Question

The force between the plates of a parallel plate capacitor of capacitance $C$ and distance of separation of the plates $d$, with a potential difference $V$ between the plates, is:

• A. $\frac{C{V}^2}{2d}$
• B. $\frac{C^2{V}^2}{2d^2}$
• C. $\frac{C^2{V}^2}{d^2}$
• D. $\frac{V^{2}d}{C}$

Answer 1

Answer: (A)

$\frac{C{V}^2}{2d}$

Total electric field between the plates of the capacitor:

$E_T=\frac{V}{d}$
Then, electric field due to only one plate:

$E_1=\frac{V}{2d}$
$\therefore$ force of one plate on another:
$F=E_1\times Q$$F=E_1\times CV$$=\frac{V}{2d}\times CV$$=\frac{C{V}^2}{2d}$