Question

What is the capacitance of a parallel plate capacitor with metal

plates, each of area 1.50 m2, separated by 2.01 mm? What charge is

stored in this capacitor if a voltage of 2.50 × 103 V is applied to it?

Answer 1

The capacitance of the parallel plates capacitor

$$C=\frac{\epsilon_0 A}{d}$$ $$C=\frac{8.85*10^{-12}*1.50}{2.01*10^{-3}}=6.60*10^{-9}\:\rm F$$

The charge stored in capacitor

$$q=CV\\ =6.60*10^{-9}*2.50*10^3=1.65*10^{-5}\:\rm C$$