# Electric field problems

1. Jan 24, 2004

### 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!

2. Jan 24, 2004

### Staff: Mentor

Find the field for a segment (d&Theta;) 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.

3. Jan 24, 2004

### 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

Last edited: Jan 24, 2004
4. Jan 24, 2004

### eku_girl83

still not getting it :)

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

5. Jan 24, 2004

### Staff: Mentor

Re: still not getting it :)

$$dQ = \frac {Q}{\pi a} a d\Theta = \frac{Q d\Theta}{\pi}$$
$$dE = \frac {k}{a^2} \sin \Theta dQ = \frac{Qk}{\pi a^2} \sin \Theta d\Theta$$

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.

6. Jan 24, 2004

### eku_girl83

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

7. Jan 24, 2004