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Physics-

General

Easy

Question

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

$\frac{q^{2}}{2 ε_{0} A K}$
$\frac{q^{2}}{ε_{0} A K}$
$\frac{q}{2 ε_{0} A}$
$\frac{q^{2}}{2 ε_{0} A^{2} K}$

The correct answer is: $fraction numerator q to the power of 2 end exponent over denominator 2 epsilon subscript 0 end subscript A K end fraction$

Force on one plate due to another is
F = qE = $q cross times fraction numerator sigma over denominator 2 epsilon subscript 0 end subscript K end fraction equals q open parentheses fraction numerator q over denominator 2 A K epsilon subscript 0 end subscript end fraction close parentheses equals fraction numerator q to the power of 2 end exponent over denominator 2 A K epsilon subscript 0 end subscript end fraction$
(where $fraction numerator sigma over denominator 2 epsilon subscript 0 end subscript K end fraction$ is the electric field produced by one plate at the location of other).

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