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

Answer:

F= 5.5 x 10⁻⁸ N

Explanation:

Given that

A= 5 cm²

d= 2 cm

εr= 4

V= 50 V

We know that force between capacitor plate given as

$F=\dfrac{\varepsilon AE^2}{2}$

The electric field given as

$E=\dfrac{V}{d}$

$F=\dfrac{\varepsilon AV^2}{2d^2}$

Now by putting the values

$F=\dfrac{4\times \times 10^{-12}\times 5\times 10^{-4}\times 50^2}{2\times 0.02^2}\ N$

F= 5.5 x 10⁻⁸ N