A parallel plate capacitor with plate separat | vedclass

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(B) Let the initial capacitance be $C_0 = \frac{A \epsilon_0}{d}$,where $d = 5 \ mm$.Initial charge $Q_0 = C_0 V = \frac{A \epsilon_0 V}{d}$.When a di

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$2$

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(B) Let the initial capacitance be $C_0 = \frac{A \epsilon_0}{d}$,where $d = 5 \ mm$.Initial charge $Q_0 = C_0 V = \frac{A \epsilon_0 V}{d}$.When a dielectric sheet of thickness $t = 2 \ mm$ and dielectric constant $K$ is introduced,the new capacitance $C'$ is given by $C' = \frac{A \epsilon_0}{d - t + \frac{t}{K}}$.The new charge is $Q' = C' V = \frac{A \epsilon_0 V}{d - t + \frac{t}{K}}$.Given that the capacitor draws $25 \%$ more charge,$Q' = 1.25 Q_0$.Substituting the expressions: $\frac{A \epsilon_0 V}{d - t + \frac{t}{K}} = 1.25 \frac{A \epsilon_0 V}{d}$.This simplifies to $\frac{1}{d - t + \frac{t}{K}} = \frac{1.25}{d} = \frac{5}{4d}$.Thus,$4d = 5(d - t + \frac{t}{K})$.Given $d = 5 \ mm$ and $t = 2 \ mm$,we have $4(5) = 5(5 - 2 + \frac{2}{K})$.$20 = 5(3 + \frac{2}{K}) \Rightarrow 4 = 3 + \frac{2}{K}$.$1 = \frac{2}{K} \Rightarrow K = 2$.

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