Question

A capacitor has a charge of 4.6 μC. An E-field of 1.8 kV/mm is desired between the plates. There's no dielectric. What must be the area of each plate?

Answer 1

Answer:

$A = 0.2875 m^2$

Explanation:

As we know that

$Q = 4.6 \mu C$

E = 1.8 kV/mm

now we know that electric field between the plated of capacitor is given as

$E = \frac{\sigma}{\epsilon_0}$

now we will have

$1.8 \times 10^6 = \frac{\sigma}{\epsilon_0}$

$\sigma = (1.8 \times 10^6)(8.85 \times 10^{-12})$

$\sigma = 1.6 \times 10^{-5} C/m^2$

now we have

$\sigma = \frac{Q}{A}$

now we have area of the plates of capacitor

$A = \frac{Q}{\sigma}$

$A = \frac{4.6 \times 10^{-6}}{1.6 \times 10^{-5}}$

$A = 0.2875 m^2$