How Is the Electric Field Calculated at the Center of a Semicircle of Charge?

In summary, the problem involves finding the electric field at the center of curvature P caused by a uniformly distributed positive charge Q around a semicircle of radius a. By dividing the semicircle into infinitesimal pieces and integrating, the electric field in the y-direction is found to be -2kQ/a^2pi, with the direction being negative.
  • #1
espen180
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Homework Statement



Positive charge Q is uniformly distributed around a semicircle of radius a (See fig). Find the electric field (magnitude and direction) at the center of curvature P.

http://img132.imageshack.us/img132/8278/electricfield1.png

Homework Equations



Electric field of point charge:


The Attempt at a Solution



Since I suspect the linear charge distribution will be troublesome to wirk with here, I will define the angular charge distribution , divide the semicircle into infinitisimal pieces, calculate the electric field for each one and integrate.

Since each piece is a distance away from point , the electric field caused by each piece is . , so .

By symmetry, the total electric field in the x-direction is zero, so I only have to find the field in the y-direction, which is given by .

Integrating from to , the result is
.

Since Q is positive, the field is in the negative y-direction.

 
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  • #2
What you've done looks perfect to me... any particular reason for asking about it?
 
  • #3
I've done this sort of problem a number of times with straight lines, and the book did not have an answer key for this paticular problem. I was just wondering if I had the correct answer.
 
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