Work Done in Introducing Dielectric Slab – Rankers Physics

Capacitor With Dielectrics: Practice Problem & Solution

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 Explained:

To solve this problem, we apply the core principles of Capacitor With Dielectrics. Understanding the underlying formula is key to arriving at the correct answer below:

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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