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

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

Step-by-step explanation:

It is given that,

Charge on the parallel plate capacitor,
`q = 5.7\ \mu C=5.7* 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=(q)/(A\epsilon_o)`

`A=(q)/(E\epsilon_o)`

`A=(5.7* 10^(-6)\ C)/(3100000\ V/m* 8.85* 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.