Two identical condensers M and N are connecte | vedclass

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(A) Let the initial capacitance of each condenser be $C = \frac{A \varepsilon_0}{d}$.For condenser $M$,the space is filled with a dielectric of consta

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$1 : 4$

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(A) Let the initial capacitance of each condenser be $C = \frac{A \varepsilon_0}{d}$.For condenser $M$,the space is filled with a dielectric of constant $K = 8$. The new capacitance is $C_M = K C = 8C$.For condenser $N$,a copper plate of thickness $t = d/2$ is introduced. The capacitance of a capacitor with a metal plate of thickness $t$ is $C_N = \frac{A \varepsilon_0}{d - t}$.Substituting $t = d/2$,we get $C_N = \frac{A \varepsilon_0}{d - d/2} = \frac{A \varepsilon_0}{d/2} = 2 \left( \frac{A \varepsilon_0}{d} \right) = 2C$.Since the capacitors are connected in series,the charge $Q$ on both is the same.The potential difference $V$ is given by $V = Q/C$,so $V \propto 1/C$.Therefore,the ratio of potential differences is $V_M : V_N = \frac{1}{C_M} : \frac{1}{C_N} = \frac{1}{8C} : \frac{1}{2C} = \frac{1}{8} : \frac{1}{2} = 2 : 8 = 1 : 4$.

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