# Help with Electric Potential

1. Mar 24, 2005

There is a plastic rod with a length L and a uniform positive charge lying on the x axis. With V = 0 at infinity find the electric potential at point P on the axis at distance d from one end of the rod. I got an answer but I have no clue if it is right or not. This is what I know:

$$dv = \frac{k dq}{x} \ \ \ \ \ \lambda = \frac{q}{x}$$

$$dv = \frac{k\lambda dx}{x} \ \ \ \ \ dq = \lambda dx$$

$$\int dv = k\lambda \int \frac{dx}{x}$$ Upper = d + L Lower = d

$$V = k\lambda(ln(d+L) - ln(d))$$

Is there anything I am missing?

The next problem is similar but $$\lambda = cx$$ where c is some positive constant. If I just substitute in cx then the x cancles leaving dx which gives x afer the integral. The answer contains a ln so I am missing something. Any help is appreciated.

2. Mar 24, 2005

### Gamma

Seems right to me. For part ii, V = KCL. Answer does not have an ln.