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Homework Help: Two point charges. Given one find the other?

  1. Mar 21, 2012 #1
    1. The problem statement, all variables and given/known data

    Two point charges are situated at A and B a distance of 35cm apart in a vacuum. The charge A is a positive charge of 16 μC and the field strength at X, a distance of 12 cm from A, is zero. Calculate:

    (i) The field strength at X due to the charge at A. (I found this to be 9.9 x 10^6 NC^-1. Just in case this is relevant to the next question.)

    (ii) The charge at B.

    2. Relevant equations

    Permittivity of free space ε0 = 8.9 x 10^-12 F m^-1

    F = Q.q / 4πεd^2

    3. The attempt at a solution
    It seems like I don't have enough information but here's an attempt anyway

    F = (16 x 10^-6) / 4π(8.9 x 10^-12)(0.35)^2 = 1.2 x 10^6

    The answer is 58.8 NC so im way off.
  2. jcsd
  3. Mar 21, 2012 #2


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    field at X is zero.
    so,q(A)/(AX)^2 =q(B)/(XB)^2
    you know q(A),AX,XB.
    solve it to get q(B)=58.8*10^-6
  4. Mar 21, 2012 #3
    How do I find XB if I dont know B?

    F = q(B) / 4πε0(0.23)^2 = q(B) / 5.91637

    With this I get a cancellation of q(B) and can't find it.

    Thanks for the formula, I have never seen it before.
  5. Mar 21, 2012 #4
    A and B are lying on a line at a distance of 35 cms.

    Think of A lying on 0 on the number line and B at 35.

    Now X can be at 12 or -12.

    Field strength at a point is given by kq/r^2( You have written the equation of force.) You can guess the direction depending on sign of charge.

    At X field is 0.

    So Q(A)/144 = Q(B)/R^2

    where R can have two values.

    You know Q(A).
    Plug in and find two values of Q(B)
  6. Mar 21, 2012 #5
    (9.9 x 10^6) / 144 = Q(B) / 529

    Q(B) = 36.37 x 10^6

    It's definitely +12, there's a diagram that shows X between the two points. Sorry I forgot to mention.

    I thought your explanation was crystal clear so I'm really at a loss as to why this is yielding the wrong answer.
    Last edited: Mar 21, 2012
  7. Mar 21, 2012 #6


    (9.9 x 10^6) / 144 = Q(B) / 529

    This step is wrong.

    Instead of the 9.9x10^6 you have to actually plug in Q (A) which is charge of A.

    See Coulombs law tells us Force is kq(1)q(2)/r^2 ( with appropriate direction).

    Now electric field strength of a charged particle having charge Q at a distance r is


    Its like plugging in q(2) as 1 in force equation if we want to find electric field due to q(1).
  8. Mar 21, 2012 #7
    Oh, I see. I really should have known that... I think I need some sleep.

    Thank you for your patience and very clear explanation.
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