# Homework Help: Fun problem in calculating the electric field in the center of a charged ring

1. Feb 24, 2006

### Lisa...

This needs to be done with Coulomb's Law:

A ring of radius R has a charge distribution on it that goes as $$\lambda (\theta)= \lambda_0 sin \theta$$ as shown in the figure below:
http://img506.imageshack.us/img506/8898/naamloos2iz.th.gif [Broken]

In what direction does the field at the center of the ring point & what is the magnitude of the field at the center of the ring?

My first reaction was: there is no field in the center, cause all the field lines cancel (take a point on the ring and another one facing it: the first one has lambda as charge distribution and the other one has - lambda (cause it's described by the -sin of the same angle theta as the first one)

So what do I actually need to do? Can anybody please give me a little start?

Last edited by a moderator: May 2, 2017
2. Feb 24, 2006

### topsquark

Taking $$\lambda_0$$ as positive the upper semi-circle has a positive charge and the lower semi-circle has a negative charge. ( $$sin(- \theta)=-sin(\theta)$$ ) This means that the E-field lines will (more or less) be pointing downward.

To attack this what you want to do is consider a small portion of the circle of arc length ds. (Recall that $$s=r \theta$$ so you can convert this to a $$d \theta$$.) ds has a charge of $$dq=\lambda_0 sin \theta d \theta$$. How do you find the electric field element dE from knowing dq and its location? How then would you find the total E? (Don't forget that E is a vector!)

-Dan

Last edited by a moderator: May 2, 2017
3. Feb 24, 2006

### Tom Mattson

Staff Emeritus
Your thinking is right, but you can actually show your answer mathematically by integrating Coulomb's law over the ring. You start with the differential version of Coulomb's law. I would actually use the electric potential so that I wouldn't have to worry about the different directions of the vectors between points on the ring and the center.

$$dV=\frac{kdq}{R}$$

Note that $dq=\lambda ds$ and go from there. Once you have the potential you can find the field from $\vec{E}=-\vec{\nabla}V$.

4. Feb 25, 2006

### Lisa...

Thanks! What I've done is the following (could you please tell me if I'm right ?)

So I did what you told me to: took a small portion of arc length $$ds_1$$ on the upper semicircle and because $$s_1=r \theta$$ this leads to $$ds_1 = r d \theta$$. Therefore: $$dq_1=\lambda ds_1= r \lambda_0 sin \theta d \theta$$. ($$\theta$$ is positive because it's the upper semicircle, therefore the sine is positive too and the charge as well).
I also took a small portion of arc length $$ds_2$$ facing $$ds_1$$ which has a charge of $$dq_2=\lambda ds_2= - r \lambda_0 sin \theta d \theta$$ ($$\theta$$ is negative because it's the lower semicircle, therefore the sine is also negative and the charge as well).

Next I split the problem of finding E in four parts:
first I'll find $$d E_x_1$$ for $$dq_1$$ on the upper semicircle and integrate over the whole semi circle in order to find the total $$E_x_1$$ caused by the upper part.(1)
Then I'll find $$d E_x_2$$ for $$dq_2$$ on the lower semicircle and integrate over the whole semi circle in order to find the total $$E_x_2$$ caused by the upper part. Then I'll add them in order to find the total $$E_x$$ in the center of the circle. (2)
I'll do the same for $$d E_y_1$$ (3) and $$d E_y_2$$ (4) in order to get to $$E_y_1$$ and $$E_y_2$$ providing $$E_y$$ when added.

(1) $$dE_x_1 = \frac{k dq_1}{r^2} cos \theta = \frac{k r \lambda_0 sin \theta d \theta}{r^2} cos \theta =\frac{k \lambda_0}{r} sin \theta cos \theta d \theta$$

$$E_x_1= \int_{-\pi/2}^{\pi/2} \frac{k \lambda_0}{r} sin \theta cos \theta d \theta = \frac{k \lambda_0}{r} \int_{-\pi/2}^{\pi/2} sin \theta cos \theta d \theta = \frac{k \lambda_0}{r} \left[ \frac{1}{2}(sin\theta)^2 \right]_{-\pi/2}^{\pi/2}= \frac{k \lambda_0}{r} (\frac{1}{2} -\frac{1}{2})=0$$

(2) $$dE_x_2 = \frac{k dq_2}{r^2} cos (-\theta) = \frac{-k r \lambda_0 sin \theta d \theta}{r^2} cos \theta =\frac{-k \lambda_0}{r} sin \theta cos \theta d \theta$$

$$E_x_2= \int_{-\pi/2}^{\pi/2} \frac{-k \lambda_0}{r} sin \theta cos \theta d \theta = \frac{-k \lambda_0}{r} \int_{-\pi/2}^{\pi/2} sin \theta cos \theta d \theta = \frac{-k \lambda_0}{r} \left[ \frac{1}{2}(sin\theta)^2 \right]_{-\pi/2}^{\pi/2}= \frac{-k \lambda_0}{r} (\frac{1}{2} -\frac{1}{2})=0$$

Therefore $$E_x_1 + E_x_2 = 0+0 = 0$$ so there is no x component of E in the center of the ring.

(3) $$dE_y_1 = \frac{k dq_1}{r^2} sin \theta = \frac{k r \lambda_0 sin \theta d \theta}{r^2} sin \theta =\frac{k \lambda_0}{r} sin^2 \theta d\theta$$

$$E_y_1= \int_{-\pi/2}^{\pi/2} \frac{k \lambda_0}{r} sin^2 \theta d \theta = \frac{k \lambda_0}{r} \int_{-\pi/2}^{\pi/2} sin^2 \theta d \theta = \frac{k \lambda_0}{r} \left[ \frac{1}{2} \theta - \frac{1}{4}(sin 2 \theta) \right]_{-\pi/2}^{\pi/2}= \frac{k \lambda_0}{r} (\frac{1}{4} +\frac{1}{4})= \frac{k \lambda_0 \pi}{2r}$$

(4) $$dE_y_2 = \frac{k dq_2}{r^2} sin -\theta = - \frac{k dq_2}{r^2} sin \theta = - \frac{- k r \lambda_0 sin \theta d \theta}{r^2} sin \theta = \frac{k r \lambda_0 sin \theta d \theta}{r^2} sin \theta = \frac{k \lambda_0}{r} sin^2 \theta d\theta$$

$$E_y_2= \int_{-\pi/2}^{\pi/2} \frac{k \lambda_0}{r} sin^2 \theta d \theta = \frac{k \lambda_0}{r} \int_{-\pi/2}^{\pi/2} sin^2 \theta d \theta = \frac{k \lambda_0}{r} \left[ \frac{1}{2} \theta - \frac{1}{4}(sin 2 \theta) \right]_{-\pi/2}^{\pi/2}= \frac{k \lambda_0}{r} (\frac{1}{4} +\frac{1}{4})= \frac{k \lambda_0 \pi}{2r}$$

Therefore $$E_y_1 + E_y_2 = \frac{k \lambda_0 \pi}{2r} + \frac{k \lambda_0 \pi}{2r} = \frac{k \lambda_0 \pi}{r}$$ which is the total $$E_y$$ component of E.

To shorten all of this: E has only a y-component (as expected) in the center of the ring... but is the formula I found the correct one?

Last edited: Feb 25, 2006
5. Feb 25, 2006

### topsquark

One teensy little technicality: In step 3), one line down, near the end, 2nd to the last term: You dropped a pi. However, you picked it back up in the last term, so it's okay.

It looks very good. To be honest I wouldn't have calculated the x components because we can make a symmetry argument to cancel them (I'm lazy! ), but it never hurts to work it out.

The only thing I would add is to remind you to make E a vector in your final answer. (Professors LOVE taking points off for that kind of thing!)

-Dan

6. Feb 25, 2006

### Lisa...

Whoops hehe I missed the pi while puzzeling in LaTeX ;), but it I did write it down properly on my paper :D. As for E being a vector, thanks for reminding me :D I'll be sure to add a j and little arrows here and there ;)