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

by Lisa...
Tags: charged, electric, field, ring
 P: 189 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: 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?
P: 335
 Quote by 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: 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?
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
 Emeritus Sci Advisor PF Gold P: 5,532 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 $\vec{E}=-\vec{\nabla}V$.
P: 189
Fun problem in calculating the electric field in the center of a charged ring

 Quote by 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
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?
P: 335
 Quote by 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?
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
 P: 189 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 ;) Thanks for your help!!!

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