

$\frac{3 V}{4}$

$q = C V$
$C^{'} = \frac{A ε _{0}}{d - \frac{d}{2} ( 1 - \frac{1}{2} )}$
$= \frac{4 A ε _{0}}{3 d} = \frac{4 C}{3}$
$q = \frac{4 C V ^{'}}{3}$
$C V = \frac{4 C V ^{'}}{3}$
$V^{'} = \frac{3 V}{4}$

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

V

2V

34V​

43V​

q=CV C′=d−2d​(1−21​)Aε0​​ =3d4Aε0​​=34C​ q=34CV′​ CV=34CV′​ V′=43V​

Q. A capacitor with plate separation $d$ is charged to $V$ volts. The battery is disconnected and a dielectric slab of thickness $\frac{d}{2}$ and dielectric constant ' $2$ ' is inserted between the plates. The potential difference across its terminals becomes

$V$

$2 V$

$\frac{4 V}{3}$

$\frac{3 V}{4}$

$q = C V$
$C^{'} = \frac{A ε _{0}}{d - \frac{d}{2} ( 1 - \frac{1}{2} )}$
$= \frac{4 A ε _{0}}{3 d} = \frac{4 C}{3}$
$q = \frac{4 C V ^{'}}{3}$
$C V = \frac{4 C V ^{'}}{3}$
$V^{'} = \frac{3 V}{4}$