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.......$pJ$
JEE MAIN 2019, Diffcult
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$W = \frac{{{Q^2}}}{{2c}} - \frac{{{Q^2}}}{{2ck}}$
$ = \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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