Question

Determine the force of attraction in a parallel-plate capacitor with A = 5 cm2 , d = 2 cm, and r = 4 if the voltage across it is 50 V.

Answer 1

F= 5.5 x 10⁻⁸ N

Step-by-step explanation:

Given that

A= 5 cm²

d= 2 cm

εr= 4

V= 50 V

We know that force between capacitor plate given as

`F=(\varepsilon AE^2)/(2)`

The electric field given as

`E=(V)/(d)`

`F=(\varepsilon AV^2)/(2d^2)`

Now by putting the values

`F=(4* * 10^(-12)* 5* 10^(-4)* 50^2)/(2* 0.02^2)\ N`

F= 5.5 x 10⁻⁸ N