Question

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.
http://session.masteringphysics.com/problemAsset/1261340/1/YF-21-096.jpg

Answer 1

use integration

Answer 2

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?

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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