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Electrostatics problem: solve for where E=0 given a configuration of charges

  1. Jan 29, 2012 #1
    Hey everyone, thanks for taking a look at this. I was hoping you could look over this to make sure I solved this correctly. Thanks for you time.

    1. The problem statement, all variables and given/known data

    ---- "+" ---- "+" ---- "-" ----

    Given this configuration of charges where the plus and minus indicate the magnitude of charge and the distance between each charge on the line is 1 unit, solve for the point where E=0.

    2. Relevant equations

    [itex]\Phi = \frac{q}{r}[/itex]

    [itex]E = - \nabla \Phi [/itex]

    [itex]F = [/itex][itex]q[/itex][itex]E[/itex]

    3. The attempt at a solution

    I'm going to place my coordinate system on the middle charge, then use the principle of superposition to add together each individual potential, then take the negative gradient to solve for E. I'll label the charge to the left 1, the middle 2, and the right 3.

    [itex]\Phi_1 = \frac{q}{x+1}[/itex]

    [itex]\Phi_2 = \frac{q}{x}[/itex]

    [itex]\Phi_3 = \frac{-q}{x-1}[/itex]

    [itex]\Phi=\frac{q}{x+1}+\frac{q}{x}+\frac{-q}{x-1}[/itex]

    [itex]E = -\frac{d}{dx}\Phi[/itex]

    [itex]E = -\frac{d}{dx}(\frac{q}{x+1}+\frac{q}{x}+\frac{-q}{x-1})[/itex]

    [itex]E = \frac{q}{(x+1)^2}+\frac{q}{x^2}+\frac{-q}{(x-1)^2}[/itex]

    I made a plot of E(x) using Wolfram Alpha:

    http://www.wolframalpha.com/input/?i=plot+1/(x+1)^2+++1/x^2+-+1/(x-1)^2,+x=-1+to+x=1

    It looks like if I placed a test charge at x~0.5, it wouldn't experience a force, since E=0 there. x=0.5 is between the the positive charge in the middle and the negative charge on the right given I set x=0 on top of the middle charge.

    Apparently the correct answer for E=0 is all the way past the negative charge to the right and I'm not sure what I did wrong. This is what the teacher said.

    "If +1C is to the right from everything, when electron is pulling it
    to the left and the protons are both pushing +1 to the right, the
    protons have more charge (+2e), but the electron is closer, so it
    could balance the protonic force, giving E=0."

    Thanks for the help everyone.
     
  2. jcsd
  3. Jan 29, 2012 #2

    ehild

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    In the formula for the electric potential, [itex]\Phi = \frac{q}{r}[/itex] "r" means the distance from the charge. So the potential terms 1-3 should be written using absolute value.

    [itex]\Phi_1 = \frac{q}{|x+1|}[/itex]

    [itex]\Phi_2 = \frac{q}{|x|}[/itex]

    [itex]\Phi_3 = \frac{-q}{|x-1|}[/itex]

    Your equation is valid only for x>1, so Wolframalpha's other solution x≈0.5 is not valid, as |x-1|=1-x in this case.

    You need to investigate all cases 0<x<1, -1<x<0, x<-1 in addition to the case x>1. You will find that there is a solution for E=0 also at x≈-0.5. It can be explained as the equal charges act with opposite forces and the negative charge is far away.

    ehild
     
  4. Jan 29, 2012 #3
    So you're saying the distance from the charge to where you're measuring the potential can be positive or negative?

    e.g. if I'm standing to the right of the 3rd charge, [itex]\Phi_3 = \frac{-q}{|x-1|} = \frac{-q}{x-1}[/itex]

    BUT, if I'm standing to the left, [itex]\Phi_3 = \frac{-q}{|x-1|} = \frac{-q}{(-(x-1))} = \frac{q}{x-1}[/itex]

    And the same goes for all the others.

    I'm still kind of confused though. Given a new situation where there's only one charge, if I stand +a on one side of it and -a on the other side, I should be standing on the same equipotential in both cases, so the potential shouldn't change, but it changes by a sign.

    It seems like that fact is required though since the E field must point radially toward it or away from it.


    Thank again.
     
    Last edited: Jan 29, 2012
  5. Jan 29, 2012 #4

    ehild

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    NO. The distance can not be negative. Can you say that the other town is -50 km away?

    The potential at distance r from a point charge q is kq/r.
    If there is a point charge q=1 C at x=2 m, and you stand at x=3 m, what distance you are from the charge? And what is your distance from the charge if you stand at x=1 m? And what is the potential in both cases?

    ehild
     
  6. Jan 29, 2012 #5
    so from above

    [itex]\Phi_{right}=\frac{q}{|3-2|}=\frac{q}{1}[/itex]
    [itex]\Phi_{left}=\frac{q}{|1-2|}=\frac{q}{1}[/itex]

    The absolute value ensures that the distance is positive and I get the same potential on each side, which makes sense since I'm standing on a equipotential.

    What I don't understand then is why the potentials change sign depending on what region of the charge configuration I'm standing in from my original post. I can't see it in my head.

    thanks again.
     
  7. Jan 29, 2012 #6

    SammyS

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    The potentials don't change sign. Eliminating the absolute value function makes the potential appear to change sign.

    If you're 1/2 unit away from the 3rd charge, and you're on the axis, then you're either at x = 0.5 , or you're at x = 1.5.

    The following potentials are from Post #2 of this thread.

    (On the left): [itex]\displaystyle \Phi_3(x=0.5) = \left.\frac{q}{x-1}\,\right|_{x=0.5}= \frac{q}{0.5-1}=\frac{q}{-0.5}=-2q\,.[/itex]

    (On the right): [itex]\displaystyle \Phi_3(x=1.5) = \left.\frac{-q}{x-1}\,\right|_{x=1.5}= \frac{-q}{1.5-1}=\frac{-q}{0.5}=-2q\,.[/itex]

    So, when you write the potential in the form you did in Post #2, the potentials appear to be different, but you get the same result once you put in the specific x values.
     
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