# Force on a point charge due to a rod

1. Apr 25, 2012

### dk321

1. The problem statement, all variables and given/known data

A rod of length L and total charge Q is a distance D from a point charge q which lies along the
perpendicular bisector of the rod. Find the force of the rod on the point charge

2. Relevant equations

E = ∫ dE

dq = λdx

F = qE
3. The attempt at a solution

I'm trying to integrate this does the second integral of j go to zero?

$$(kQ/L) [i\hat D \int _0L dy/(D^2+y^2)^(3/2) -j\hat \int _0^L ydy/ (D^2+y^2)^(3/2)]$$

sorry for the equation I'm still learning to use latex

2. Apr 25, 2012

### collinsmark

The second integral should go to zero due to symmetry. But I don't think you're setting up the limits of integration correctly. The problem statement says the test charge q "lies along the perpendicular bisector of the rod."

In other words, set up the system in your mind such that the test charge is on one of the axis, a distance D from the origin, and the the rod lies on a different (perpendicular) axis such that the center of the rod is at the origin. Then integrate along the rod from -L/2 to L/2.
See the link in my signature for a primer in $\LaTeX$ btw.

Last edited: Apr 26, 2012
3. Apr 25, 2012

### collinsmark

By the way, as a $\LaTeX$ example, I think the equation you were trying to type in, with limits of integration corrected, is (Right lick on it to see the LaTeX source):

$$\vec F = \frac{kQq}{L} \left[\int_{-L/2}^{L/2} {\frac{D \ dy}{\left(D^2+y^2 \right)^{\frac{3}{2}}} \hat \imath} + \int_{-L/2}^{L/2} {\frac{y \ dy}{\left(D^2+y^2 \right)^{\frac{3}{2}}} \hat \jmath} \right]$$