A) \[\frac{{{q}^{2}}}{2{{\varepsilon }_{0}}AK}\]
B) \[\frac{{{q}^{2}}}{{{\varepsilon }_{0}}AK}\]
C) \[\frac{q}{2{{\varepsilon }_{0}}A}\]
D) \[\frac{{{q}^{2}}}{2{{\varepsilon }_{0}}{{A}^{2}}K}\]
Correct Answer: A
Solution :
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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