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.

 

[sardar]

=0

The electric field of a ring with radius r, at a distance z from the center along the axis of the ring, can be calculated using the integral form of Coulomb's law:

E = ∫dE = ∫ kλdl/r^2

Where k is the Coulomb constant, λ is the uniform line charge, and dl is the element of length along the ring.

In cylindrical coordinates, dl = rdθ and r^2 = z^2 + r^2. Substituting these values into the integral, we get:

E = ∫ kλrdθ/(z^2 + r^2)

Since the ring is symmetric around the z-axis, the integral limits are from 0 to 2π. This gives:

E = kλ/(z^2 + r^2) ∫0^2πrdθ

Using the unit vector rˆ in cylindrical coordinates, we can rewrite the integral as:

E = kλ/(z^2 + r^2) ∫0^2πrˆrdθ

This integral can be easily evaluated to give:

E = kλ/(z^2 + r^2) (2πrˆ)

Simplifying further, we get:

E = 2πkλrˆ/(z^2 + r^2)

From this expression, we can see that the radial component of the electric field, given by Erad = 2πkλr/(z^2 + r^2), is dependent on the distance from the center of the ring (r) and the distance along the axis (z). However, as z approaches 0, the radial component becomes 0, which means that the radial component of the electric field is canceled out due to symmetry.

This can also be shown mathematically by taking the limit as z approaches 0:

lim(Erad) = lim(2πkλr/(z^2 + r^2))

= 2πkλr/(r^2)

= 0

Therefore, the radial component of the electric field is indeed canceled out at the center of the ring, proving the statement mathematically.