1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Voltage from Ring of Charge

  1. Mar 9, 2008 #1
    [SOLVED] Voltage from Ring of Charge

    1. The problem statement, all variables and given/known data
    A plastic rod has been bent into a circle of radius R = 6.40 cm. It has a charge Q1 = +2.40 pC uniformly distributed along one-quarter of its circumference and a charge Q2 = -6Q1 uniformly distributed along the rest of the circumference (Figure 24-39). Take V = 0 at infinity. What is the electric potential at the center C of the circle?

    2. Relevant equations
    [tex]dV = \frac{dq}{4\pi\epsilon_0r}[/tex]
    [tex]\lambda = \frac{q}{L}[/tex]

    3. The attempt at a solution
    [tex]dq = \lambda r d \theta[/tex]
    [tex]\lambda = \frac{q}{2 \pi r}[/tex]
    [tex]V = \int_{0}^{\phi}{\frac{\lambda d \theta}{4\pi\epsilon_0}} = \frac{\lambda \phi}{4 \pi \epsilon_0} = \frac{q \phi}{8 \pi^2 \epsilon_0 r}[/tex]
    [tex]V_1 = \frac{(2.4*10^-12)(\pi / 2)}{8 \pi^2 \epsilon_0 (0.064)} = 0.0843V[/tex]
    [tex]V_2 = \frac{-6(2.4*10^-12)(3\pi / 2)}{8 \pi^2 \epsilon_0 (0.064)} = -1.517V[/tex]
    [tex]V_{net} = V_1 + V_2 = -1.433V[/tex] (Incorrect)
  2. jcsd
  3. Mar 9, 2008 #2
    You should start from integrating the Differential electric field strength that a point charge would exert at that point
  4. Mar 9, 2008 #3
    [tex]\lambda = \frac{q}{L} = \frac{q}{2 \pi r} = \frac{dq}{ds}[/tex]
    [tex]ds = r d\theta[/tex]
    [tex]dq = \lambda r d\theta[/tex]
    [tex]dE = \frac{cos\theta dq}{4 \pi \epsilon_0 r^2}[/tex]
    [tex]E = \int_{\frac{-\phi}{2}}^{\frac{\phi}{2}}{\frac{\lambda cos\theta d\theta}{4 \pi \epsilon_0 r}} = \frac{\lambda(sin\frac{\phi}{2} - sin\frac{-\phi}{2})}{4 \pi \epsilon_0 r} = \frac{qsin\frac{\phi}{2}}{4 \pi^2 \epsilon_0 r^2}[/tex]

    [tex]E = \frac{dV}{dr}[/tex]
    [tex]V = \int{\frac{qsin\frac{\phi}{2}dr}{4 \pi^2 \epsilon_0 r^2}} = \frac{q sin\frac{\phi}{2}}{4 \pi^2 \epsilon_0} \int{\frac{dr}{r^2}} = \frac{-q sin\frac{\phi}{2}}{4 \pi^2 \epsilon_0 r}[/tex]

    [tex]V_1 = \frac{-2.4*10^{-12} sin\frac{\pi/2}{2}}{4 \pi^2 \epsilon_0 (0.064)} = -0.076V[/tex]
    [tex]V_2 = \frac{6 * 2.4*10^{-12} sin\frac{3\pi/2}{2}}{4 \pi^2 \epsilon_0 (0.064)} = 0.455V[/tex]
    [tex]V_{net} = V_1 + V_2 = 0.379 V[/tex] (Incorrect)
  5. Mar 9, 2008 #4
    your expression for [tex]\lambda[/tex] is wrong. the charge isn't distributed along the whole circle, but along a secion of 1/4 or 3/4 of the circle. The rest is OK.

    The whole problem gets much simpler, if you see that V doesn't depend on where the charge is, but only on the distance. Since all the charges are at the same distance from the center, you can move them all to a single point at distance r from the cednter
  6. Mar 9, 2008 #5
    Hadn't even considered that the density equation could be wrong - I figured the mistake was somewhere in the following page of calculus. Thanks for pointing that out, that must be why the answers are incorrect.

    Realized this and got the answer correct just prior to returning here to mark this thread solved. Thanks for your help!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Voltage from Ring of Charge