Two large conducting plates of a parallel plate capacitor are given charges $Q$ and $3 Q$ respectively. If the electric field in the region between the plates is $E_{0}$ , then the force of interaction between the plates is

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The net electric field between the plates due to both the plates is

$E_{0} = \frac{σ}{ε_{0}} = \frac{Q}{A ε_{0}}$

So the electric field due to each plate is

$E = \frac{E_{0}}{2} = \frac{Q}{2 A ε_{0}}$

So the force on one plate due to the other one is

$F = Q E = 3 Q \times [\frac{Q}{2 A ε_{0}}] = \frac{3 Q^{2}}{2 A ε_{0}}$
$= \frac{3 Q E_{0}}{2}$

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