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Electric field of a charged ring.

  1. Sep 1, 2011 #1
     
  2. jcsd
  3. Sep 2, 2011 #2
    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.
    [​IMG]

    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:

    [​IMG]

    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
     
    Last edited: Sep 2, 2011
  4. Sep 2, 2011 #3
    Thanks for the great explanation, I got it!
     
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