# Homework Help: Electric field of a non-uniformly charged ring

1. Feb 6, 2016

### zapman345

1. The problem statement, all variables and given/known data
Calculate the magnitude and direction of the electric field due to the ring in the point P on the z-axis.

2. Relevant equations
Charge distribution on the ring is given by $λ(φ)=λ_0\cdot sin(φ)$. This should result in the total charge on the ring being zero.
Electric field is given by $E=\frac{1}{4\pi\epsilon_{0}}\int_{Path}\frac{\lambda\left(r\right)}{r^{2}}\hat{r}dl$
The distance $r$ from any point on the ring to the point P: $r=\sqrt{z_0^2+R^2}$

3. The attempt at a solution
Due to symmetry and the non-uniformity of the charge distribution we can say that the electric field in the z-direction is 0 ($E_z=0$) but there will be an x-component as seen in the drawing I made.

So now to select the x-component we say $dE=\frac{dE_{x}}{\sin\left(\theta\right)}$ so $dE_x=dE\cdot\sin\left(\theta\right)$.
$$\begin{eqnarray*} dE_{x} & = & dE\cdot\sin\left(\theta\right)\\ & = & \frac{1}{4\pi\epsilon_{0}}\cdot\frac{1}{r^{2}}\sin\left(\theta\right)dq\\ & = & \frac{1}{4\pi\epsilon_{0}}\cdot\frac{1}{r^{2}}\sin\left(\theta\right)\lambda dl\\ & = & \frac{1}{4\pi\epsilon_{0}}\cdot\frac{1}{r^{2}}\sin\left(\theta\right)\lambda_{0}\sin\left(\varphi\right)dl\\ & = & \frac{1}{4\pi\epsilon_{0}}\cdot\frac{1}{r^{2}}\sin\left(\theta\right)\lambda_{0}\sin\left(\varphi\right)Rd\varphi\\ & = & \frac{1}{4\pi\epsilon_{0}}\cdot\frac{1}{r^{2}}\frac{R}{r}\lambda_{0}\sin\left(\varphi\right)Rd\varphi\\ & = & \frac{\lambda_{0}R^{2}}{4\pi\epsilon_{0}}\cdot\frac{1}{r^{3}}\sin\left(\varphi\right)d\varphi\\ & = & \frac{\lambda_{0}}{4\pi\epsilon_{0}}\cdot\frac{R^2}{\left(z_{0}^{2}+R^{2}\right)^{\frac{3}{2}}}\sin\left(\varphi\right)d\varphi \end{eqnarray*}$$

Then you should integrate to get the final answer $$\begin{eqnarray*} E_{x} & = & \frac{\lambda_{0}}{4\pi\epsilon_{0}}\cdot\frac{R^{2}}{\left(z_{0}^{2}+R^{2}\right)^{\frac{3}{2}}}\int_{0}^{2\pi}\sin\left(\varphi\right)d\varphi\\ & = & 0 \end{eqnarray*}$$

But as you can see that results in 0 so I'm not really sure where the mistake is but I think it's either in selecting the x-component or I'm doing the integration wrong.

2. Feb 6, 2016

### haruspex

What makes you think the field will be nonzero in the x direction? Have you looked at the symmetry for that? Any other directions to consider?

3. Feb 6, 2016

### zapman345

Because of the charge distribution there will be a negatively charged lower ring ($φ=π$ to $φ=2π$) and a positively charged upper ring ($φ=0$ to $φ=π$). The negative semi-ring will have a negative z-component; a positive x-component and a negative y-component (?) and the positive semi-ring will have a positive z-component; a positive x-component and a negative y-component on the z-axis thus the z-components cancel out and we're left with the x and y components?
Or because of the fact $λ=λ_0\cdot \sin(φ)$ which is 0 at both $φ=0$ and $φ=π$ there can be no component in the x-direction and we're only left with the y-component?

4. Feb 6, 2016

### haruspex

Let S' be the reflection of S in the XZ plane, so in terms of the diagram its angle is -φ. What is the direction of the net field at P due to the elements at S and S'?

5. Feb 6, 2016

### zapman345

Okay so an element at $S$ produces an electric field in point P like $E=-E_x \hat{x}-E_y \hat{y}+E_z \hat{z}$ and an element at $S'$ produces $E=+E_x \hat{x}-E_y \hat{y}-E_z \hat{z}$ thus adding both of those we have a total electric field of $E=-2E_y \hat{y}$?

6. Feb 6, 2016

Yes.

7. Feb 7, 2016

### zapman345

Now when I try to calculate the electric field in the y-direction I get $E_{y} = \frac{\lambda_{0}}{4\pi\epsilon_{0}}\cdot\frac{R^{2}}{\left(z_{0}^{2}+R^{2}\right)^{\frac{3}{2}}}\int_{0}^{\pi}\sin\left(\varphi\right)d\varphi$ (following the same procedure as I did in the first post) my question would be what to pick for the integration limits. If I chose $φ=[0,π]$ then the integral gives me $2$ but if I pick $φ=[π,2π]$ then I get $-2$. But I'm guessing since the field has to be in the negative y-direction at the end we have to pick $φ=[0,π]$ as our limits such that when we multiply by the -2 we got from symmetry arguments the final answer is negative?

8. Feb 7, 2016

### haruspex

I had not previously studied the integral you did for the x component in the initial post. Your expression for Ex there is not correct. Esin(θ) gives you the component of E in tne XY plane, but for the component in specifically the X or Y direction there must be some dependence on φ.

9. Feb 7, 2016

### zapman345

So it should be $dE_y=\sin(φ)\cos(θ)dE \hat{y}$ then? Because that would be the y-component of a vector in spherical coordinates.

10. Feb 7, 2016

### haruspex

Not sure I understand your notation there, but yes, the component factor you need is cos(θ)sin(φ).

11. Feb 7, 2016

### zapman345

Oh what I tried to say was that an element of the electric field in the y direction was equal to the y-component of the electric field vector.

Sorry if I'm a bit slow to understand but the resulting electric field would be $dE_y=\frac{1}{4\pi\epsilon_{0}}\cdot\frac{1}{r^{2}}\cos\left(\theta\right)\cdot \lambda_{0}\sin^2\left(\varphi\right)\cdot Rd\varphi$ which then leads to $E_y=\frac{\lambda_{0}}{4\pi\epsilon_{0}}\cdot\frac{z_{0}R}{\left(z_{0}^{2}+R^{2}\right)^{\frac{3}{2}}}\int_{0}^{2\pi}\sin^{2}\left(\varphi\right)d\varphi$ and $E=-2\cdot E_y$ as per symmetry arguments.
But then another question arises from this, if $z_0\gg R$ this configuration should reduce to a dipole which has an electric field that is proportional to $z_0^{-3}$ in this case but the electric field I found is proportional with $z_0^{-2}$ for $z_0\gg R$.

12. Feb 7, 2016

### haruspex

Sorry, I copied your error without noticing. Originally you wrote sin(θ), which was correct, not cos.

13. Feb 7, 2016

### zapman345

Okay so then it becomes $dE_y=\frac{1}{4\pi\epsilon_{0}}\cdot\frac{1}{r^{2}}\sin\left(\theta\right)\lambda_{0}\sin^{2}\left(\varphi\right)Rd\varphi$ which then results in $$\begin{eqnarray*} E_{y} & = & \frac{\lambda_{0}}{4\pi\epsilon_{0}}\cdot\frac{R^{2}}{\left(z_{0}^{2}+R^{2}\right)^{\frac{3}{2}}}\int_{0}^{2\pi}\sin^{2}\left(\varphi\right)d\varphi\\ & = & \frac{\lambda_{0}}{4\pi\epsilon_{0}}\cdot\frac{R^{2}}{\left(z_{0}^{2}+R^{2}\right)^{\frac{3}{2}}}\cdot\pi\\ E & = & -2\cdot E_{y}\hat{y}\\ & = & -\frac{\lambda_{0}}{2\epsilon_{0}}\cdot\frac{R^{2}}{\left(z_{0}^{2}+R^{2}\right)^{\frac{3}{2}}}\hat{y} \end{eqnarray*}$$

This does properly reduce to a $z_0^{-3}$ proportionality for large $z_0$ so I'm guessing this is correct?

14. Feb 7, 2016

### haruspex

Yes, I think it's right.

15. Feb 7, 2016

### zapman345

Thank you so much for all your help!