Force of attraction between the plates of a parallel plate capacitor is
A$\frac{{{q^2}}}{{2{\varepsilon _0}AK}}$
B$\frac{{{q^2}}}{{{\varepsilon _0}AK}}$
C$\frac{q}{{2{\varepsilon _0}A}}$
D$\frac{{{q^2}}}{{2{\varepsilon _0}{A^2}K}}$
Medium
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A$\frac{{{q^2}}}{{2{\varepsilon _0}AK}}$
a (a) Force on one plate due to another is
$F = qE =$ $q \times \frac{\sigma }{{2{\varepsilon _0}K}}$$ = q\,\left( {\frac{q}{{2AK{\varepsilon _0}}}} \right) = \frac{{{q^2}}}{{2AK{\varepsilon _0}}}$
(where $\frac{\sigma }{{2{\varepsilon _0}K}}$ is the electric field produced by one plate at the location of other).
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