- #1
venkman1080
- 3
- 0
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
<br /> E = \int_{charge} \frac{kdq}{r_\delta ^2}\hat{r_\delta}<br />
<br /> r_\delta = -R cos(\phi)\hat{i} - R sin(\phi)\hat{j} + x\hat{k}<br />
<br /> \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}}<br />
<br /> dq = \lambda Rd\phi<br />
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.
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
<br /> E = \int_{charge} \frac{kdq}{r_\delta ^2}\hat{r_\delta}<br />
<br /> r_\delta = -R cos(\phi)\hat{i} - R sin(\phi)\hat{j} + x\hat{k}<br />
<br /> \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}}<br />
<br /> dq = \lambda Rd\phi<br />
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.