Question

A cylindrical capacitor of inner radii $R$ and outer radii $2R$ is filled with two dielectrics of constant $K_1$ and $K_2$ respectively. Each dielectric occupies half the length of the cylinder. Find the capacitance of the system between the inner and outer cylinders.

• A. $\frac{2\pi {\epsilon }_{0}L}{\ln \left(2\right)}\left(\frac{K_1{K}_2}{K_1+K_2}\right)$
• B. $\frac{\pi {\epsilon }_{0}L}{\ln R}(K_1+K_2)$
• C. $\frac{\pi {\epsilon }_{0}L}{\ln \left(2\right)}(K_1+K_2)$
• D. $\frac{\pi {\epsilon }_{0}L}{2\ln \left(2\right)}(K_1+K_2)$

Answer 1

The correct option is C $\frac{\pi {\epsilon }_{0}L}{\ln \left(2\right)}(K_1+K_2)$

Let the capacitance of the regions filled with dielectrics $K_1$ and $K_2$ be $C_1$ and $C_2$ respectively.

From the figure, it is clear that $C_1$ and $C_2$ are in parallel combination.

As we know, the capacitance of a cylindrical capacitor is given by,$$C=\frac{2\pi {\epsilon }_{0}L}{\ln \left(\frac{R_{out}}{R_{in}}\right)}$$
Given, the total length of cylinder is $L$ and inner and outer radii as $R$ and $2R$ respectively.

$C_1=K_{1}C=\frac{2\pi {\epsilon }_0{K}_1\left(\frac{L}{2}\right)}{\ln \frac{2R}{R}}=\frac{\pi {\epsilon }_0{K}_{1}L}{\ln 2}$

Similarly,

$C_2=K_{2}C=\frac{2\pi {\epsilon }_0\left(\frac{L}{2}\right)}{\ln \left(\frac{2R}{R}\right)}=\frac{\pi {\epsilon }_0{k}_{2}L}{\ln 2}$

So, the total capacitance of the cylinder,

$C_{eff}=C_1+C_2=\frac{\pi {\epsilon }_0{K}_{1}L}{\ln 2}+\frac{\pi {\epsilon }_0{K}_{2}L}{\ln 2}$

$\therefore C_{eff}=\frac{\pi {\epsilon }_{0}L}{\ln \left(2\right)}(K_1+K_2)$

Hence, option $\left(c\right)$ is correct answer.

Why this Question ? Note: The capacitance of a cylindrical capacitor of length $L$ and inner and outer radii $R_{in}$ and $R_{out}$ respectively, is given by$$C=\frac{2\pi {\epsilon }_{0}L}{\ln \left(\frac{R_{out}}{R_{in}}\right)}$$