# Find the force of the rod on the point charge

## Homework Statement

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

E = ∫ dE

dq = λdx

F = qE

## 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

Related Introductory Physics Homework Help News on Phys.org
collinsmark
Homework Helper
Gold Member
I'm trying to integrate this does the second integral of j go to zero?
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.
sorry for the equation I'm still learning to use latex
See the link in my signature for a primer in $\LaTeX$ btw.

Last edited:
collinsmark
Homework Helper
Gold Member
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):

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