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Initially, when there is a vacuum between the two plates, the capacitance of the capacitor is

$C_{0}=dε_{0}A $

where A is the area of parallel plates

Suppose that the capacitor is connected to a battery, an electric field $E_{0}$ is produced

if we insert the dielectric slab of thickness t=d/2 the electric field is reduced to E

Now, the gap between plates is divided in two parts, for distance t=$2d $there is electric field E and for the remaining distance

(d-t) the electric field is $E_{0}$

If V be the potential difference between the plates of the capacitor, then

$V=E_{t}+E_{0}(d−t)V=2Ed +2E_{0}d =2d (E+E_{0})(t=2d )$

$V=2d (KE_{0} +E_{0})=2KdE_{0} (K+1)(asEE_{0} =K)$

$nowE_{0}=ε_{0}σ =ε_{0}Aq ⇒V=2Kd ε_{0}Aq (K+1)$

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