1. Nov 9, 2004

### stunner5000pt

There is a rod with uniform charge distribution lambda . Point P is located directly above the rod at a distance d from the centre of the rod. Find the electric potential at point P

Also in thee second part similar situation however one half of the rod has a negative charge distribution while the other is positivelambda.

For the first part

$$V = 2 \int \frac{1}{4 \pi \epsilon_{0}} \frac{\lambda dx}{r}$$ where r is the distance from the poin P to the infinitesimal point on the rod and x is the horizontal axsi upon whihc the rod sits

so then $$r = \sqrt{d^2 + x^2}$$
and then it follows that

$$V = 2 \int \frac{1}{4 \pi \epsilon_{0}} \frac{\lambda dx}{\sqrt{d^2 + x^2}}$$

For the second part

wouldnt i get a zero answer because instead of 2 times the expression you'd get one of those expresions and then the negative of it which when you add them (right?) they cancel out?