Find the force of the rod on the point charge

  • Thread starter dk321
  • Start date
  • #1
4
0

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


Homework Equations



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?


[tex](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)][/tex]

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

Answers and Replies

  • #2
collinsmark
Homework Helper
Gold Member
2,897
1,238
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 [itex] \LaTeX [/itex] btw.
 
Last edited:
  • #3
collinsmark
Homework Helper
Gold Member
2,897
1,238
By the way, as a [itex] \LaTeX [/itex] 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):

[tex] \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][/tex]
 

Related Threads on Find the force of the rod on the point charge

Replies
4
Views
1K
Replies
3
Views
5K
Replies
6
Views
3K
Replies
3
Views
3K
Replies
4
Views
2K
  • Last Post
Replies
4
Views
11K
Replies
9
Views
2K
Replies
3
Views
763
Replies
5
Views
1K
Top