Potential Difference with Unequal Charges – Rankers Physics
Topic: Capacitors
Subtopic: Parallel Plate Capacitor

Potential Difference with Unequal Charges

A charge of \( + 2.0 \times 10^{-8}\text{ C} \) is placed on the positive plate and a charge of \( -1.0 \times 10^{-8}\text{ C} \) on the negative plate of a parallel-plate capacitor of capacitance \( 1.2 \times 10^{-3}\ \mu\text{F} \). Calculate the potential difference developed between the plates.
\( 25\text{ V} \)
\( 7.5\text{ V} \)
\( 12.5\text{ V} \)
\( 50\text{ V} \)

Solution:

The potential difference depends on the charge on the inner facing surfaces, which is given by \( q = \frac{q_1 - q_2}{2} = \frac{2.0 \times 10^{-8} - (-1.0 \times 10^{-8})}{2} = 1.5 \times 10^{-8}\text{ C} \). Using \( V = \frac{q}{C} \), we get \( V = \frac{1.5 \times 10^{-8}}{1.2 \times 10^{-9}} = 12.5\text{ V} \).

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