Combination of Capacitors: Practice Problem & Solution
The capacitance of a a parallel plate capacitor is (C) when the region between the plate has air. This region is now filled with a dielectric slab of dielectric constant (k). The capacitor is connected to a cell of emf (E), and the slab is taken out
Solution Explained:
To solve this problem, we apply the core principles of Combination of Capacitors. Understanding the underlying formula is key to arriving at the correct answer below:
Concept: Work done by external agent in removing dielectric from capacitor connected to battery. Formula: (W_{\text{ext}} = \Delta U - W_{\text{cell}}\), where (W_{\text{cell}} = E\Delta Q\). Solution: Initial stored energy (U_1 = \frac{1}{2} kCE^2\) and charge (Q_1 = kCE\). Final stored energy (U_2 = \frac{1}{2} CE^2\) and charge (Q_2 = CE\). Change in stored energy (Delta U = U_2 - U_1 = -\frac{1}{2} CE^2(k-1)\). Charge returned to cell (Delta Q = Q_1 - Q_2 = CE(k-1)\). Work done by cell (W_{\text{cell}} = E (Q_2 - Q_1) = -E^2C(k-1)\). Work done by external agent (W_{\text{ext}} = \Delta U - W_{\text{cell}} = -\frac{1}{2} CE^2(k-1) - (-E^2C(k-1)) = \frac{1}{2} E^2C(k-1)\).
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