Question

A cylindrical capacitor has two coaxial cylinders of length 20 cm and radii 2r and r respectively. Inner cylinder is given a charge of 10 $$\mu C$$ and the outer cylinder a charge of -10 $$\mu C$$. The potential difference between the two cylinders will be?
A. $${\displaystyle \frac{0.1\ln 2}{4\pi {\epsilon }_0}}$$mV
B. $${\displaystyle \frac{\ln 2}{4\pi {\epsilon }_0}}$$ mV
C. $${\displaystyle \frac{10\ln 2}{4\pi {\epsilon }_0}}$$mV
D. $${\displaystyle \frac{0.01\ln 2}{4\pi {\epsilon }_0}}$$mV

Answer 1

Hint: We are given with cylindrical capacitors; a cylindrical capacitor consists of a hollow or a solid cylindrical conductor surrounded by another concentric hollow spherical cylinder. Here the length is the same for both but the radius is different. The charges are equal in magnitude but opposite in polarity. We need to find the potential difference between the two.

Complete step by step answer:
Length, l = 20 cm
The inner radius is r and the outer radius is 2r. The inner cylinder is given a charge 10μC and outer cylinder a charge of -10μC
To find the potential difference, we use the concept, the capacitance of the given system is given by,
$$C={\displaystyle \frac{2\pi {\epsilon }_{0}l}{\ln [{\displaystyle \frac{r_2}{r_1}}]}}$$
The length is given as 20 cm that is 0.2 m
So, capacitance comes out to be $$C={\displaystyle \frac{0.4\pi {\epsilon }_0}{\ln r}}$$
Now we need to find the potential difference, the Inner cylinder is given a charge 10μC and outer cylinder a charge of -10μC, so the net charge is
Using Q=CV
$$V={\displaystyle \frac{Q}{C}}={\displaystyle \frac{Q}{{\displaystyle \frac{0.4\pi {\epsilon }_0}{\ln 2}}}}={\displaystyle \frac{Q\ln 2}{0.4\pi {\epsilon }_0}}={\displaystyle \frac{\ln 2}{4\pi {\epsilon }_0}}\times 10\times 10\times {10}^{-6}=[{\displaystyle \frac{\ln 2}{4\pi {\epsilon }_0}}\times {10}^{-4}]V={\displaystyle \frac{0.1\ln 2}{4\pi {\epsilon }_0}}$$

So, the correct answer is “Option A”.

Note:
The capacitance of a cylindrical capacitor can be found out by using Gauss law as a charge enclosed by the gaussian cylinder. The final answer is given in millivolt using the conversion factor, 1V=1000 mV.