Find E field for a ring of charge - Charge per length non-uniform

In summary: So the E field at every point is the same.In summary, the ring of charge has a uniform charge distribution, but when the charge is non-uniform the E field at every point is different.
  • #1
venkman1080
3
0
I have to find the E field at all points on the z-axis for a ring of charge with radius = R. [tex] \lambda(\phi) = \lambda_0 cos(\phi) [/tex] where [tex] 0 \leq \phi < 2 \pi [/tex]

I know how to do the problem when it is the charge per length is uniform but when I do the calculation for the non-uniform case I get [tex] E = \frac{-kR\lambda_0\pi}{(R^2 + z^2)^{3/2}} \hat{i} [/tex] The integral for the j_hat part goes to zero because sin(phi)cos(phi) from 0 to 2 Pi and the z_hat part also goes to zero because the integral of cos(phi) from 0 to 2 Pi is zero.

I think my calculations are right, but I'm not totally sure. I just find it strange that its only in the i_hat. Any help/suggestions is greatly appreciated.

I used the formulas
[tex]
E = \int_{charge} \frac{kdq}{r_\delta ^2}\hat{r_\delta}
[/tex]

[tex]
r_\delta = -R cos(\phi)\hat{i} - R sin(\phi)\hat{j} + x\hat{k}
[/tex]

[tex]
\hat{r_\delta} = \frac{-R cos(\phi)\hat{i}}{\sqrt{R^2 + z^2}}
- \frac{R sin(\phi)\hat{j}}{\sqrt{R^2 + z^2}} + \frac{z\hat{k}}{\sqrt{R^2 + z^2}}
[/tex]

[tex]
dq = \lambda Rd\phi
[/tex]
Sorry if there is any typos with the latex formulas. I kept doing the preview post and it wouldn't change what I had entered the first time.
 
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  • #2
venkman1080 said:
The integral for the j_hat part goes to zero because sin(phi)cos(phi) from 0 to 2 Pi and the z_hat part also goes to zero because the integral of cos(phi) from 0 to 2 Pi is zero.

Sounds like you're multiplying every component by an extra factor of [itex]\cos\phi[/itex] before integrating...why are you doing that?
 
  • #3
I thought because [tex] \lambda(\phi) = \lambda_0 cos(\phi) [/tex] so when you do the integral the cos doesn't pull out.
 
  • #4
venkman1080 said:
I thought because [tex] \lambda(\phi) = \lambda_0 cos(\phi) [/tex] so when you do the integral the cos doesn't pull out.

Yes, I overlooked that. Your answer looks good to me.
 
  • #5
I'm just having trouble visualizing why the answer is only in the i_hat. Can you help clarify this for me?
 
  • #6
Think symmetry. For every spot on the ring, there's a spot on the opposite side with the same charge in magnitude but with opposite sign.
 

1. How do you calculate the electric field for a ring of non-uniform charge per length?

The formula for the electric field at a point on the axis of a ring of non-uniform charge is given by E = (kλz/R^2) * ∫(1/√(R^2 + z^2 - 2Rzcosθ) * dq), where k is the Coulomb's constant, λ is the charge per length, z is the distance from the center of the ring, R is the radius of the ring, and θ is the angle between the radius vector and the horizontal axis.

2. How does the electric field vary along the axis of a ring of non-uniform charge per length?

The electric field varies along the axis of a ring of non-uniform charge per length due to the non-uniform distribution of charge. As the distance from the center of the ring increases, the electric field decreases due to the inverse square relationship between distance and electric field strength.

3. What is the difference between a ring of non-uniform charge and a ring of uniform charge?

A ring of non-uniform charge has a varying charge density along its circumference, whereas a ring of uniform charge has the same charge density at every point along its circumference. This difference affects the calculation of the electric field, as a ring of non-uniform charge requires integration to take into account the varying charge density.

4. Can the electric field for a ring of non-uniform charge per length be negative?

Yes, the electric field for a ring of non-uniform charge per length can be negative. This occurs when the distance from the center of the ring is greater than the radius of the ring, resulting in a negative value for the electric field. It is important to pay attention to the direction of the electric field, as it can be either positive or negative depending on the point of observation.

5. How can the electric field for a ring of non-uniform charge per length be used in practical applications?

The electric field for a ring of non-uniform charge per length can be used in practical applications such as studying the electric potential and forces on a charged particle placed on the axis of the ring. It can also be used in the design and analysis of electrical circuits and devices.

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