Force of attraction between the plates of a parallel plate capacitor is

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Q.

Force of attraction between the plates of a parallel plate capacitor is

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a

$\frac{q^{2}}{2 ε_{0} A}$

b

$\frac{q^{2}}{ε_{0} A}$

c

$\frac{q}{2 ε_{0} A}$

d

$\frac{q^{2}}{2 ε_{0} A^{2}}$

answer is A.

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Detailed Solution

$F o r c e b e t w e e n t h e p l a t e s Q E = q \frac{σ}{2 ε_{0}} = \frac{q q}{A} \frac{1}{2 ε_{0}}$

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