Question

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

Answer 1

Capacitance

$C=\frac{A\epsilon_0}{d}$

Put value

$C=\frac{1\times8.85\times10^{-12}}{2\times10^{-3}}=4.43n F$

Charge

$Q=CV$

$$Q=3\times10^3\times4.43\times10^{-9}=13.27\times10^{-6}C$$

$Q=13.27\mu C$