Question

Two 4.137 cm by 4.137 cm plates that form a parallel-plate capacitor are charged to +/- 0.531 nC.What is the electric field strength inside the capacitor if the spacing between the plates is 2.198 mm?

Answer 1

The electric field of two parallel plates is given by:

$\begin{gathered} E\cdot d=\frac{Q}{d\cdot A\epsilon o} \\ so: \\ E=\frac{Q}{A\epsilon o} \end{gathered}$

Where:

$\begin{gathered} \epsilon o=8.8542\times10^{-12} \\ A=0.04137^2=1.71\times10^{-3}m^2 \\ Q=0.531\times10^{-9}C \end{gathered}$

Therefore:

$\begin{gathered} E=\frac{0.531\times10^{-9}}{(8.8542\times10^{-12})(1.71\times10^{-3})} \\ E\approx35071V/m \end{gathered}$