A capacitor with plate separation d is charged to V volts. battery is disconnected and a dielectric

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2. Electric Potential and Capacitance

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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

A

$V$

B

$2 V$

C

$\frac{4 V}{3}$

D

$\frac{3 V}{4}$

Solution

(d)

$q=C V$

$C^{\prime}=\frac{A \varepsilon_0}{d-\frac{d}{2}\left(1-\frac{1}{2}\right)}=\frac{4 A \varepsilon_0}{3 d}=\frac{4 C}{3}$

$q=\frac{4 C V^{\prime}}{3}$

$C V=\frac{4 C V^{\prime}}{3}$

$V^{\prime}=\frac{3 V}{4}$

Standard 12

Physics

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