A parallel plate capacitor having capacitance 12, pF is charged by a battery to a potential differen

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2. Electric Potential and Capacitance

hard

A parallel plate capacitor having capacitance $12\, pF$ is charged by a battery to a potential difference of $10\, V$ between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant $6.5$ is slipped between the plates. The work done by the capacitor on the slab is.......$pJ$

A

$692$

B

$508$

C

$560$

D

$600$

(JEE MAIN-2019)

Solution

$W = \frac{{{Q^2}}}{{2c}} – \frac{{{Q^2}}}{{2ck}}$

$ = \frac{{{Q^2}}}{{2c}}\left[ {1 – \frac{1}{k}} \right]$

$ = \frac{1}{2} \times 12 \times 100\,pJ\left( {1 – \frac{1}{{6.5}}} \right)$

$ = \frac{{12 \times 100 \times 11}}{{2 \times 13}}\,{\text{pJ}}$

$ = 507.69\,{\text{pJ}}$

Standard 12

Physics

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