Electric field on a ring's axis

[ranger281]

Homework Statement

Find the electric field above ring's center

Relevant Equations

$d\vec{E}=\frac{dQ}{r^2}cos\theta \hat{z}$

The contribution coming from a little segment of the ring is $d\vec{E}=\frac{dQ}{r^2}cos\theta \hat{z}$, assuming that the horizontal components cancel out. But how can we show that?

 

[TSny]

Hello. Just to be clear, are you asking why the horizontal components of all of the $d\vec E$'s cancel out?

 

[ranger281]

Yes, the horizontal (radial) component of the field above the center.

 

[TSny]

OK. A good diagram is essential. Pick a small segment of the ring at the left side of the ring and draw the electric field vector produced by this segment at the location of the point above the center. Repeat for another segment at the right side of the ring (diametrically opposite to the segment at the left).

 

[ranger281]

This I understand. But how can we obtain that by calculation? In this case, the total field (counting only the vertical components) is $\vec{E} = \int_{S}^{}\frac{\lambda* cos\theta\hat{z}}{R^2+z^2}ds$ (S being the ring), where we can simply multiply the integrated function by $2\pi R$. Why does not a similar calculation with $sin\theta\hat{x}$ instead of $cos\theta\hat{z}$ yield 0?

 

[TSny]

The x-component of $dE$ for an arbitrary segment is not given by using a factor of just $\sin \theta$

Show that the x-component of $dE$ shown above is $-dE\sin\theta \cos \phi$. The net x-component is obtained by integrating with respect to $\phi$.

 

[ranger281]

I have one more (related) question. In the attached exercise, to obtain $E_{z}$, $E$ is multiplied by $cos\psi$. Why is it so?

 

[TSny]

Note that for the element of charge shown, the angle between $d\vec E$ and the z-axis is $\psi$. So, you get the z-component of $d \vec E$ by multiplying $dE$ by $\cos \psi$.

 

[ranger281]

How can we determine the direction of dE (or E) in this example?

 

[haruspex]

assuming that the horizontal components cancel out. But how can we show that?

The usual way is by a symmetry argument. The arrangement has circular symmetry about the vertical axis, so the field must have that symmetry too.

 

[TSny]

How can we determine the direction of dE (or E) in this example?

The element of charge can be considered a point charge.
$d \vec E$ points radially away from the element of charge.