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Electric field of a charged ring.

by NeverSummer
Tags: charged, electric, field, ring
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NeverSummer
#1
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
E=kQ/(r^2)

3. The attempt at a solution

dE=kdQ/(r^2)=kdQ/(sqrt(h^2+r^2))=∫(k/4.206)dQ

=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
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DougUTPhy
#2
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
NeverSummer
#3
Sep2-11, 09:40 PM
P: 4
Thanks for the great explanation, I got it!


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