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Point charge equilibrium w/ line charge problem

  1. Oct 7, 2011 #1
    1. The problem statement, all variables and given/known data
    There is a charge located at the origin of magnitude 5nC. There exists a vertical line of charge, 2 meters in length, that runs from (2m,-1m) to (2m,1m).
    a) What uniform linear charge density must this line of charge have in order that the net force on a charge placed at (5.0m, 0m) has no net force acting on it?
    b) What would be the voltage at this point?
    c) aside from x = +/- (infty), where (if at all) would the voltage be zero?

    Q1= 5nC. I called the line charge Q2. This could be very wrong. Q3 is the unknown.

    2. Relevant equations
    F = (Q1Q2)/r^2
    (delta)V = Ke(integral) dq/r
    dq = (lambda)(dx)

    3. The attempt at a solution
    I treated the line charge as a point charge by saying:
    F13 - F23 = 0 (the force of 1 on three and the force of the line charge acting on 3 must be zero)

    This meant that the total charge for Q2 was -1.8nC. To find rho (charge density) I just took 1.8nC and divided by 2, since that is the length of the line of charge.

    part b is a little trickier but I think it must be something akin to integrating from -1 to 1 for the integral of (Ke)(dq)/(y), but that doesn't seem quite right since we have Q3 to think about and also because "r" (which in this equation is "y") changes as we go from -1 to 1, so r must be something besides just "y". Obviously, dq is now (lambda)(dy). Ke and lambda are constants, so it should just be dy/r, whatever r is. could r = [(1-y)^2 + 4]?

    I have seen (and know how to solve) point charge problems on an X axis that look a lot like this, but the fact that the Q3 is unknown (and my method for solving for Q2 could be wrong) and the y integration is kind of throwing me. Help? (but not too much)


  2. jcsd
  3. Oct 8, 2011 #2
    First of all it is important to recognise that you have just one unknown and that is q2. Think of q3 as a unit test charge. You do not need its value in this problem.

    Also I think you method of calculating q2 is not quite right -- you need to perform an integration and cannot treat the line s a point charge.

    Tell me how you get on.
  4. Oct 8, 2011 #3
    I saw that yesterdayafternoon. So we use the superposition formula and account for all charges, in this case 2, and integrate q2. The integral should look like this?
    (Lambda)*(int) dy/r^2, where r is the square root of ([1-y)^2+2^2], yes?


  5. Oct 8, 2011 #4
    Superposition principle is the correct approach as you point out and q1 is easier. For the integral, the expression in terms of r is correct (ignoring some constants) but not r itself which should be:

    y^2 + 3^2

    Check that you agree by relooking at the geometry (the coordinates are in terms of (x,y) )
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