Question

What area must the plates of a capacitor be if they have a charge of 5.7uC and an electric field of 3.1 kV/mm between them? O 0.151 m O 0.147m2 0.169 ? O 0.208 m e

Answer 1

Answer:

Area of the plates of a capacitor, A = 0.208 m²

Explanation:

It is given that,

Charge on the parallel plate capacitor, $q = 5.7\ \mu C=5.7\times 10^{-6}\ C$

Electric field, E = 3.1 kV/mm = 3100000 V/m

The electric field of a parallel plates capacitor is given by :

$E=\dfrac{q}{A\epsilon_o}$

$A=\dfrac{q}{E\epsilon_o}$

$A=\dfrac{5.7\times 10^{-6}\ C}{3100000\ V/m\times 8.85\times 10^{-12}\ F/m}$

A = 0.208 m²

So, the area of the plates of a capacitor is 0.208 m². Hence, this is the required solution.