A parallel plate capacitor having capacitance \(12 \mathrm{pF}\) is charged by a battery to a potential…
A parallel plate capacitor having capacitance \(12 \mathrm{pF}\) is charged by a battery to a potential difference of \(10 \mathrm{~V}\) 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
\(508 \mathrm{pJ}\)
\(692 \mathrm{pJ}\)
\(560 \mathrm{pJ}\)
\(600 \mathrm{pJ}\)
Solution
Initial Energy of the capacitor, \(U_{i}=(1 / 2) \mathrm{CV}^{2}\)
\(=(1 / 2) \times 12 \mathrm{pF} \times 10 \times 10\)
\(=600 \mathrm{pJ}\)
After the slab, the energy of the slab, \(U_{f}=(1 / 2) Q^{2} / C^{\prime}\)
\(Q=C V=(12 \mathrm{pF})(10 \mathrm{~V})=120 \mathrm{p} \mathrm{C}\)
\(C^{\prime}=k C=6.5 \times 120 \times 10^{-12} \mathrm{~F}\)
Therefore, \(U_{f}=\left[(1 / 2)\left(120 \times 10^{-2}\right)^{2}\right] /\left[6.5 \times 120 \times 10^{-12}\right]\)
\(U_{f}=92 \mathrm{pJ}\)
\(W+U_{f}=U_{i}\)
\(\Rightarrow W=U_{i}-U_{f}\)
\(=600 \mathrm{pJ}-92 \mathrm{pJ}\)
\(=508 \mathrm{pJ}\)
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