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Separation between plates is $d$ and area of plates is $A$.

From gauss's theorem one can show electric field inside the capaciton plates $Q$,

$E=ϵ_{o}σ $, where $σ=AQ $ (charge / Area)

$⇒E=Aϵ_{o}Q $

Now, voltage across the plates is related by :

$V=Ed$ [In general , $dV=−E.dr$ but we take as $E$ is uniform]

$⇒V=Aϵ_{o}Qd ⇒VQ =dAϵ_{o} $

Now, capacitance is defined by $C=VQ $, thus we get :-

$C=dAϵ_{o} $

Solve any question of Electrostatic Potential and Capacitance with:-

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