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(D) The capacitance of a cylindrical capacitor is given by $C = \frac{2\pi \varepsilon_0 L}{\ln(b/a)}$,where $L$ is the length,$b$ is the outer radius

Vedclass · Accepted Answer

$1.3$

Vedclass · Answer

(D) The capacitance of a cylindrical capacitor is given by $C = \frac{2\pi \varepsilon_0 L}{\ln(b/a)}$,where $L$ is the length,$b$ is the outer radius,and $a$ is the inner radius.Since the capacitance $C$ remains constant,we have $L \propto \ln(b/a)$.Let the initial radii be $a_1$ and $b_1$,where $b_1/a_1 = 10$.For the second case,the new inner radius $a_2 = a_1/2$. The outer radius $b_2$ remains the same as $b_1$ (i.e.,$b_2 = b_1 = 10a_1$).Thus,the new ratio is $b_2/a_2 = (10a_1) / (a_1/2) = 20$.Setting the capacitances equal: $L_1 \ln(b_1/a_1) = L_2 \ln(b_2/a_2)$.Therefore,the ratio of the lengths is $L_2/L_1 = \frac{\ln(b_1/a_1)}{\ln(b_2/a_2)} = \frac{\ln(10)}{\ln(20)}$.Using $\ln(10) \approx 2.302$ and $\ln(20) = \ln(10) + \ln(2) \approx 2.302 + 0.693 = 2.995$.$L_2/L_1 = 2.302 / 2.995 \approx 0.768$. Wait,re-evaluating the ratio: $L_2/L_1 = \frac{\ln(20)}{\ln(10)} = \frac{2.995}{2.302} \approx 1.301$.Thus,the length must be increased by a factor of approximately $1.3$.

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