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

[Davidllerenav]

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

Hi! I need help with this problem. I tried to solve it like this:
First I calculated the electric field of each ring:

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]

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]

...

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?

 

[Davidllerenav]

...

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]

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.

 

[Davidllerenav]

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]

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.

 

[Davidllerenav]

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]

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?

 

[Davidllerenav]

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]

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?)

 

[Davidllerenav]

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]

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.

 

[Davidllerenav]

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]

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.

 

[Davidllerenav]

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]

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?

 

[Davidllerenav]

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]

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.

 

[Davidllerenav]

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]

Glad I could help some.