Separation between the plates of a parallel plate capacitor is $d$ and the area of each plate is $A$. When a slab of material of dielectric constant $k$ and thickness $t(t < d)$ is introduced between the plates, its capacitance becomes
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(c) Potential difference between the plates $V$ = $V_{air}$ + $V_{medium}$
$ = \frac{\sigma }{{{\varepsilon _0}}} \times (d - t) + \frac{\sigma }{{K{\varepsilon _0}}} \times t$
$==>$ $V = \frac{\sigma }{{{\varepsilon _0}}}(d - t + \frac{t}{K})$
$ = \frac{Q}{{A{\varepsilon _0}}}(d - t + \frac{t}{K})$
Hence capacitance $C = \frac{Q}{V} = \frac{Q}{{\frac{Q}{{A{\varepsilon _0}}}(d - t + \frac{t}{K})}}$
$ = \frac{{{\varepsilon _0}A}}{{(d - t + \frac{t}{k})}} = \frac{{{\varepsilon _0}A}}{{d - t\,\left( {1 - \frac{1}{k}} \right)}}$
$ = \frac{\sigma }{{{\varepsilon _0}}} \times (d - t) + \frac{\sigma }{{K{\varepsilon _0}}} \times t$
$==>$ $V = \frac{\sigma }{{{\varepsilon _0}}}(d - t + \frac{t}{K})$
$ = \frac{Q}{{A{\varepsilon _0}}}(d - t + \frac{t}{K})$
Hence capacitance $C = \frac{Q}{V} = \frac{Q}{{\frac{Q}{{A{\varepsilon _0}}}(d - t + \frac{t}{K})}}$
$ = \frac{{{\varepsilon _0}A}}{{(d - t + \frac{t}{k})}} = \frac{{{\varepsilon _0}A}}{{d - t\,\left( {1 - \frac{1}{k}} \right)}}$
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