Question
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
A.
$$\frac{{{C^2}{V^2}}}{{2d}}$$
B.
$$\frac{{C{V^2}}}{{2d}}$$
C.
$$\frac{{C{V^2}}}{d}$$
D.
$$\frac{{{C^2}{V^2}}}{{2{d^2}}}$$
Answer :
$$\frac{{C{V^2}}}{{2d}}$$
Solution :
Force between plates of parallel capacitor,
$$F = qE = q\left[ {\frac{\sigma }{{2{\varepsilon _0}}}} \right]$$
$$\because $$ Surface charge density $$\sigma = \frac{q}{A}$$
$$\therefore F = q\left[ {\frac{q}{{2A{\varepsilon _0}}}} \right] \Rightarrow F = \frac{{{q^2}}}{{2A{\varepsilon _0}}}$$
So, net charge across a capacitor, $$q = CV$$
$$F = \frac{{{C^2}{V^2}}}{{2A{\varepsilon _0}}}\,\,\,\,\,\left[ {C = \frac{{A{\varepsilon _0}}}{d}} \right]$$
$$ \Rightarrow F = \frac{{\left( {\frac{{A{\varepsilon _0}}}{d}} \right) \times C{V^2}}}{{2A{\varepsilon _0}}} = \frac{{C{V^2}}}{{2d}}$$