question_answer

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: [JEE Main 10-Jan-2019 Evening]

A) 508 pJ

B) 692 pJ

C) 560 pJ

D) 600 pJ

Correct Answer: A

Solution :

[a]

Internal energy = u

\[\begin{align}   & Q=CV\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{u}_{i}}=\frac{1}{2}\times 12\times {{10}^{-12}}\times 100 \\  & =12\times {{10}^{-12}}\,\times 10\,\,\,\,\,\,=600\,\times \,10-12\,\, \\  & =12\times {{10}^{-11}}\,J\,\,\,\,\,\,\,\,\,\,\,\,\,=6\times {{10}^{-10}}\,J \\  &  \\ \end{align}\]

After insertion

\[C=KC=6.5\times 12\times {{10}^{-12}}\]

Final energy \[{{u}_{f}}=\frac{{{Q}^{2}}}{2C'}\]

\[\begin{align}   & =\,\,\frac{12\times 12\times {{10}^{-11}}\,\times \,{{10}^{-11}}}{2\,\times \,6.5\,\times 12\times {{10}^{-12}}} \\  & \,\,\,\,\,\,\,\,\,\, \\ \end{align}\]

So energy dissipated = \[=\,\,{{u}_{i}}\,=\,{{u}_{f}}\]

\[\Rightarrow \text{ }508\text{ }pJ\]