Question

What is the capacitance of a parallel plate capacitor with metal plates, each of area 1.00 m2, separated by 2.00 mm? What charge is stored in this capacitor if a voltage of 3.00 × 103 V is applied to it?

Answer 1

The capacitance of a parallel plate capacitor is $C = \frac{\varepsilon \varepsilon_0 S}{d} = \frac{1 \cdot 8.85 \cdot 10^{-12} \frac{F}{m} \cdot 1 m^2 }{2 \cdot 10^{-3} m} \approx 4.4 nF$.

The charge, stored in the capacitor is $q = C U = 4.4 nF \cdot 3 \cdot 10^3 V \approx 13.3 \cdot 10^{-6} C$