Electric field of a non-uniformly charged ring

[zapman345]

Homework Statement

Calculate the magnitude and direction of the electric field due to the ring in the point P on the z-axis.

Homework 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}

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).
<br /> \begin{eqnarray*}<br /> <br /> dE_{x} &amp; = &amp; dE\cdot\sin\left(\theta\right)\\<br /> <br /> &amp; = &amp; \frac{1}{4\pi\epsilon_{0}}\cdot\frac{1}{r^{2}}\sin\left(\theta\right)dq\\<br /> <br /> &amp; = &amp; \frac{1}{4\pi\epsilon_{0}}\cdot\frac{1}{r^{2}}\sin\left(\theta\right)\lambda dl\\<br /> <br /> &amp; = &amp; \frac{1}{4\pi\epsilon_{0}}\cdot\frac{1}{r^{2}}\sin\left(\theta\right)\lambda_{0}\sin\left(\varphi\right)dl\\<br /> <br /> &amp; = &amp; \frac{1}{4\pi\epsilon_{0}}\cdot\frac{1}{r^{2}}\sin\left(\theta\right)\lambda_{0}\sin\left(\varphi\right)Rd\varphi\\<br /> <br /> &amp; = &amp; \frac{1}{4\pi\epsilon_{0}}\cdot\frac{1}{r^{2}}\frac{R}{r}\lambda_{0}\sin\left(\varphi\right)Rd\varphi\\<br /> <br /> &amp; = &amp; \frac{\lambda_{0}R^{2}}{4\pi\epsilon_{0}}\cdot\frac{1}{r^{3}}\sin\left(\varphi\right)d\varphi\\<br /> <br /> &amp; = &amp; \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<br /> <br /> \end{eqnarray*}<br />

Then you should integrate to get the final answer <br /> \begin{eqnarray*}<br /> <br /> E_{x} &amp; = &amp; \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\\<br /> <br /> &amp; = &amp; 0<br /> <br /> \end{eqnarray*}<br />

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.

 

[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?

 

[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?

 

[haruspex]

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?

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'?

 

[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&#039; 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}?

 

[haruspex]

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&#039; 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}?

Yes.

 

[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?

 

[haruspex]

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?

Your integral is wrong.
I had not previously studied the integral you did for the x component in the initial post. Your expression for E_x 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.

 

[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.

 

[haruspex]

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.

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

 

[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.

 

[haruspex]

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.

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

 

[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 <br /> \begin{eqnarray*}<br /> <br /> E_{y} &amp; = &amp; \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\\<br /> <br /> &amp; = &amp; \frac{\lambda_{0}}{4\pi\epsilon_{0}}\cdot\frac{R^{2}}{\left(z_{0}^{2}+R^{2}\right)^{\frac{3}{2}}}\cdot\pi\\<br /> <br /> E &amp; = &amp; -2\cdot E_{y}\hat{y}\\<br /> <br /> &amp; = &amp; -\frac{\lambda_{0}}{2\epsilon_{0}}\cdot\frac{R^{2}}{\left(z_{0}^{2}+R^{2}\right)^{\frac{3}{2}}}\hat{y}<br /> <br /> \end{eqnarray*}

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

 

[haruspex]

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 <br /> \begin{eqnarray*}<br /> <br /> E_{y} &amp; = &amp; \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\\<br /> <br /> &amp; = &amp; \frac{\lambda_{0}}{4\pi\epsilon_{0}}\cdot\frac{R^{2}}{\left(z_{0}^{2}+R^{2}\right)^{\frac{3}{2}}}\cdot\pi\\<br /> <br /> E &amp; = &amp; -2\cdot E_{y}\hat{y}\\<br /> <br /> &amp; = &amp; -\frac{\lambda_{0}}{2\epsilon_{0}}\cdot\frac{R^{2}}{\left(z_{0}^{2}+R^{2}\right)^{\frac{3}{2}}}\hat{y}<br /> <br /> \end{eqnarray*}

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

Yes, I think it's right.

 

[zapman345]

Thank you so much for all your help!