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=(Q)/(d\cdot A\epsilon o) \\ so: \\ E=(Q)/(A\epsilon o) \end{gathered}`

Where:

`\begin{gathered} \epsilon o=8.8542*10^(-12) \\ A=0.04137^2=1.71*10^(-3)m^2 \\ Q=0.531*10^(-9)C \end{gathered}`

Therefore:

`\begin{gathered} E=(0.531*10^(-9))/((8.8542*10^(-12))(1.71*10^(-3))) \\ E\approx35071V/m \end{gathered}`