Question

The charge on a parallel plate capacitor is 200 μC and its capacitance is 4 μF. If the distance between the plates of the capacitor is 2 mm, then the electric force between the plates of the capacitor is

• A. 2.5 N
• B. 5 N
• C. 10 N
• D. 1.25 N

Hint:

- For a parallel plate capacitor, $F = \frac{Q^2}{2C} = \frac{1}{2} Q V$ (alternative formula using potential). - Make sure to convert μC to C and μF to F. - Use $\epsilon_0$ formula only if area and distance given; otherwise, the $F = Q^2/2C$ shortcut is very handy.

Answer 1

The Correct Option is B

1. Force between capacitor plates: $F = \frac{Q^2}{2 \epsilon_0 A}$ or using $F = \frac{1}{2} \frac{Q^2}{C}$ (since $C = \epsilon_0 A / d$).
2. Given: $Q = 200~\mu C = 2 \times 10^{-4}$ C, $C = 4~\mu F = 4 \times 10^{-6}$ F.
3. Electric force: \[ F = \frac{1}{2} \frac{Q^2}{C} = \frac{1}{2} \frac{(2 \times 10^{-4})^2}{4 \times 10^{-6}} = \frac{1}{2} \cdot \frac{4 \times 10^{-8}}{4 \times 10^{-6}} = \frac{1}{2} \cdot 10^{-2} = 5~\text{N} \]