Question

A hollow cylindrical shell of length L and radius R has charge Q uniformly distributed along its length. What is the electric potential at the center of the cylinder ?

Short answer

The electric potential is $V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{L}\ln \left(\frac{\sqrt{L^2+4R^2}+L}{\sqrt{L^2+4R^2}-L}\right)$

Step-by-step solution

Step 1. Given information and concept used

A hollow cylindrical shell with a radius of R and a length of L has a uniform charge distribution along its length, resulting in a total charge of Q.

The electric potential at a distance r from a point charge q is calculated as follows:

$V\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}$

Potential along the central axis of a ring of radius R having charge q, at a distance $X$ from the center,

$$\begin{gathered} V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{\sqrt{R^2+x^2}} \\ \int \frac{dx}{\sqrt{R^2+x^2}}=\ln \left(x+\sqrt{x^2+R^2}\right) \end{gathered}$$

Step 2. Calculate the electric potential in the middle of a charged cylindrical shell. 

Consider a ring with a width of $dx$and a distance of x from the cylinder's center. The charge per square meter of cylinder area,

$\sigma =\frac{Q}{2\pi RL}$

The ring element's charge,

$\begin{array}{rcl}dq& =& \sigma \times 2\pi R\times dx\\ & =& \frac{Q}{L}dx\end{array}$

Therefore, the potential at the center of the cylinder due to ring element

$\begin{array}{rcl}dV& =& \frac{1}{4\pi {\epsilon }_0}\frac{dq}{\sqrt{R^2+x^2}}\\ & =& \frac{1}{4\pi {\epsilon }_0}\frac{Q}{L}\frac{dx}{\sqrt{R^2+x^2}}\end{array}$

As a result of the whole charge distribution, the potential at the cylinder's core,

$\begin{array}{rcl}V& =& \int dV\\ & =& \frac{1}{4\pi {\epsilon }_0}\frac{Q}{L}{\int }_{-4/2}^{+1/2}\frac{q}{\sqrt{R^2+x^2}}\\ & =& \frac{1}{4\pi {\epsilon }_0}\frac{Q}{L}\ln \left(\frac{\sqrt{L^2+4R^2}+L}{\sqrt{L^2+4R^2}-L}\right)\end{array}$

The potential at the center of the cylinder is found to be

$V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{L}\ln \left(\frac{\sqrt{L^2+4R^2}+L}{\sqrt{L^2+4R^2}-L}\right)$