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Homework Help: External force work on a particle

  1. Sep 13, 2009 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    3. The attempt at a solution

    To be honest, I have no idea how to start this problem.
  2. jcsd
  3. Sep 13, 2009 #2


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    Have you been told about the potential energy for two charges? That would be the key equation to use here.
  4. Sep 13, 2009 #3
    yeah. the potential energy U=qV=(kQ1Q2)/d. I don't know what to do with it.
  5. Sep 14, 2009 #4


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    You can think of the ring as being composed of many small pieces of charge. Let's call each small bit of charge Δq.

    So, what is the potential energy due to the -5.5 pC point charge and one of the Δq charges?
  6. Sep 14, 2009 #5
    ΔU = k(ΔQ)Q2/d
  7. Sep 17, 2009 #6
    The only thing I can work out is that if you break the ring up into differential elements, compute the Potential at the initial point due to that dV element, integrate over 2pi, you'll get the total potential at the initial point due to the ring.

    [tex]V_i = \frac{kQ}{r}[/tex]
    [tex]dV= \frac{kdQ}{r}[/tex]
    [tex]dQ = \lambda ds[/tex]
    [tex]ds = r d\Theta[/tex]
    [tex]dV= \frac {k \lambda r d \Theta}{r}[/tex]
    [tex]dv=k \lambda d\Theta[/tex]

    [tex]V_i=\int_0^{2\pi}{k \lambda d\Theta}[/tex]
    [tex]V_i= k\lambda \int_0^{2\pi} d\Theta[/tex]
    [tex]V_i=k\lambda 2\pi[/tex]

    [tex]\lambda = \frac{Q_1}{L}[/tex]
    [tex]\lambda = \frac {Q_1}{2\pi r}[/tex]

    [tex]V_i= k 2\pi\frac {Q_1}{2\pi r}[/tex]
    [tex]r=\sqrt{R^2 + d^2}[/tex]
    [tex]V_i= \frac{KQ_1}{\sqrt{R^2 + d^2}}[/tex]

    And if we do the same for the charge at the origin [itex]V_f[/itex], we get:

    [tex]V_f= \frac{KQ_1}{R}[/tex]

    So, if the potential at the starting point is [tex]V_i= \frac{KQ_1}{\sqrt{R^2 + d^2}}[/tex] then the potential energy at that point is [itex]Q_2V_i[/itex] or [tex]U_i= \frac{KQ_1Q_2}{\sqrt{R^2 + d^2}}[/tex]

    And the potential energy at the ending point would be [itex]Q_2V_f[/itex] or [tex]U_f= \frac{KQ_1Q_2}{R}[/tex]

    So the change in potential energy would be [itex]U_f-U_i[/itex], correct? And this is equal to the work being done by the electric field present ([itex]U_f-U_i=W_{e}[/itex])?? If so, then the work done by ME would be the negative of that work? ([itex]W_{app}=-W_{e}[/itex])

    Am I even close??
  8. Sep 17, 2009 #7


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    That's the right idea :smile:

    Also, the work done on the charge would equal the change in potential energy. It's not clear from what you say if you are throwing an extra minus sign in here. In other words, positive work is done by the external force if the potential energy increases.
  9. Sep 17, 2009 #8
    I have been trying to learn how potential energy and work are related for months now and can't understand it. I've had other texts try to teach me, teachers, peers, and I can't understand it. I have absolutely no idea how to visualize work, or potential energy. I just can't understand it.

    Like, when the question asks "how much work would an external force do..." I have absolutely no idea what that means, or how to visualize it, or even how work and the force are relating to the object in question.
  10. Sep 18, 2009 #9


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    Yeah, it can take a while for this stuff to sink in.

    This might help get a more intuitive feel of what energy is (read from p. 69 to the 1st paragraph of p. 72):

    I like to picture potential energy in terms of going uphill or downhill. Doing work is a way of adding to the energy of something, for example rolling a ball up a hill will raise it's potential energy. Notice that you must push the ball in the same direction it is moving in order to do this -- the work done (F·Δx) is positive and the ball's energy increases when it is rolled uphill.

    On the other hand, if you hold a ball as it rolls downhill, and don't allow it's speed to change as it rolls, then you must pull back on the ball as it rolls forward. The force on the ball is opposite to the direction it moves -- the work done (F·Δx) is negative, and the ball's energy becomes lower when it's allowed to roll down a hill in this way.
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