How to Calculate the Electric Field at the Center of Curvature?

[eku_girl83]

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!

 

[Doc Al]

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.

 

[himanshu121]

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

 

[eku_girl83]

still not getting it :)

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

 

[Doc Al]

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.

 

[eku_girl83]

answer

I get (Qk)/(Pi a^2) directed down?
Thank you so much for helping me!

 

[Doc Al]

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 Θ?

 

[himanshu121]

Yes u are off by a factor 2