A parallel plate capacitor has 1 \(\mu\)F capacitance. One of its two plates is given \(+2 \mu C\) charge and the other plate, \(+4 \mu C\) charge. The potential difference developed across the capacitor is:
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- The charge on the inner plates of a capacitor is given by \( \frac{|Q_1 - Q_2|}{2} \).
- The potential difference across a capacitor is calculated using \(V = \frac{Q}{C}\).
- Always ensure proper sign conventions when dealing with capacitors in circuits.