# Electric Potential of a Finite Rod

1. Mar 1, 2009

### eckz59

1. The problem statement, all variables and given/known data
The figure shows a thin rod of length L and charge Q. It lies directly along the x axis with its center at the origin. Find an expression for the electric potential a distance x away from the center of the rod on the axis of the rod. (Give your answer in terms of x, L, Q and appropriate constants.)

2. Relevant equations
$$V=Q/(4\pi\epsilon_{0}x)$$
$$\lambda=Q/L$$

3. The attempt at a solution
I know that I have to integrate each infinitesimal potential, but I am not sure what bounds to integrate over? Here is what I have done so far:

$$dQ=\lambda*dx$$
$$dV=dQ/(4\pi\epsilon_{0}x)=\lambda*dx/(4\pi\epsilon_{0}x)$$

Now, when both sides are integrated the result is:

$$V=\lambda/(4\pi\epsilon_{0})\int(1/x)dx$$

The integral becomes a natural log. I have tried with several combinations of lower boundaries (0, -L/2, +L/2) and upper boundaries (L/2, x) but none of my results are correct (I get instant feedback on whether my answer is correct).

Thanks!

Last edited: Mar 1, 2009
2. Mar 1, 2009

### LowlyPion

Welcome to PF.

Isn't the x of your integral the distance from the discrete element charges to the point X, so the range of the distances is x-L/2 and x-3L/2 for E field along the + x-axis?

3. Mar 1, 2009

### eckz59

$$\left(\frac{1}{4{\cdot}{\pi}{\cdot}{\epsilon}_{0}}\right){\cdot}\left(\frac{Q}{L}\right){\cdot}{\ln}\left(\frac{2{\cdot}x+L}{2{\cdot}x-L}\right)$$

This is the correct answer! I'm not sure if what you said is correct but it set off a light bulb in my mind that made me realise that the integration variable had to run from x-L/2 to x+L/2! Thanks for your insight, and your welcome to the PF.

4. Mar 1, 2009

### LowlyPion

Right. I was sloppy in the 3/2 because that distance is x+L/2 to the other end. Glad you caught my error and it didn't mislead but rather served to possibly inspire.