## electric field problems

Here an electric field problems I'm struggling with:
Positive charge Q is uniformly distributed around a semicircle of radius a. Find the electric field at the center of curvature P. Answer in terms of Q, k, and a.
When I worked this out on my own, I calculated the magnitude of the electric field to be (2Qk)/(a^2), but this is wrong. Could someone give me a hint on how to do this correctly?
Thanks!
 Mentor Blog Entries: 1 Find the field for a segment (dΘ) of the semi-circle and integrate. Luckily it's an easy integral to do (since the distance is the same for all points on the curve). Try and set it up.
 Use symmetry Consider an axis passing through the Centre and dividing the semicircle into two halves So charge density(lambda)=Q/(pi)r. Now consider dq charge symmetrical to axis u will see one component of Field produced cancels and other adds up in one direction. Can u show what u get from above hint Anyway it is going to be moved to HW

## electric field problems

Does mean dE=(kdQ)/(a^2)?
dQ=Pi(a)(lamda) da?
Is this remotely close?

Mentor
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 Originally posted by eku_girl83 Does mean dE=(kdQ)/(a^2)? dQ=Pi(a)(lamda) da? Is this remotely close?
$$dQ = \frac {Q}{\pi a} a d\Theta = \frac{Q d\Theta}{\pi}$$
$$dE = \frac {k}{a^2} \sin \Theta dQ = \frac{Qk}{\pi a^2} \sin \Theta d\Theta$$

I did neglect to mention, as himanshu points out, that you will need to take advantage of symmetry. Imagine the semicircle intersecting points (-a,0) (0,a) and (a,0). The only component of field you need to worry about is the y-component: by symmetry, the x-components (from opposite sides) cancel.
 I get (Qk)/(Pi a^2) directed down? Thank you so much for helping me!

Mentor
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 Originally posted by eku_girl83 I get (Qk)/(Pi a^2) directed down? Thank you so much for helping me!
Check your work. I think you are off by a factor of 2. (Unless I made a mistake.) Did you integrate over the full range of &Theta;?
 Yes u are off by a factor 2