# electric field problems

by eku_girl83
Tags: electric, field
 P: 90 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 P: 40,263 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.
 P: 662 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
P: 90

## electric field problems

Does mean dE=(kdQ)/(a^2)?
dQ=Pi(a)(lamda) da?
Is this remotely close?
Mentor
P: 40,263
 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.
 P: 90 I get (Qk)/(Pi a^2) directed down? Thank you so much for helping me!
Mentor
P: 40,263
 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;?
 P: 662 Yes u are off by a factor 2

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