A parallel plate capacitor has area of each plate as \(A\), the separation between the plates as \(d\) and it is charged to potential \(V\), and then disconnected from the battery. If a dielectric slab, completely filling the capacitor is introduced, how much work will be done in doing so
\(\frac{1}{2}\frac{V^2\varepsilon_0 A}{kd}\)
\(\frac{1}{2}\frac{V^2\varepsilon_0 A}{k^2 d}\)
\(\frac{1}{2}\frac{\varepsilon_0 A V^2}{d}\left(1-\frac{1}{k}\right)\)
\(\frac{1}{2}\frac{\varepsilon_0 A V^2}{d}\left(1-\frac{1}{k^2}\right)\)
Solution:
The initial energy of the isolated capacitor is \(U_i = \frac{1}{2} C V^2 = \frac{\varepsilon_0 A V^2}{2d}\). After the dielectric is introduced, capacitance becomes \(kC\) and energy becomes \(U_f = \frac{U_i}{k}\). The work done by the system is \(-\Delta U = U_i - U_f = \frac{1}{2}\frac{\varepsilon_0 A V^2}{d}\left(1 - \frac{1}{k}\right)\).
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