Electric field of a non-uniformly charged ring

In summary, the conversation discusses the calculation of the magnitude and direction of the electric field due to a ring in a specific point on the z-axis. The charge distribution on the ring is given and the electric field is calculated using an integral formula. The conversation then explores the symmetry of the charge distribution and possible components of the electric field.
  • #1
zapman345
11
2

Homework Statement


Calculate the magnitude and direction of the electric field due to the ring in the point P on the z-axis.
06b46043ae.jpg


Homework Equations


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

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 ([itex]E_z=0[/itex]) but there will be an x-component as seen in the drawing I made.
64baaffdd6.jpg

So now to select the x-component we say [itex]dE=\frac{dE_{x}}{\sin\left(\theta\right)}[/itex] so [itex]dE_x=dE\cdot\sin\left(\theta\right)[/itex].
[tex]
\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*}
[/tex]

Then you should integrate to get the final answer [tex]
\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*}
[/tex]

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.
 
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  • #2
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
Because of the charge distribution there will be a negatively charged lower ring ([itex]φ=π[/itex] to [itex]φ=2π[/itex]) and a positively charged upper ring ([itex]φ=0[/itex] to [itex]φ=π[/itex]). 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 [itex]λ=λ_0\cdot \sin(φ)[/itex] which is 0 at both [itex]φ=0[/itex] and [itex]φ=π[/itex] there can be no component in the x-direction and we're only left with the y-component?
 
  • #4
zapman345 said:
Because of the charge distribution there will be a negatively charged lower ring ([itex]φ=π[/itex] to [itex]φ=2π[/itex]) and a positively charged upper ring ([itex]φ=0[/itex] to [itex]φ=π[/itex]). 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 [itex]λ=λ_0\cdot \sin(φ)[/itex] which is 0 at both [itex]φ=0[/itex] and [itex]φ=π[/itex] 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'?
 
  • #5
Okay so an element at [itex]S[/itex] produces an electric field in point P like [itex]E=-E_x \hat{x}-E_y \hat{y}+E_z \hat{z} [/itex] and an element at [itex]S'[/itex] produces [itex]E=+E_x \hat{x}-E_y \hat{y}-E_z \hat{z} [/itex] thus adding both of those we have a total electric field of [itex]E=-2E_y \hat{y} [/itex]?
 
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  • #6
zapman345 said:
Okay so an element at [itex]S[/itex] produces an electric field in point P like [itex]E=-E_x \hat{x}-E_y \hat{y}+E_z \hat{z} [/itex] and an element at [itex]S'[/itex] produces [itex]E=+E_x \hat{x}-E_y \hat{y}-E_z \hat{z} [/itex] thus adding both of those we have a total electric field of [itex]E=-2E_y \hat{y} [/itex]?
Yes.
 
  • #7
Now when I try to calculate the electric field in the y-direction I get [itex]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[/itex] (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 [itex]φ=[0,π][/itex] then the integral gives me [itex]2[/itex] but if I pick [itex]φ=[π,2π][/itex] then I get [itex]-2[/itex]. But I'm guessing since the field has to be in the negative y-direction at the end we have to pick [itex]φ=[0,π][/itex] as our limits such that when we multiply by the -2 we got from symmetry arguments the final answer is negative?
 
  • #8
zapman345 said:
Now when I try to calculate the electric field in the y-direction I get [itex]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[/itex] (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 [itex]φ=[0,π][/itex] then the integral gives me [itex]2[/itex] but if I pick [itex]φ=[π,2π][/itex] then I get [itex]-2[/itex]. But I'm guessing since the field has to be in the negative y-direction at the end we have to pick [itex]φ=[0,π][/itex] 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 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
So it should be [itex]dE_y=\sin(φ)\cos(θ)dE \hat{y}[/itex] then? Because that would be the y-component of a vector in spherical coordinates.
 
  • #10
zapman345 said:
So it should be [itex]dE_y=\sin(φ)\cos(θ)dE \hat{y}[/itex] 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).
 
  • #11
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 [itex]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[/itex] which then leads to [itex]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[/itex] and [itex]E=-2\cdot E_y[/itex] as per symmetry arguments.
But then another question arises from this, if [itex]z_0\gg R[/itex] this configuration should reduce to a dipole which has an electric field that is proportional to [itex]z_0^{-3}[/itex] in this case but the electric field I found is proportional with [itex]z_0^{-2}[/itex] for [itex]z_0\gg R[/itex].
 
  • #12
zapman345 said:
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 [itex]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[/itex] which then leads to [itex]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[/itex] and [itex]E=-2\cdot E_y[/itex] as per symmetry arguments.
But then another question arises from this, if [itex]z_0\gg R[/itex] this configuration should reduce to a dipole which has an electric field that is proportional to [itex]z_0^{-3}[/itex] in this case but the electric field I found is proportional with [itex]z_0^{-2}[/itex] for [itex]z_0\gg R[/itex].
Sorry, I copied your error without noticing. Originally you wrote sin(θ), which was correct, not cos.
 
  • #13
Okay so then it becomes [itex]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[/itex] which then results in [tex]
\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*}[/tex]

This does properly reduce to a [itex]z_0^{-3}[/itex] proportionality for large [itex]z_0[/itex] so I'm guessing this is correct?
 
  • #14
zapman345 said:
Okay so then it becomes [itex]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[/itex] which then results in [tex]
\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*}[/tex]

This does properly reduce to a [itex]z_0^{-3}[/itex] proportionality for large [itex]z_0[/itex] so I'm guessing this is correct?
Yes, I think it's right.
 
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  • #15
Thank you so much for all your help!
 

1. What is the formula for calculating the electric field of a non-uniformly charged ring?

The electric field of a non-uniformly charged ring can be calculated using the formula E = kqz/(2πε0R2), where k is the Coulomb's constant, q is the total charge on the ring, z is the distance from the ring's center to the point of interest, ε0 is the permittivity of free space, and R is the radius of the ring.

2. How is the electric field of a non-uniformly charged ring different from that of a uniformly charged ring?

In a uniformly charged ring, the charge is evenly distributed along the circumference of the ring, resulting in a constant electric field at all points around the ring. However, in a non-uniformly charged ring, the charge is unevenly distributed, causing the electric field to vary at different points around the ring.

3. How does the distance from the ring affect the strength of the electric field?

The strength of the electric field is inversely proportional to the square of the distance from the ring. This means that as the distance increases, the electric field decreases, and vice versa.

4. Can the electric field of a non-uniformly charged ring be negative?

Yes, the electric field of a non-uniformly charged ring can be negative. This occurs when the charge on one side of the ring is greater than the charge on the other side, resulting in an overall negative electric field in that region.

5. How can the electric field of a non-uniformly charged ring be visualized?

The electric field of a non-uniformly charged ring can be visualized using electric field lines. These lines represent the direction and strength of the electric field at different points around the ring. Closer lines indicate a stronger electric field, while farther lines indicate a weaker electric field.

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