# Electric field of a charged ring.

1. Sep 1, 2011

2. Sep 2, 2011

### DougUTPhy

I kind of follow your work, but I'm afraid I don't see the complete equation needed, so I will go ahead and derive it for you.
Alrighty, here's a little picture to help us out.

Let's keep this in mind throughout the whole problem.

We know by symmetry that the electric field due to the ring at point P is going to be straight up in the Z direction. In the picture, I have the electric field element $$$dE$$$ due to a charge element $$$dq$$$.
Let us define that charge element. Because we have a ring, it will have a linear charge density (charge per unit length), $\lambda = Q/l = Q/2\pi R$, where L (lower case in the equation) is the circumference of the ring, it's length. Let's look at another picture for substitution reasons:

This picture shows us that $$$dl=Rd\theta$$$.
So now we can say (using the equations above) $$$dq=\lambda dl = \frac{Q}{2\pi R} Rd\theta = \frac{Q}{2\pi} d\theta$$$

Let's not forget any of this as we move on!

We know from the equation of an electric field that $$$dE = k \frac{dq}{r^2}$$$. Where $$$r=\sqrt{R^2+x^2}$$$ - from the first picture.
Like said before, we know that the electric field is going to be vertically up in the Z direction, so we want to find the electric field straight up ($dE_z$) due to our charge element, we can use trig from our first picture to say this: $dE_z = dE \cos{\theta}$
And we know from our first picture again that $\cos{\theta} = R/r$

Now let's do some substitution using A LOT of what we have done before:

$dE_z = dE\cos{\theta}$

$dE_z = k \frac{dq}{r^2} \frac{x}{r}$

$dE_z = k \frac{Q/(2\pi)}{x^2+R^2} \frac{x}{\sqrt{x^2+R^2}} d\theta$

$\int dE_z = \frac{k}{2\pi} \frac{Qx}{(x^2+R^2)^{3/2}} \int_{0}^{2\pi} d\theta$ <-- integrating from 0 to 2pi (the whole ring) = 2*pi

And now we have our solution:

$E_z = \frac{kQx}{(x^2+R^2)^{3/2}}$

Now all you do is plug your numbers in

By the way this is not the only way to reach this solution, just the way I was taught. If you google "electric field due to a ring of charge" - you may find some other ways

Last edited: Sep 2, 2011
3. Sep 2, 2011

### NeverSummer

Thanks for the great explanation, I got it!