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

In summary, the conversation is about a problem involving calculating the electric field of two rings at a certain point. The individual electric fields of each ring are calculated and then the total electric field is found by taking the sum of both. However, the answer is incorrect and the person is seeking help to understand why. Through further discussion, it is discovered that the expression for the electric field of the upper ring is incorrect, and the correct direction for the field is determined by considering a point charge at the same location. After correcting for some sign errors, the correct direction for the field is found to be upward in the positive z-direction.
  • #1
Davidllerenav
424
14
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?
Relevant 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.
 
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  • #2
Davidllerenav said:
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.)
 
  • #3
...
Davidllerenav said:
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?
 
  • #4
TSny said:
...

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.
 
  • #5
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.
 
  • #6
TSny said:
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?
 
  • #7
Davidllerenav said:
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.
 
  • #8
TSny said:
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.
 
  • #9
Davidllerenav said:
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?
 
  • #10
TSny said:
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.
 
  • #11
Davidllerenav said:
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?)
 
  • #12
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?
 
  • #13
Davidllerenav said:
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.
 
  • #14
TSny said:
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?
 
  • #15
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.
 
  • #16
TSny said:
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?
 
  • #17
Davidllerenav said:
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?
 
  • #18
TSny said:
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.
 
  • #19
Davidllerenav said:
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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  • #20
TSny said:
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!
 
  • #21
Glad I could help some.
 

1. What is the formula for calculating the electric field at a point on the axis of two rings?

The formula for calculating the electric field at a point on the axis of two rings is given by:

E = k*q*(z1-z2)/((z1^2 + d^2)^1.5) + k*q*(z2-z1)/((z2^2 + d^2)^1.5)

Where E is the electric field, k is the Coulomb's constant, q is the charge on the rings, z1 and z2 are the distances from the point to the center of each ring, and d is the distance between the two rings.

2. What is the direction of the electric field at a point on the axis of two rings?

The direction of the electric field at a point on the axis of two rings depends on the relative charges and positions of the rings. If the rings have opposite charges, the electric field will point towards the positive ring. If the rings have the same charge, the electric field will point away from both rings. Additionally, the direction of the electric field will be parallel to the axis of the two rings.

3. How do I determine the distance between the two rings?

The distance between the two rings, denoted as d in the formula, can be found by measuring the distance from the center of one ring to the center of the other ring. This distance should be measured along the axis of the rings, which is the line passing through the centers of both rings.

4. Can I use this formula for rings with non-uniform charge distribution?

Yes, this formula can be used for rings with non-uniform charge distribution. However, the charge q in the formula should be replaced with the charge density, which is the amount of charge per unit length on the ring. Additionally, the distances z1 and z2 should be measured from the point to the center of mass of each ring, rather than the center of the ring.

5. Are there any simplifications or approximations that can be made when calculating the electric field at a point on the axis of two rings?

Yes, there are a few simplifications or approximations that can be made when calculating the electric field at a point on the axis of two rings. For example, if the distance d between the two rings is much larger than the distances z1 and z2 from the point to the rings, the formula can be simplified to:

E = k*q*z1/(d^2) + k*q*z2/(d^2)

This is known as the small angle approximation and is often used when the rings are very far apart compared to the distance from the point to the rings.

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