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
[tex]F=\dfrac{\varepsilon AE^2}{2}[/tex]
The electric field given as
[tex]E=\dfrac{V}{d}[/tex]
[tex]F=\dfrac{\varepsilon AV^2}{2d^2}[/tex]
Now by putting the values
[tex]F=\dfrac{4\times \times 10^{-12}\times 5\times 10^{-4}\times 50^2}{2\times 0.02^2}\ N[/tex]
F= 5.5 x 10⁻⁸ N
