Question

Two 4.634 cm by 4.634 cm plates that form a parallel-plate capacitor are charged to +/- 0.947 nC.What is the electric field strength inside the capacitor if the spacing between the plates is 1.624 mm?

Answer 1

The electric field between two parallel plate is given by:

$E=4\pi k\frac{Q}{A}$

where k is the Coulomb's constant, Q is the charge on the plates and A is the area of each plate. In this case we have that:

$\begin{gathered} Q=0.947\times10^{-9} \\ A=(4.634\times10^{-2})^2 \end{gathered}$

Then we have:

$\begin{gathered} E=4\pi(9\times10^9)\frac{(0.947\times10^{-9})}{(4.634\times10^{-2})^2} \\ E=4.99\times10^4 \end{gathered}$

Therefore, the electric field is:

$E=4.99\times10^4\text{ }\frac{N}{C}$