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  4. 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

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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

Electrostatic Potential and Capacitance

Solution:

$q=C V$
$C'=\frac{A \varepsilon_{0}}{d-\frac{d}{2}\left(1-\frac{1}{2}\right)}$
$=\frac{4 A \varepsilon_{0}}{3 d}=\frac{4 C}{3}$
$q=\frac{4 C V'}{3}$
$C V=\frac{4 C V'}{3}$
$V'=\frac{3 V}{4}$