Positive charge Q is uniformly distributed around a semicircle of radius a. Find the magnitude of the electric field at the center of curvature P.
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Question

anonymous · Answer

use integration

anonymous · Answer

I have so far, but I'm not sure I'm going the right way.
$$dE_x=kQ/(\pi a^3)sin(\theta)d\theta$$ from 0 to pi?

anonymous · Answer

Ex=[(kQ/R*pi)/R]integral from 0to pi of cos(t)dt=[(kQ/R*pi)/R][sin t]0topi=0   ans....  and
Ey=-[(kQ/R*pi)/R]integral from 0to pi of sin(t)dt=[(kQ/R*pi)/R[cos t]0topi=-2[(kQ/R*pi)/R]  ans

anonymous · Answer

or you can use lambda L=Q/R*pi
Ex=[kL/R]integral from 0to pi of cos(t)dt=[(kL/R][sin t]0topi=0 ans.... and 
Ey=-[kL/R]integral from 0to pi of sin(t)dt=[kL/R[cos t]0topi=-2[kL/R] ans

anonymous · Answer

note that your a=radius R,   w/c i used there

anonymous · Answer

?? you don't need to do all of that though right, because of symmetry?