Question

In a parallel plate capacitor, if \(10^{12}\) electrons pass from one plate to another, a potential difference of 10 V is developed across the plates. The capacitance of the capacitor is

• A. \(0.16 \times 10^{-6}\ \text{F}\)
• B. \(1.6 \times 10^{-8}\ \text{F}\)
• C. \(1.6 \times 10^{-9}\ \text{F}\)
• D. \(0.8 \times 10^{-8}\ \text{F}\)

Hint:

Use \(C = \frac{Q}{V}\) where charge \(Q\) is found from \(n \cdot e\).

Answer 1

The Correct Option is B

Charge transferred: \[ Q = n \cdot e = 10^{12} \cdot 1.6 \times 10^{-19} = 1.6 \times 10^{-7}\ \text{C} \] Capacitance: \[ C = \frac{Q}{V} = \frac{1.6 \times 10^{-7}}{10} = 1.6 \times 10^{-8}\ \text{F} \]