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(D) The initial capacitance of the parallel plate capacitor is given by $C = \frac{A\varepsilon_0}{d} = 10\,\mu F$.When a dielectric slab of thickness

Vedclass · Accepted Answer

$13.33$

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(D) The initial capacitance of the parallel plate capacitor is given by $C = \frac{A\varepsilon_0}{d} = 10\,\mu F$.When a dielectric slab of thickness $t$ and dielectric constant $K$ is inserted between the plates,the new capacitance is given by $C' = \frac{A\varepsilon_0}{d - t + \frac{t}{K}}$.Here,the dielectric fills half the distance,so $t = \frac{d}{2}$ and $K = 2$.Substituting these values: $C' = \frac{A\varepsilon_0}{d - \frac{d}{2} + \frac{d/2}{2}} = \frac{A\varepsilon_0}{\frac{d}{2} + \frac{d}{4}} = \frac{A\varepsilon_0}{\frac{3d}{4}} = \frac{4}{3} 	imes \frac{A\varepsilon_0}{d}$.Since $\frac{A\varepsilon_0}{d} = 10\,\mu F$,we have $C' = \frac{4}{3} 	imes 10 = 13.33\,\mu F$.

The capacity of a parallel plate capacitor is $10\,\mu F$ without a dielectric. If a dielectric of constant $K = 2$ is used to fill half the distance between the plates,the new capacitance in $\mu F$ is:

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