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Homework Help: Work required to move a charged ball

  1. Feb 1, 2009 #1
    1. The problem statement, all variables and given/known data
    A ring of diameter 7.10 cm is fixed in place and carries a charge of 5.00 μC uniformly spread over its circumference. How much work does it take to move a tiny 3.40 μC charged ball of mass 1.70 g from very far away to the center of the ring?

    2. Relevant equations
    I tried using: [tex]W=-\Delta U[/tex]

    3. The attempt at a solution
    We know that when the ball is very far away U=0J

    Then to find the potential energy when the ball is in the center of the ring I did:
    Change in charge is [tex]5.00X10^{-9}C/2\pi(0.0710m)=1.12X10^{-8}C/m[/tex]

    [tex]\int \frac{k*q*dq}{r} ds[/tex] from 0 to 2pi.

    [tex]\frac{k*q*dq}{r} \int ds[/tex] from 0 to 2pi.

    [tex]\frac{8.988X10^{9}N*m^2/C^2*3.40X10^{-9}C*1.12*10^{-8}C/m}{0.0710m} \int ds[/tex] from 0 to 2pi [tex]=3.031X10^-5J[/tex].

    [tex]W=-\Delta U=-(3.031X10^-5J-0J)=-3.031X10^-5J[/tex]
     
  2. jcsd
  3. Feb 1, 2009 #2

    Delphi51

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    Homework Helper

    I'm having trouble following your integral from the first line. I'm thinking V = kq/r for a point charge so dV = k/r*dq, where r is the radius of the ring.
    Ah, you are doing energy E = QV so dE = QdV = kQ/r*dq
    I'm using q for the charge on the ring, Q for the charge on the particle.
    Your s must be something to do with the arc length going round the ring.
    So dq = q/(2*pi*r)*ds
    Using s = r*A where A is the angle running from 0 to 2pi, you get ds = r*dA and
    dq = q/(2*pi)dA
    so the integral is dE = kQq/(2*pi*r)*dA from A = 0 to 2*pi.
    The nice thing about this integral is that nothing depends on the angle so
    integral 0 to 2*pi of dA is just 2*pi.
    E = kQq/r.
    Well, I guess we didn't need to integrate at all! Because all the charge is the same distance from the center point.
     
  4. Feb 1, 2009 #3
    I'm still slightly confused. So I can find E at the center of the ring using E=kQq/r. But, since W=-(U_f-U_i) how do I get U from E? Since E=kq/r^2 and U=kq_1*q_2/r would you just multiply the E you get by qr and if so what q would you use: the charge on the ring or the charge on the ball? Thanks for your help.
     
  5. Feb 2, 2009 #4
    Try this approach. Find E as a function of r along the axis of the ring using integration. Then

    [tex]V=-\int_{\infty}^{r}\mbox{Edr}[/tex]

    The work done is

    [tex]W=Vq[/tex]

    Is the mass in a gravitational field also and if so what is the orientation of the ring axis with the field?
     
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