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Coulomb's Law Problem and charged particles

  1. Jan 20, 2005 #1
    I cannot figure out this problem. I mean, I can't really even figure out what it is asking. I am not expecting anyone to solve it for me or anything. I just need some help as to how to get started.

    Particles A and B Figure 22-25a shows charged particles A and B that are fixed in place on an x axis. Particle A has an amount of charge of |qA| = 9.00e. Particle C, with a charge of qC = +9.00e, is initially on the x axis near particle B. Then particle C is gradually moved in the positive direction of the x axis. As a result, the magnitude of the net electrostatic force net, B on particle B due to particles A and C changes. Figure 22-25b gives the x-component of that net force as a function of the position x of particle C. The plot has an asymptote of net, B = 1.553 10-25 N as x . As a multiple of e, what is the charge qB of particle B?

    Figure 22-25a is basically a plot of three points, with point (a) on the negative x axis, point (b) at the origin, and point (c) on the positive x axis. Then there is also a graph of Fnet-->b. I don't know if you can figure out the problem without the graph, but any help on how to do it would be greatly appreciated!!!!!

    Thanks,
    Beverly
     
    Last edited: Jan 20, 2005
  2. jcsd
  3. Jan 21, 2005 #2

    Andrew Mason

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    Try writing out the Coulomb force of C on B and A on B. The vector sum of those two forces will be the net force. F should be a function in the form:

    [tex]F = kq_b(q_a/A^2 + q_c/x^2)[/tex]

    Now the first issue is: is q_a negative or positive? Then determine if q_b is positive or negative.

    I can't figure out from your information what the graph looks like. You should be able to figure the kind of charge for q_a and q_b from just looking at the direction of F(x). If the F(x)<0 as x approaches 0 then q_b would be negative. If it is greater than 0, q_b would be positive.

    I can't tell from your description what the equation of the asymptote is. Can you provide it?

    AM
     
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