The distance between the plates of an isolated charged parallel plate condenser is \(4 \text{ mm}\) and potential difference is \(60 \text{ volts}\). If the distance between the plates is increased to \(12 \text{ mm}\), then
The potential difference of the condenser will become \(180 \text{ volts}\)
The P.D. will become \(20 \text{ volts}\)
The P.D. will remain unchanged
The charge on condenser will reduce to one third
Solution:
For an isolated capacitor, the charge \(Q\) remains constant. \(Q = C_1 V_1 = C_2 V_2\). Since \(C = \frac{\epsilon_0 A}{d}\), we have \( \frac{\epsilon_0 A}{d_1} V_1 = \frac{\epsilon_0 A}{d_2} V_2 \implies \frac{V_1}{d_1} = \frac{V_2}{d_2} \). Thus, \(V_2 = V_1 \frac{d_2}{d_1} = 60 \text{ V} \times \frac{12 \text{ mm}}{4 \text{ mm}} = 180 \text{ V}\).
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