# Homework Help: Electric Field of a Charged Ring

1. Oct 15, 2011

### Tater

Hi everyone,

1. The problem statement, all variables and given/known data

Let a charge Q be uniformly distributed on a circular ring defined by a < $\rho$ < b. Find D at (0,0,h).

2. Relevant equations
E = kQ/r2 ar
D = $\epsilon$o E

3. The attempt at a solution

Well I thought I had this figured out, but I was wrong and I still can't figure it out after 2 hours :(

Since I have a range for the radius (a and b), I thought that I could calculate the E-field for both a and b separately then simply apply b-a to the field.

Finding both r1 and r2:
cos ∅1 = h / r1 = h / $\sqrt{a^2 + h^2}$
cos ∅2 = h / r2 = h / $\sqrt{b^2 + h^2}$

Then, E = kQh / r^2 [ cos ∅1 - cos ∅2 ]

I just don't know how to tackle this one. I have a large gap that I need to make note of.

The solution in the back of the book is D = $\frac{Qh}{2\pi(b^2-a^2)}$ [$\frac{1}{\sqrt{a^2+h^2}}$ - $\frac{1}{\sqrt{b^2+h^2}}$ ]

I would greatly appreciate any help!! I kind of think I had the right idea, I just don't understand how to fix what I've done wrong.

Any help is greatly appreciated!! :)

2. Oct 16, 2011

### runnergirl

First you need to calculate the charge using the charge density. We know that:
$Q = \int_{a}^{b}\int_{0}^{2\pi}\rho_lrd\phi$

From there you should find the potential of a disk along the z-axis. The integral should be set up as:
$d\phi = \frac{\rho_l*2*\pi*rdr}{4*pi*\epsilon_o\sqrt{z^2+r^2}}$.

If you integrate correctly it should come out to be in terms of both b and a. Then using the fact that $E =-\nabla\phi$ in the z-direction should get you to the answer. Let me know if you need more explicit help.

3. Oct 17, 2011

### Tater

A few questions though,

Charge enclosed: Qenc= ρsdS
I don't understand how I can solve for this when I don't know what ρs or Qenc are.

Also I don't understand your last comment: E=−∇ϕ
Did you mean to use E=−∇V (where V is the potential)?

As always, thanks for the help :)

4. Oct 17, 2011

### runnergirl

That is true you don't know what $\rho_s$ or $Q_{enc}$ is but you do know that there is going to be a charge distribution on the ring. You know that it will have a total charge Q enclosed. If you recall from your solution, it is indeed in terms of Q which is the charge on the ring.

So you want to integrate $Q = \int_{a}^{b}\int_{0}^{2\pi}\rho_srd\phi$ and solve for $\rho_s$ in terms of Q.

And sorry about the notation for $E =-\nabla\phi$ in this case I did mean that $\phi$ was the potential and that's just the notation I'm used to. So yes, $\phi$ is the same as V.

Also, when you do $E =-\nabla\phi$, you want to solve for when z = h.

If you have any more questions let me know.

5. Oct 17, 2011

### Tater

Qenc= ρsdS
Qencs2∏*$\frac{1}{2}$(b2-a2)
ρs=Qenc/∏(b2-a2)

So now I'd have to find resultant vector R (I'm bad at this). Here goes though:
If I drew a triangle with a radius of r and a height of h I would get: R=(r2+h2)1/2. But now I'd have to convert this to cylindrical coordinates, so I would replace h with z and r with ??? (I don't exactly know how to do this here because I don't have a value for the radius - I have 2 possible values. I'm over-complicating it in my head. I don't really understand what radius I need/want.)

Then after that, I can simply plug it into the formula for potential.

Because it's a point charge, the potential is simply:

V=$\frac{Q}{4\piεR}$

After I find the potential, I can apply $E =-∇V$, then apply $D =εE$

Also, why is it that we sometimes use h (the length) and sometimes we refer to it as z? This really confuses me when I try to write my vectors going from cartesian to cylindrical.

Thank you so much for your help runnergirl :D

6. Oct 17, 2011

### runnergirl

So you're close. In this case we're integrating over r so instead of $R = (r^2+h^2)^{1/2}$ you'll use $R = (r^2+z^2)^{1/2}$. This is because you're not solving for z = h yet. So when it's written like this, it is in cylindrical coordinates (a function of r and z). So yes, use a triangle with r being the change in the radial direction, which I'll show in a minute is what you integrate over, and z since that too is a variable in which you will use later.

So now if you plug it all in:
$V = \int_{a}^{b}\frac{Q}{2\pi\epsilon_o(b^2-a^2)}\frac{rdr}{\sqrt{z^2+r^2}}$

It's a fairly nice integral with a integral table. From there you do indeed apply $E = -\nabla V$ where in this case it will be only in the z-direction and solving for when z = h. So more explicitly:
$E = -\frac{\partial V}{\partial z}\mid_{z=h}$.

Well typically it is usually z that is used because most of the time in problems like these we're looking for the E-field along the z-axis which z is a variable. In this case, the reason it is h at the end is because you're evaluating the E(0,0,h) on the z-axis, so you're looking for the E-field at a particular point along the z-axis.

Hope that helps.

7. Oct 18, 2011

### Tater

So out of curiosity, since this was a point charge:

V=$\frac{Q}{4\piεR}$

But is it also fair to say that we could have used V = - ∫E dl ? I'm just curious as to why we're integrating to find potential because technically it is a point charge and we don't have to.

This problem was a disaster and you've cleared it up for me. Thank you so much.

8. Oct 18, 2011

### runnergirl

Well since this is a charge Q distributed over the entire ring, it cannot be thought of as a point charge, unless you're looking at the far field. If z>>a,b the potential will due to the disk charge Q will behave like a point charge. So in this case, it is easier to find the potential, V (and we have to integrate over the entire ring), and then use $E = -\nabla V$ in the z-direction because the field is expected to be along the z-axis (this is due to the cylindrical symmetry of the problem). Now if we were given the E-field right away and asked to find the potential, then yes applying $V = -\int E\cdot dl$ would be a good method.

Glad it makes more sense to you.