A capacitor has a charge of 4.6 Î¼C. 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?

Question

asked Feb 27, 2020 215k views

1 Answer

Rosarito · Answer 1 · 2020-03-04T02:18:45+0000

Answer:

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