Question

An electric field of 8.30 x 10^5 V/m is desired between two parallel plates, each of area 31.5 cm^2 and separated by 2.45 mm. There's no dielectric. What charge must be on each plate?

Answer 1

Answer:

Charge on each plate = 2.31 x 10⁻⁸ C

Explanation:

We have the equations

           $E=\frac{V}{d}\texttt{ and }V=\frac{Q}{C}$

Combining both equations

           $E=\frac{\left (\frac{Q}{C}\right )}{d}=\frac{Q}{Cd}$

We also have the equation for capacitance

           $C=\frac{\epsilon A}{d}$

That is

          $E=\frac{Q}{\frac{\epsilon A}{d}\times d}=\frac{Q}{\epsilon A}\\\\Q=\epsilon AE$

Substituting

           $Q=8.85\times 10^{-12}\times 31.5\times 10^{-4}\times 8.30\times 10^5=2.31\times 10^{-8}C$

Charge on each plate = 2.31 x 10⁻⁸ C