Parallel Plate Capacitor: Practice Problem & Solution
If \(2 \mu C\) charge is given to the one plate of a parallel plate capacitor of capacitance \(1 \mu F\) (initial charge on capacitor \(= 0\)). Potential difference across the plates of capacitor is
Solution Explained:
To solve this problem, we apply the core principles of Parallel Plate Capacitor. Understanding the underlying formula is key to arriving at the correct answer below:
When a total charge \(Q_{\text{total}} = 2 \mu C\) is given to one plate, the effective charge stored on the capacitor plates is \(Q = \frac{Q_{\text{total}}}{2} = \frac{2 \mu C}{2} = 1 \mu C\). Given capacitance \(C = 1 \mu F\). Potential difference \(V = \frac{Q}{C} = \frac{1 \mu C}{1 \mu F} = 1 \text{ V}\).
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