# Electric field of a uniformly distributed rod

1. Sep 25, 2007

### tongpu

1. The problem statement, all variables and given/known data
rod -a-------|---------a on x-axis, for x>-a has -Q charge, for X<a has +Q charge, find the electric field at point (x,0) x is positive

2. Relevant equations
lambda = Q/a linear charge density dQ = (lambda)(ds)
E=KQ/r^2

3. The attempt at a solution
i make dE=kdQ/(x-s)^2
integrate kdQ/(x-s)^2 between -a and a ( i can make it 0 to a and multiply by two, symmetry)
i get the result 2KQ/a[1/x^2 - 1/(x-a)^2] for E sub-x

now if i want to find x>>a i need o somehow change [1/x^2 - 1/(x-a)^2] to the 2nd taylor expansion term but how? And is my integration correct?

2. Sep 25, 2007

### Staff: Mentor

I suggest checking the assumption about symmetry.

3. Sep 25, 2007

### learningphysics

I don't understand how you multiplied by 2... you have the integral:

$$\int_{-a}^{0}\frac{k\sigma ds}{(x-s)^2} + \int_{0}^{a}\frac{-k\sigma ds}{(x-s)^2}$$

where sigma is Q/a