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Work required to bring charge from infinity

  1. Apr 16, 2015 #1
    1. The problem statement, all variables and given/known data
    Two point charges are located on the x-axis, q1 = -e at x=0 and q2 = +e at x=a. Find the work that must be done by an external force to bring a third point charge q3 = +e from infinity to x = 2a.

    2. Relevant equations
    W = -ΔU

    3. The attempt at a solution
    W = -(Ufinal - Uinitial)
    Uinitial = 0 because it is at infinity.
    So W = -Ufinal = -kq3(q1/r1 + q2/r2). After doing this I get:
    -(e^2)/(8πεa) which has the correct magnitude but wrong sign. My book uses W = U instead of W = -U. Is there a reason for this? Work is equal to change in NEGATIVE potential energy, so why would it be positive?
     
  2. jcsd
  3. Apr 16, 2015 #2

    H Smith 94

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    This is a issue with conventions stemming from the fact you've used a negative Coulomb potential. By convention the Coulomb force ##\vec{F}(\vec{r})## is negative to imply attraction: [tex] \vec{F}(\vec{r}) = -\frac{q_1q_2}{4\pi\epsilon_0}\frac{1}{r}\cdot \hat{r} [/tex] which gives rise to the Coulomb (or electrostatic) potential ##U(\vec{r})## found by, [tex] U(r) = \int F(r) \, dr [/tex] and so [tex] U(r) = -\frac{q_1q_2}{4\pi\epsilon_0} \int \frac{1}{r^2} dr = -\frac{q_1q_2}{4\pi\epsilon_0} \left( - \frac{1}{r} \right) = \frac{q_1q_2}{4\pi\epsilon_0} \frac{1}{r}.[/tex]
    Either that, or [tex] \Delta U = U_\text{final}-U_\text{initial}.[/tex]
     
  4. Apr 17, 2015 #3

    rude man

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    You know from the basic charge layout that it will take positive work to bring the positive charge to x=2a. Try to think that way in general instead of relying on the math to give you the correct signs.
     
  5. Apr 17, 2015 #4

    ehild

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    The work of the electric field is equal to change in negative potential energy, but the work that must be done by an external force is opposite.
     
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