Register to reply

Electric field of a charged ring.

by NeverSummer
Tags: charged, electric, field, ring
Share this thread:
Sep1-11, 10:06 PM
P: 4
[QUOTE=bionut;3480950]1. The problem statement, all variables and given/known data

As part of the optics in a prototype scanning electron microscope, you have a uniform circular ring of charge Q=6.10 microCoulombs and radius R=1.30 cm located in the x-y plane, centered on the origin as shown in the figure.

What is the magnitude of the electric field, E at point P located at z=4.00 cm ?

2. Relevant equations

3. The attempt at a solution


=2.14*10^9 Q

Now, I'm not sure I've made it thus far correctly, so if I've gone wrong, please let me know. Either way, I have no idea what my limits of integration would be for this, so I'm pretty much stuck here.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
Phys.Org News Partner Science news on
Scientists develop 'electronic nose' for rapid detection of C. diff infection
Why plants in the office make us more productive
Tesla Motors dealing as states play factory poker
Sep2-11, 12:39 AM
P: 11
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 [itex]\begin{equation} dE \end{equation}[/itex] due to a charge element [itex]\begin{equation} dq \end{equation}[/itex].
Let us define that charge element. Because we have a ring, it will have a linear charge density (charge per unit length), [itex]\lambda = Q/l = Q/2\pi R[/itex], 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 [itex]\begin{equation} dl=Rd\theta \end{equation}[/itex].
So now we can say (using the equations above) [itex]\begin{equation} dq=\lambda dl = \frac{Q}{2\pi R} Rd\theta = \frac{Q}{2\pi} d\theta \end{equation}[/itex]

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

We know from the equation of an electric field that [itex]\begin{equation} dE = k \frac{dq}{r^2} \end{equation}[/itex]. Where [itex]\begin{equation} r=\sqrt{R^2+x^2}\end{equation}[/itex] - 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 ([itex]dE_z[/itex]) due to our charge element, we can use trig from our first picture to say this: [itex]dE_z = dE \cos{\theta}[/itex]
And we know from our first picture again that [itex]\cos{\theta} = R/r[/itex]

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

[itex]dE_z = dE\cos{\theta}[/itex]

[itex]dE_z = k \frac{dq}{r^2} \frac{x}{r}[/itex]

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

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

And now we have our solution:

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

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
Sep2-11, 09:40 PM
P: 4
Thanks for the great explanation, I got it!

Register to reply

Related Discussions
Electric Field of a Uniformly Charged Ring Advanced Physics Homework 3
Electric Field of a Uniformly Charged Ring Introductory Physics Homework 3
Electric field of a charged ring Introductory Physics Homework 2
Charged ring and electric field problem Introductory Physics Homework 0
Electric field due to a charged ring off-axis Classical Physics 1