# Homework Help: Electric field of a semi-circular rod

1. Mar 14, 2006

### twiztidmxcn

hey

just looking for some help on an electric field question involving a rod of charge. heres the problem:

You have a charge, Q, uniformly distributed along a thin, flexible rod with length L. The rod is then bent into a semi-circle.

a) Find expression for electric field at center of semicircle
b) Evaluate field strength if L = 10cm, Q = 30nC.

The rod starts out straight and is then bent into a half circle.

We are also given the hint that: A small piece of arc length delta-s spans a small angle delta-theta = delta-s / R , where R is the radius.

Now, I realize that this problem has lots of symmetry, mostly where the x and y components of the electric field are concerned. I know that all the y components will cancel due to this symmetry and all that we're left with are the x components.

I'm attempting to use the equation of a rod of charge to derive something for the circle, but I am a bit stuck. Basically, I'm stuck at E = kq/r, r hat.

I believe that I can just use that equation, find r in terms of x and y (using triangles, pythagorean theorem) and then integrate in terms of x. I'm not quite sure about this though...

Any sort of help in the right direction would be much appreciated.

thanks
twiztidmxcn

2. Mar 14, 2006

### eep

Well, I'm thinking r should be constant. If you extended the semi-circle to a full circle, would the point be at the center of the circle? If so, you're going to want an expression for the field in terms of theta.

3. Mar 14, 2006

### twiztidmxcn

The radius is constant yes, and yes, the point is at the center of the circle.

It's a semi-circle (half circle) with the electric field needing to be determined for a point at the 'center' of this circle.

So if in terms of theta, we have something like kQ/r^2, where r is something like, delta-s / delta-theta? How would I relate that theta to the equation I need?

4. Mar 14, 2006

### BerryBoy

http://www.soton.ac.uk/~sab304/elec.gif [Broken]
Can you go from here?

Greek-Symbol "Lamda" is my notation for charge per unit length.

Regards,
Sam

Last edited by a moderator: May 2, 2017
5. Mar 14, 2006

### twiztidmxcn

I took what you gave me and here's what I got:

Since my semi-circle has different orientation (up/down with the axis on y-axis), my y-component was zero.

Ex = integral (0->pi) of (kQ)/(LR) * sin(theta) d\theta.

I integrated this equation with the bounds and ended up with:

Ex = (2kQ)/(LR) = (2k*lambda)/(R)

As far as I can tell, I've done it right, but since I have no answer or anything of the sort to reference, I have no clue if I came to the right solution. I am a bit perplexed though as to where you got that negative sign in front of the k*lambda/R^2, it doesn't cancel out the way you have it and you're left with -(2kQ)/LR)

For the second part of the problem, we were given L = 10cm (0.10m) and Q = 30nC (3x10^-8 C). Knowing that L = pi*R, I solved for R from the given L, then plugged all my numbers into the equation and ended up with 1.70x10^5 N/C. I think this may be right, but I have a tiny niggling feeling that this number is too big. However, I realize this is a physics question from a textbook so it doesn't have to be particularly realistic but....any kind of confirmation on what I've come up with would be nice.

Last edited: Mar 14, 2006
6. Mar 14, 2006

### BerryBoy

The negative sign was because of my axis, it shows that when cos(0) the field is in the negative x direction.

It cancels out nicely because:
$$\int_0^{\pi} \cos \theta = 0$$

Ahh I see, I have made a mistake on the diagram, there isn't supposed to be an integral sign before $$\sin \theta$$

Cheers,
Sam

Last edited: Mar 14, 2006