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Optimization of Coulomb's Law

  1. Sep 3, 2011 #1
    Of the charge Q initially on a tiny sphere, a portion q is to be transferred to a second, nearby sphere. Both spheres can be treated a particles. For what value of q/Q>0.5 will the electrostatic force between the two parts have 1/5 of the maximum possible value?

    F = [ k(Q-q)(q) ] / r^2
    dF/dq = (k/r^2) d/dq(Qq - q^2)
    = (k/r^2)(Q-2q) = 0
    Q - 2q = 0
    q = (1/2)Q --> Thus, this will give me the maximum electrostatic force between the two particles. However, the question is asking about the situation where it is 1/5 of the maximum value.

    F = [ k(Q-q)(q) ] / r^2, and using q = (1/2)Q or 2q = Q,
    F/5 = [k(2q - q)(q)] / 5r^2
    = (kq^2) / 5r^2 --> This is 1/5 of the maximum force.

    Thus, to find q/Q:
    F = [ k(Q-q)(q) ] / r^2 = (kq^2) / 5r^2 and I found q/Q to be 0.83333. This answer is wrong, however. Can anyone point me in the right direction?
  2. jcsd
  3. Sep 3, 2011 #2


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    Homework Helper

    Mixing symbols a bit

    F = [ k(Q-q)(q) ] / r^2 should be 1/5 of k(Q/2)^2) / r^2

    The force with the different charges, is only 1/5th of the force with the equal [half the original] charges
  4. Sep 3, 2011 #3


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    Staff: Mentor

    By writing 2q = Q you're forcing a particular value upon q rather than solving for a new one.

    Suppose that fmax is the maximum force, and f is the force you're looking for. Then the desired condition is:
    [tex]\frac{f}{f_{max}} = \frac{1}{5}[/tex]
    But you've already determined that
    [tex] f = k \frac{(Q - q)(q)}{r^2} [/tex]
    [tex] f_{max} = k \frac{Q^2}{4 r^2}[/tex]
    Use both of these expressions to set up the ratio and solve for the 'new' q.
  5. Sep 4, 2011 #4
    I did F / Fmax and still got the answer to be 0.833333.
  6. Sep 4, 2011 #5


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    Staff: Mentor

    Can you show your work?
  7. Sep 4, 2011 #6

    Ray Vickson

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    You want q*(Q-q) = (1/5)*{max of q*(Q-q)} = (1/5)*(1/2 Q)^2 = Q^2/20, so if x = q/Q you want x(1-x) = 1/20.

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