Posted by **peter** on Monday, September 13, 2010 at 9:49pm.

A thin nonconducting rod is bent to form a semi-circle of radius R. A charge Q is uniformly distributed along the upper half and a charge -Q is uniformly distributed along the lower half, as shown int he sketch at the right.find E vector at P, the center of the semi-circle. the semi circle looks like that, plus at the top and minuses at the bottom. P is the center and the radius arrow point to the minuses

- physics 2 -
**bobpursley**, Monday, September 13, 2010 at 10:23pm
So I assume you have calculus based physics.

integrate the charge thru and angle. let theta begin at the bottom rotate 90 deg for all the neg charge, then from 90 to 180 for the positive charge.

E=INT kdq/r^2 where dq=Q/(pi*r/2)dTheta

integrate from 0 to PI/2, that is E for the lower charge, and it points to the center point of the quadrant (45 deg, by symettry).

Do the same for the + charge, it points away from the center of the upper quadrant.

Now add the two E components. Neat, it point downward, and has value of E*sqrt2

work that out.

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