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Charges in 1-D (homework help please!)

  1. Jan 25, 2006 #1
    Hi, here is a homework problem which i'm absolutely stuck on.

    A charge of -3.2 × 10-9 C is at the origin and a charge of 7.3 × 10-9 C is on the x-axis at x = 3 m. At what location on the x-axis is the electric field zero?

    I've asked my professor for help on this, and his explanation wasn't very good. What I got out of it was the following. The location will exist somewhere to the left of the origin. Also, this may be wrong, but am I supposed to solve for r in the following equation: kQ1\r^2 = kQ2 \ (r^2 +3)? I've tried that, but a bunch of the answers I come up w/ seem not to be right. Any help would be greatly appreciated. Thanks in advance.
     
  2. jcsd
  3. Jan 25, 2006 #2

    robphy

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    Check your equation... first, for consistency with units.
    What is the reasoning behind that equation?
     
  4. Jan 25, 2006 #3

    berkeman

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    Well, I could be wrong, but it looks to me like E=0 only at +/- infinity for that set of charges.

    Remember that E=-GRAD(V), and that for point charges, V = Q/[4PI*epsilon0*r] for each charge. So you can plot V=f(x) for the two point charges in your problem... You will get V = negative infinity at x=0, and V= positive infinity at x=3m. Between x=0 and x=3m, V comes up from the negative assymptote and heads for the positive assymptote at x=3. Past x=3, V comes down from the positive assymptote and heads to 0 as x-->infinity. For x<0, V comes up from the negative assymptote at x=0 and heads to -0 as X-->negative infinity.

    Since E=-GRAD(V), V has to stop changing with respect to x for E=0 to be true. That only seems to happen at +/- infinity..... I wonder what I'm missing here... Anybody?
     
  5. Jan 25, 2006 #4

    robphy

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    Note that the two charges have different magnitudes.
     
  6. Jan 25, 2006 #5
    I honestly have no idea what the reasoning behind the equation is. That's just what I got out of my professors explanation. What I got out of it was that I needed to set the two equal to each other as I did and solve for 'r'. Whether or not that's right, I have no idea. I'm completely lost w/ this question. I don't really even know where to begin to be honest.
     
  7. Jan 25, 2006 #6

    robphy

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    By now, you should know that an electric charge has an associated electric field, directed radially away from the charge (assumed positive, otherwise it is directed towards. When you have two charges in space, their electric field patterns are added vectorially... yielding the "electric field due to those two charges". Your job is to locate a certain point [on the line through those charges] where the "electric field due to those two charges" is zero. How does this point have electric field zero? Well... it must be that the electric field vector from one charge plus that of the other charge is the zero vector. How can two vectors add up to be the zero vector?
     
  8. Jan 25, 2006 #7

    berkeman

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    Interesting. Yeah, I considered that at first, but dismissed it too early. The potential V actually goes positive for parts of the negative x axis. To the OP -- now that you know that V(x) goes positive for part of the negative x axis, how can you use a differentiation of V(x) to find where E=0?
     
  9. Jan 25, 2006 #8

    robphy

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    It's likely that the OP has not studied the electric potential yet.
    Using the electric field alone, the problem can be solved with simple algebra... calculus is not necessary.

    Some good bookkeeping may be needed because of the three regions created by the two charges on this line. With a little physical intuition, one can pick out the region where E=0.
     
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