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(D) When a dielectric slab is introduced into a capacitor after disconnecting the battery,the charge $(Q)$ on the plates remains constant.Initial capa

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The $P.D.$ across the capacitor will be $45\,V$

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(D) When a dielectric slab is introduced into a capacitor after disconnecting the battery,the charge $(Q)$ on the plates remains constant.Initial capacitance $C_0 = \frac{\varepsilon_0 A}{d}$.Initial potential difference $V_0 = 120\,V$.When the dielectric slab of thickness $t = 6\,mm$ and dielectric constant $K = 6$ is inserted into the gap $d = 8\,mm$,the new capacitance $C'$ is given by $C' = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}$.Since the charge $Q$ is constant,$Q = C_0 V_0 = C' V'$,where $V'$ is the new potential difference.Therefore,$V' = V_0 	imes \frac{C_0}{C'} = V_0 	imes \frac{d - t + \frac{t}{K}}{d}$.Substituting the values: $V' = 120 	imes \frac{8 - 6 + \frac{6}{6}}{8} = 120 	imes \frac{2 + 1}{8} = 120 	imes \frac{3}{8} = 15 	imes 3 = 45\,V$.

The distance between the plates of a parallel plate capacitor is $8\,mm$ and the potential difference $(P.D.)$ is $120\,V$. If a $6\,mm$ thick slab of dielectric constant $K = 6$ is introduced between its plates,then:

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