Find E field for a ring of charge - Charge per length non-uniform

[venkman1080]

I have to find the E field at all points on the z-axis for a ring of charge with radius = R. $$\lambda(\phi) = \lambda_0 cos(\phi)$$ where $$0 \leq \phi < 2 \pi$$

I know how to do the problem when it is the charge per length is uniform but when I do the calculation for the non-uniform case I get $$E = \frac{-kR\lambda_0\pi}{(R^2 + z^2)^{3/2}} \hat{i}$$ The integral for the j_hat part goes to zero because sin(phi)cos(phi) from 0 to 2 Pi and the z_hat part also goes to zero because the integral of cos(phi) from 0 to 2 Pi is zero.

I think my calculations are right, but I'm not totally sure. I just find it strange that its only in the i_hat. Any help/suggestions is greatly appreciated.

I used the formulas
$$E = \int_{charge} \frac{kdq}{r_\delta ^2}\hat{r_\delta}$$

$$r_\delta = -R cos(\phi)\hat{i} - R sin(\phi)\hat{j} + x\hat{k}$$

$$\hat{r_\delta} = \frac{-R cos(\phi)\hat{i}}{\sqrt{R^2 + z^2}}<br /> - \frac{R sin(\phi)\hat{j}}{\sqrt{R^2 + z^2}} + \frac{z\hat{k}}{\sqrt{R^2 + z^2}}$$

$$dq = \lambda Rd\phi$$
Sorry if there is any typos with the latex formulas. I kept doing the preview post and it wouldn't change what I had entered the first time.

 

[gabbagabbahey]

The integral for the j_hat part goes to zero because sin(phi)cos(phi) from 0 to 2 Pi and the z_hat part also goes to zero because the integral of cos(phi) from 0 to 2 Pi is zero.

Sounds like you're multiplying every component by an extra factor of $\cos\phi$ before integrating...why are you doing that?

 

[venkman1080]

I thought because $$\lambda(\phi) = \lambda_0 cos(\phi)$$ so when you do the integral the cos doesn't pull out.

 

[gabbagabbahey]

I thought because $$\lambda(\phi) = \lambda_0 cos(\phi)$$ so when you do the integral the cos doesn't pull out.

Yes, I overlooked that. Your answer looks good to me.

 

[venkman1080]

I'm just having trouble visualizing why the answer is only in the i_hat. Can you help clarify this for me?

 

[vela]

Think symmetry. For every spot on the ring, there's a spot on the opposite side with the same charge in magnitude but with opposite sign.