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

by Lisa...
Tags: charged, electric, field, ring
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Lisa...
#1
Feb24-06, 12:43 PM
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 [tex] \lambda (\theta)= \lambda_0 sin \theta [/tex] 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?
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topsquark
#2
Feb24-06, 03:03 PM
P: 335
Quote 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 [tex] \lambda (\theta)= \lambda_0 sin \theta [/tex] 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 [tex]\lambda_0[/tex] as positive the upper semi-circle has a positive charge and the lower semi-circle has a negative charge. ( [tex]sin(- \theta)=-sin(\theta)[/tex] ) 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 [tex]s=r \theta[/tex] so you can convert this to a [tex]d \theta[/tex].) ds has a charge of [tex]dq=\lambda_0 sin \theta d \theta[/tex]. 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
Tom Mattson
#3
Feb24-06, 03:18 PM
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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.

[tex]dV=\frac{kdq}{R}[/tex]

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

Lisa...
#4
Feb25-06, 06:20 AM
P: 189
Fun problem in calculating the electric field in the center of a charged ring

Quote Quote by topsquark
Taking [tex]\lambda_0[/tex] as positive the upper semi-circle has a positive charge and the lower semi-circle has a negative charge. ( [tex]sin(- \theta)=-sin(\theta)[/tex] ) 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 [tex]s=r \theta[/tex] so you can convert this to a [tex]d \theta[/tex].) ds has a charge of [tex]dq=\lambda_0 sin \theta d \theta[/tex]. 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 [tex]ds_1[/tex] on the upper semicircle and because [tex]s_1=r \theta[/tex] this leads to [tex]ds_1 = r d \theta[/tex]. Therefore: [tex]dq_1=\lambda ds_1= r \lambda_0 sin \theta d \theta[/tex]. ([tex]\theta[/tex] 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 [tex]ds_2[/tex] facing [tex]ds_1[/tex] which has a charge of [tex]dq_2=\lambda ds_2= - r \lambda_0 sin \theta d \theta[/tex] ([tex]\theta[/tex] 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 [tex]d E_x_1[/tex] for [tex]dq_1[/tex] on the upper semicircle and integrate over the whole semi circle in order to find the total [tex]E_x_1[/tex] caused by the upper part.(1)
Then I'll find [tex]d E_x_2[/tex] for [tex]dq_2[/tex] on the lower semicircle and integrate over the whole semi circle in order to find the total [tex]E_x_2[/tex] caused by the upper part. Then I'll add them in order to find the total [tex]E_x[/tex] in the center of the circle. (2)
I'll do the same for [tex]d E_y_1[/tex] (3) and [tex]d E_y_2[/tex] (4) in order to get to [tex]E_y_1[/tex] and [tex]E_y_2[/tex] providing [tex]E_y[/tex] when added.

(1) [tex]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 [/tex]

[tex] 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 [/tex]

(2) [tex]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 [/tex]

[tex] 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 [/tex]

Therefore [tex]E_x_1 + E_x_2 = 0+0 = 0[/tex] so there is no x component of E in the center of the ring.

(3) [tex]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 [/tex]

[tex] 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} [/tex]

(4) [tex]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 [/tex]

[tex] 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} [/tex]

Therefore [tex]E_y_1 + E_y_2 = \frac{k \lambda_0 \pi}{2r} + \frac{k \lambda_0 \pi}{2r} = \frac{k \lambda_0 \pi}{r} [/tex] which is the total [tex]E_y[/tex] 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?
topsquark
#5
Feb25-06, 07:09 AM
P: 335
Quote 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 [tex]ds_1[/tex] on the upper semicircle and because [tex]s_1=r \theta[/tex] this leads to [tex]ds_1 = r d \theta[/tex]. Therefore: [tex]dq_1=\lambda ds_1= r \lambda_0 sin \theta d \theta[/tex]. ([tex]\theta[/tex] 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 [tex]ds_2[/tex] facing [tex]ds_1[/tex] which has a charge of [tex]dq_2=\lambda ds_2= - r \lambda_0 sin \theta d \theta[/tex] ([tex]\theta[/tex] 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 [tex]d E_x_1[/tex] for [tex]dq_1[/tex] on the upper semicircle and integrate over the whole semi circle in order to find the total [tex]E_x_1[/tex] caused by the upper part.(1)
Then I'll find [tex]d E_x_2[/tex] for [tex]dq_2[/tex] on the lower semicircle and integrate over the whole semi circle in order to find the total [tex]E_x_2[/tex] caused by the upper part. Then I'll add them in order to find the total [tex]E_x[/tex] in the center of the circle. (2)
I'll do the same for [tex]d E_y_1[/tex] (3) and [tex]d E_y_2[/tex] (4) in order to get to [tex]E_y_1[/tex] and [tex]E_y_2[/tex] providing [tex]E_y[/tex] when added.

(1) [tex]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 [/tex]

[tex] 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 [/tex]

(2) [tex]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 [/tex]

[tex] 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 [/tex]

Therefore [tex]E_x_1 + E_x_2 = 0+0 = 0[/tex] so there is no x component of E in the center of the ring.

(3) [tex]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 [/tex]

[tex] 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} [/tex]

(4) [tex]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 [/tex]

[tex] 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} [/tex]

Therefore [tex]E_y_1 + E_y_2 = \frac{k \lambda_0 \pi}{2r} + \frac{k \lambda_0 \pi}{2r} = \frac{k \lambda_0 \pi}{r} [/tex] which is the total [tex]E_y[/tex] 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
Lisa...
#6
Feb25-06, 07:21 AM
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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