# Electric field of ring of charge/washer of charge/disk of charge

## Homework Statement ## Homework Equations

charge density equations, electric field equations,

## The Attempt at a Solution

My attempts are attached. The attachment labeled A, is part A. Part B took a lot of paper so there are two attachments labeled B1 and B2. It is really B and C that I am struggling with. Not sure if I am on the right track or if it is correct. Any help would be much appreciated!

A little more thought: All the problems I have done in the back of my physics textbook are not like part B. I can find the electric field of a disk without a cut out in it. My thought process of finding the electric field with a cut out in it is integrating from the smaller radius to the larger radius. However, The total area of the washer is piR2^2-piR1^2. I plugged this result into my surface charge density toward the end of my solution.

My attempt at solution C:
Washer has charge +Q, Ring has radius 2R2, Lambda=Q/L
Clearly the charge on the ring has to be negative
therefore Lambda on ring is: -Q/4piR2

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Homework Helper
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B1.JPG is correct. APG is also correct. Combine the two results and set it equal to zero with $z=3 R_2$. $\\$ In APG, you must set $Q=2 \pi (2R_2) \lambda=4 \pi R_2 \lambda$, and $R=2R_2$. (This is not the same $Q$ that you have in B1.JPG. so call it $Q'$. It turns out $Q' \neq Q$. ). $\\$ Suggestion: Call the $E_z$ at $z=3R_2$ by $E_1$ and $E_2$. You must have $E_1+E_2=0$, which means $E_2=-E_1$. Your answer above for $\lambda$ is incorrect, but the rest is really just algebra. Try again. I'll be happy to check your result. (Note: $E_2$ is an expression that has $\lambda$ in it, and $E_1$ has $Q$ in it. Using $E_2=-E_1$, solve for $\lambda$).

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• Thank you! I’ll work on it tomorrow and respond! Appreciate it

• B1.JPG is correct. APG is also correct. Combine the two results and set it equal to zero with $z=3 R_2$. $\\$ In APG, you must set $Q=2 \pi (2R_2) \lambda=4 \pi R_2 \lambda$, and $R=2R_2$. (This is not the same $Q$ that you have in B1.JPG. so call it $Q'$. It turns out $Q' \neq Q$. ). $\\$ Suggestion: Call the $E_z$ at $z=3R_2$ by $E_1$ and $E_2$. You must have $E_1+E_2=0$, which means $E_2=-E_1$. Your answer above for $\lambda$ is incorrect, but the rest is really just algebra. Try again. I'll be happy to check your result. (Note: $E_2$ is an expression that has $\lambda$ in it, and $E_1$ has $Q$ in it. Using $E_2=-E_1$, solve for $\lambda$).
Okay, So I am finally looking at your response in more detail.
What I have so far is:
E2 is electric field for concentric ring
E2=(k)(lambda)(dl)/r^2
I see what you mean that E1 + E2 has to = 0. Therefore E2=-E1. I guess my question is did I already solve for E1 in image B1JPG?

Homework Helper
Gold Member
Yes, that is $E_1$ and it is correct.$\\$ And $E_2$ is correct in A1JPG, but you need to set $R=2R_2$, and $Q =2 \pi (2R_2) \lambda$. (As previously mentioned, this is not the same $Q$ that it in B2jpg).$\\$ In your post 4 here, $E_2$ is missing the $\cos(\theta) =\frac{z}{(R^2+z^2)^{1/2} }$ factor.

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Okay, I believe I see what you are saying-
Ering= (kλ2pi2R2z)/(2R22 + z2)(3/2)
and
EDisk= (2pizQ)/((4piε0(piR22-piR12)) [1/(z2+R12)1/2-1/(z2+R22)1/2]

Then I say ERing+EDisk=0 and solve for lambda?
That seems like some really tedious algebra which I haven't tried yet. Wanted to know if this was the correct method before I try to do it.
Thanks!!

SammyS
Staff Emeritus
Homework Helper
Gold Member
Okay, I believe I see what you are saying-
Ering= (kλ2pi2R2z)/(2R22 + z2)(3/2)
and
EDisk= (2pizQ)/((4piε0(piR22-piR12)) [1/(z2+R12)1/2-1/(z2+R22)1/2]

Then I say ERing+EDisk=0 and solve for lambda?
That seems like some really tedious algebra which I haven't tried yet. Wanted to know if this was the correct method before I try to do it.
Thanks!!
Quite a bit of that "cancels out" . You should either use Coulomb's coefficient, $\ k\$, for both or $\ \displaystyle \frac{1}{4\pi \varepsilon_0}\$. I suggest the latter. Gets rid of some $\ \pi \,$s .

You have: $\ \displaystyle E_{Ring} = \frac{kλ2\pi 2 R_2\, z} { \left( 2{R_2}^2 + z^2 \right)^{3/2}}\$ but I suppose you should be squaring (2R2), not just the R2 .

Changing to ε0 form and the above correction gives:
$\ \displaystyle E_{Ring} = \frac{2\, \lambda \pi \, 2 R_2\, z} { 4\pi\,\varepsilon_0 \left( {(2R_2)}^2 + z^2 \right)^{3/2}}\$​

The expression for the disk is:
$\ \displaystyle E_{Disk} = \frac{2\pi z \,Q} {(4 \pi \varepsilon_0) \left(\pi {R_2}^2 - \pi {R_1}^2 \right)} \left(\frac{1}{( z^2 + {R_1}^2 )^{1/2} } - \frac{1}{(z^2 + {R_2}^2 )^{1/2} } \right)$
.

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• Okay! I’ll try it out! Thanks

SammyS
Staff Emeritus
Homework Helper
Gold Member
Okay! I’ll try it out! Thanks
I was editing that for a while. Just got finished.

• 