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

`A = 0.2875 m^2`

Step-by-step 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 = (\sigma)/(\epsilon_0)`

now we will have

`1.8 * 10^6 = (\sigma)/(\epsilon_0)`

`\sigma = (1.8 * 10^6)(8.85 * 10^(-12))`

`\sigma = 1.6 * 10^(-5) C/m^2`

now we have

`\sigma = (Q)/(A)`

now we have area of the plates of capacitor

`A = (Q)/(\sigma)`

`A = (4.6 * 10^(-6))/(1.6 * 10^(-5))`

`A = 0.2875 m^2`