Electric Field of a ring , mathematically

[MinaKaiser]

This is about the electric field of a ring with radius r, at a distance z from center, along the axis of the ring. The ring carries a uniform line charge
λ
. We always say that the radial component of the field cancels out due to symmetry. Can somebody tell how to prove it mathematically (using cylindrical coordinate system only)?

dErad=−14πϵ0rλdθ(r2+z2)rˆ

Erad=−14πϵ0rλ(r2+z2)∫2π0rˆdθ

 

[I like Serena]

Welcome to PF, MinaKaiser!

This is about the electric field of a ring with radius r, at a distance z from center, along the axis of the ring. The ring carries a uniform line charge
λ
. We always say that the radial component of the field cancels out due to symmetry. Can somebody tell how to prove it mathematically (using cylindrical coordinate system only)?

dErad=−14πϵ0rλdθ(r2+z2)rˆ

Erad=−14πϵ0rλ(r2+z2)∫2π0rˆdθ

First, let's clean up your formula.

I get:

$$d\vec E_{rad} = - {1 \over 4\pi \epsilon_o} \cdot {\lambda r d\theta \over r^2 + z^2} \cdot {r \over \sqrt{r^2 + z^2}} \hat r$$

$$\vec E_{rad} = - {1 \over 4\pi \epsilon_o} \cdot {\lambda r d\theta \over r^2 + z^2}\cdot {r \over \sqrt{r^2 + z^2}} \int_0^{2\pi} \hat r d\theta$$


So what we need is:
$$\int_0^{2\pi} \hat r d\theta = 0$$

Since $\hat r$ is not a constant vector, we rewrite it in a basis that is constant:

$$\hat r = \hat x \cos \theta + \hat y \sin \theta$$

So
$$\int_0^{2\pi} \hat r d\theta = \int_0^{2\pi} (\hat x \cos \theta + \hat y \sin \theta) d\theta$$

Can you see that this is zero?

 

[MinaKaiser]

I dont' get it , Why r is not a constant vector ?

 

[I like Serena]

I dont' get it , Why r is not a constant vector ?

Here's a picture of cylindrical coordinates:

At some point (r, theta, z) we have a local basis which contains $\hat r$.
But if theta is increased, the direction of $\hat r$ changes.