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erhaus01 · Jan 23, 2012, 06:12 PM

http://www.smartphysics.com/images/content/EM/02/h2_arcA.png

Aurora2000 · Jan 24, 2012, 10:53 AM

First from symmetry considerations, you have that the electric field is directed along the line y=x. So you need only to compute the total Electric field in the origin: this is

From the figure the charge Q is uniformly distributed on the cyan quarter circle, thus the linear density is $2Q/\pi a$ . Thus the electric field is
$\int_0^{\pi/2} \frac{2Q/\pi}{a^2} d\theta=Q/a^2$
and the x component is $\frac{Q}{\sqrt{2}a^2}$ .