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How to calculate the electric field at a point on the axis of two rings

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11
Homework Statement
Two rings with radius r have charge Q and −Q uniformly
distributed around them. The rings are parallel and located a
distance h apart, as shown in Fig. 1.35. Let z be the vertical
coordinate, with z = 0 taken to be at the center of the lower
ring. As a function of z, what is the electric field at points on
the axis of the rings?
Homework Equations
Coulomb Law
figura.PNG

Hi! I need help with this problem. I tried to solve it like this:
First I calculated the electric field of each ring:
WhatsApp Image 2019-09-11 at 6.23.25 PM.jpeg

Thus the electric field at a point that is at a distance z from the ring is ##E=\frac{Qz}{4\pi\epsilon_0(z^2+r^2)^{3/2}}##, Thuss for the upper ring, the electric field would be ##E_1=\frac{-Qz}{4\pi\epsilon_0(z^2+r^2)^{3/2}}## and for the lower one, it would be ##E_2=\frac{Qz}{4\pi\epsilon_0(z^2+r^2)^{3/2}}##.

Then, I choose a random point between the both rings, at a z height, so ##E_1=\frac{-Q(h-z)}{4\pi\epsilon_0((h-z)^2+r^2)^{3/2}}## and
##E_2=\frac{Qz}{4\pi\epsilon_0(z^2+r^2)^{3/2}}##, so the total electric field would be the sum of both, right? ##E_t=\frac{Q}{4\pi\epsilon_0}\left(\frac{z}{(z^2+r^2)^{3/2}}-\frac{(h-z)}{((h-z)^2+r^2)^{3/2}}\right)##.

The problem is that my answer is wroing and I don't know why. Hope someone can help me.
 

TSny

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Thuss for the upper ring, the electric field would be ##E_1=\frac{-Qz}{4\pi\epsilon_0(z^2+r^2)^{3/2}}## and for the lower one, it would be ##E_2=\frac{Qz}{4\pi\epsilon_0(z^2+r^2)^{3/2}}##.
Note that your expression for##E_1## gives zero electric field at ##z = 0## due the upper ring. That can't be right.

But your expression for ##E_2## looks good. (You should also indicate the direction of the field.)
 

TSny

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Then, I choose a random point between the both rings, at a z height, so ##E_1=\frac{-Q(h-z)}{4\pi\epsilon_0((h-z)^2+r^2)^{3/2}}##
OK. Here it looks like you have taken into account that the center of the upper ring is not at z = 0. However, does your expression give the right direction for the field of the upper ring?
 
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11
...

OK. Here it looks like you have taken into account that the center of the upper ring is not at z = 0. However, does your expression give the right direction for the field of the upper ring?
It doesn't give me any direction, right? Since is only the magnitude. But I guess that it should be upwards.
 

TSny

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You know the field of each ring has only a z component. So write expressions for ##E_z## for each ring making sure they give the correct sign for the z component.
 
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11
You know the field of each ring has only a z component. So write expressions for ##E_z## for each ring making sure they give the correct sign for the z component.
OK. ##\vec{E_1}=\frac{-Q(h-z)}{4\pi\epsilon_0((h-z)^2+r^2)^{3/2}} \hat{z}## and ##\vec{E_2}=\frac{Qz}{4\pi\epsilon_0(z^2+r^2)^{3/2}} \hat{z}##. Is it correct?
 

TSny

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OK. ##\vec{E_1}=\frac{-Q(h-z)}{4\pi\epsilon_0((h-z)^2+r^2)^{3/2}} \hat{z}## and ##\vec{E_2}=\frac{Qz}{4\pi\epsilon_0(z^2+r^2)^{3/2}} \hat{z}##. Is it correct?
Check to see if you get the right direction for the field due to the upper ring at z =0.
 
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Check to see if you get the right direction for the field due to the upper ring at z =0.
Replacing z=0 on ##E_1## made the unit vector ##\hat{z}## equat to 0, am I correct? So I guess it doesn't give me the right direction.
 

TSny

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Replacing z=0 on ##E_1## made the unit vector ##\hat{z}## equat to 0, am I correct? So I guess it doesn't give me the right direction.
Your expression for ##E_1## does not give zero at z = 0. Does it give a positive or negative result for the z -component of ##\vec E_1## at z = 0?
 
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Your expression for ##E_1## does not give zero at z = 0. Does it give a positive or negative result for the z -component of ##\vec E_1## at z = 0?
Sorry, I'm confused as you can see. It gives me a negative component of ##\vec E_1## at z=0, because of the -Q.
 

TSny

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Sorry, I'm confused as you can see. It gives me a negative component of ##\vec E_1## at z=0, because of the -Q.
A negative value of the z-component of ##\vec E_1## for z = 0 means that the field at z = 0 due to the upper, negatively charged ring would be pointing in the negative z-direction. But you can see that this is the wrong direction. (It might help to imagine replacing the negatively charged ring by a negatively charged point charge at z = h. What would be the direction of the E-field produced by this point charge at z = 0?)
 
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The point charge has a fiedl of ##\vec E=-\frac{KQ}{|\vec r- \vec r'|^2}\frac{(\vec r-\vec r')}{|\vec r- \vec r'|}##. Where ##\vec r=(0,0,h)## and ##\vec r'=(0,0,0)##, thus ##\vec r-\vec r'=(0,0,-h)\Rightarrow |\vec r-\vec r'|=h##, thus the electric field would be ##\vec E=-\frac{KQ}{h^2}\frac{(0,0,-h)}{h}=-\frac{KQ}{h^2}(-\hat k)##. Am I correct?
 

TSny

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The point charge has a fiedl of ##\vec E=-\frac{KQ}{|\vec r- \vec r'|^2}\frac{(\vec r-\vec r')}{|\vec r- \vec r'|}##. Where ##\vec r=(0,0,h)## and ##\vec r'=(0,0,0)##, thus ##\vec r-\vec r'=(0,0,-h)\Rightarrow |\vec r-\vec r'|=h##, thus the electric field would be ##\vec E=-\frac{KQ}{h^2}\frac{(0,0,-h)}{h}=-\frac{KQ}{h^2}(-\hat k)##. Am I correct?
I think you have a couple of sign errors that happen to cancel their damage:

(1) A unit vector pointing from the point charge toward the origin would be ##\frac{(\vec r'-\vec r)}{|\vec r- \vec r'|}## instead of ##\frac{(\vec r-\vec r')}{|\vec r- \vec r'|}##.

(2) The value of ##\vec r-\vec r'## would be ##(0,0,h)## instead of ##(0,0,-h)##.

These two errors cancel out and your final expression for the value of ##\vec E## at the origin is correct. Note that you get a vector that points upward in the positive z -direction. But I didn't really expect you to work this out mathematically. You can tell what the direction of the field must be at the origin by remembering that a negative point charge has a field at any point that is directed toward the point charge. So, the negative point charge at z = h must produce a field at the origin that points upward.

The same is true for the negatively charged ring located at z = h. It will produce a field at the origin that points upward. That is, the z-component of the field at the origin due to the upper ring must be positive. But your result for ##\vec E_1## gives a downward pointing field at the origin. So, you should re-examine how you got your expression for ##\vec E_1## and see why your expression has the wrong overall sign.
 
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I think you have a couple of sign errors that happen to cancel their damage:

(1) A unit vector pointing from the point charge toward the origin would be ##\frac{(\vec r'-\vec r)}{|\vec r- \vec r'|}## instead of ##\frac{(\vec r-\vec r')}{|\vec r- \vec r'|}##.
Ok, could you please explain this a bit more? I thought that the vector I needed was from the origin to the point charge, that's why I wrote it as ##\frac{(\vec r-\vec r')}{|\vec r- \vec r'|}##.

To be honest, I don't see where the error could be, since my - comes from the charge, I suppose I need the unit vector to be negative, right?
 

TSny

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The electric field of a point charge is ##\vec E = \frac{kq}{r^2} \hat r## , where the unit vector ##\hat r## points from the point charge toward the location of the point where the field is being evaluated. If the charge is located on the positive z-axis and you want the field at the origin, then ##\hat r## will point downward (in the negative z-direction). With your definitions of ##\vec r## and ##\vec r'##, you would have

##\hat r = \frac{\vec r' - \vec r}{|\vec r' -\vec r|} = (0, 0, -h) \,\,\,## which points downward.

If q is a negative charge written as q = -Q, where Q is a positive number, then ##\vec E = -\frac{kQ}{r^2} \hat r##

The negative sign from the charge makes ##\vec E## point opposite to ##\hat r##. So, ##\vec E## points upward at the origin if the negative charge sits on the positive z-axis. That is, the field points toward the charge, as expected for a negative charge.
 
371
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The electric field of a point charge is ##\vec E = \frac{kq}{r^2} \hat r## , where the unit vector ##\hat r## points from the point charge toward the location of the point where the field is being evaluated. If the charge is located on the positive z-axis and you want the field at the origin, then ##\hat r## will point downward (in the negative z-direction). With your definitions of ##\vec r## and ##\vec r'##, you would have

##\hat r = \frac{\vec r' - \vec r}{|\vec r' -\vec r|} = (0, 0, -h) \,\,\,## which points downward.

If q is a negative charge written as q = -Q, where Q is a positive number, then ##\vec E = -\frac{kQ}{r^2} \hat r##

The negative sign from the charge makes ##\vec E## point opposite to ##\hat r##. So, ##\vec E## points upward at the origin if the negative charge sits on the positive z-axis. That is, the field points toward the charge, as expected for a negative charge.
Oh, I think I see it know. I'll try to do it. I saw on Wikipedia that the magnitude of the electric force is ##F=\frac{k|q_1q_2|}{r^2}## does that also happens on electric field?
 

TSny

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I saw on Wikipedia that the magnitude of the electric force is ##F=\frac{k|q_1q_2|}{r^2}## does that also happens on electric field?
I don't understand what you are asking here. Can you rephrase the question?
 
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I don't understand what you are asking here. Can you rephrase the question?
I saw on Wikipedia that the magnitude of the electric force is ##F=\frac{k|q_1q_2|}{r^2}##, with absolute value, and I was wondering if the magnitude of the electric field also has the absolute value of the charge.
 

TSny

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I saw on Wikipedia that the magnitude of the electric force is ##F=\frac{k|q_1q_2|}{r^2}##, with absolute value, and I was wondering if the magnitude of the electric field also has the absolute value of the charge.
Yes. If you want the magnitude of the electric field of a single point charge, you would use the absolute value of the charge.
 
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Yes. If you want the magnitude of the electric field of a single point charge, you would use the absolute value of the charge.
Thanks, you've helped me a lot!
 

TSny

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Glad I could help some.
 

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