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.......\(pJ\)

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Answer & Solution

Correct Answer

(B) \(508\)

Step-by-step Solution

Detailed explanation

\(W = \frac{{{Q^2}}}{{2c}} - \frac{{{Q^2}}}{{2ck}}\)

\( = \frac{{{Q^2}}}{{2c}}\left[ {1 - \frac{1}{k}} \right]\)

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

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

\( = 507.69\,{\text{pJ}}\)

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