Rankers Physics

Capacitors: Practice Problem & Solution

A parallel plate air capacitor has capacity '$C$', distance of separation between plates is '$d$' and potential difference '$V$' is applied between the plates force of attraction between the plates of the parallel plate air capacitor is: (2015 Pre)
$\frac{C^2V^2}{2d^2}$
$\frac{C^2V^2}{2d}$
$\frac{CV^2}{2d}$
$\frac{CV^2}{d}$

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Capacitors. Understanding the underlying formula is key to arriving at the correct answer below:

The force between the plates is $F = \frac{Q^2}{2A\varepsilon_0}$. Since $C = \frac{A\varepsilon_0}{d}$ and $Q = CV$, we can write $F = \frac{(CV)^2}{2(Cd)} = \frac{C^2V^2}{2Cd} = \frac{CV^2}{2d}$.

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