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Homework Help: Work against an electric field due to a point charge

  1. May 15, 2010 #1
    1. The problem statement, all variables and given/known data
    I need help seeing if I did this right....

    This is from Fundamentals of Physics, 8th edition, volume 2. By Jearl Walker.
    Chapter 24, problem 88.

    A particle of positive charge Q is fixed at point P. A second particle of mass m and negative charge -q moves at a constant speed in a circle of radius [tex]r_{1}[/tex] centered at P. Derive an expression for the work W that must be done by an external agent on the second particle to increase the radius of the circle of motion to [tex]r_{2}[/tex].


    2. Relevant equations
    [tex]

    W_{\vec{E}} = -\int_{r_{1}}^{r_{2}} \vec{E}\cdot d\vec{r}

    [/tex]
    [tex]
    \vec{E} = \frac{1}{4\pi\epsilon_{0}} \frac{Q}{r^2} \hat{r}
    [/tex]

    3. The attempt at a solution

    Here's what I'm thinking. The work against the electric field is just the integral above. Since the particle stays on an equipotential surface, I should only have to worry about work outward, right?

    Here's what I have:

    [tex]

    W_{total} = W_{\vec{E}} = \left(\frac{1}{r_{2}} - \frac{1}{r_{1}}\right) \left(\frac{Q}{4\pi\epsilon_{0}}\right)

    [/tex]

    Please help point out if I've done something wrong!


    -Mike
     
    Last edited: May 16, 2010
  2. jcsd
  3. May 16, 2010 #2

    AEM

    User Avatar

    Your idea is correct, but your first equation is wrong. Usually gravity is ignored because the gravitational force is small compared to the electric force.
     
  4. May 16, 2010 #3
    I think I figured out where I went wrong in my equation.

    So I'm confused, where does the mass 'm' come in?
     
  5. May 16, 2010 #4

    AEM

    User Avatar

    In brief, Work = Force times distance. So your equation for work should be

    [tex]


    W_{\vec{E}} = -\int_{r_{1}}^{r_{2}} q \vec{E}\cdot d\vec{r}


    [/tex]

    Where the force is qE, q being the charge that's moved, and E is the field due to the charge Q.
     
  6. May 16, 2010 #5

    AEM

    User Avatar

    Also, I should mention that you are correct. Work is done only when the charged is moved along a radial path.
     
  7. May 16, 2010 #6

    AEM

    User Avatar

    Besides the approach that I mentioned in my previous post, you can also use the fact that Work = qV where q is the charge that's being moved, and V is the change in electric potential.

    What you calculated in your work below, was the change in electric potential, not the work. All you have to do is change the "W" to "V" and multiply through by the charge q.

    You were on the right track. Just forgot the definition, W = qV. So now you have two ways to look at it.

     
  8. May 16, 2010 #7
    Thank you so much!

    I've been nearly pulling my hair out over this.
     
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