Posted by habtamu on .
apositive cgarge Q is uniformly distributed around asemicircile of radius R.find the electric field (magneitud and direction)at the center of curvature P.

physics 
drwls,
Charge per unit length along the semicircle is Q/(pi*R). Net field E will be perpendicular to the diameter that runs from one end to the other. Use Coulomb's Law and integrate along the semicircle.
E = [kQ/(pi*R)]*[R*d(theta)/R^2]*
Integral of cos theta from pi/2 to pi/2.
k is the Coulomb constant.
I get E = 2 k Q/(pi*R^2)