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Charge Distribution.

  1. Aug 31, 2006 #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 will the electrostatic force between the two parts have 1/4 of the maximum possible value?

    q1=Q when the charge is transferred its Q - q
    q2=0 becomes q after transfer

    F = K(|q1||Q-q|)/r^2

    dF(q)/dq = 0

    0 = dF/dq = K/r^2 d/dq (qQ - q^2)
    then Q - 2q = 0
    so q = Q/2

    so Fmax is 1/2

    My question is, where do i apply the 1/4 of the max. value? im rather confused. I thought it would just be 1/4 of the max.

    any help would be appreciated. thanks.
     
  2. jcsd
  3. Aug 31, 2006 #2
    If you know the max. value, multiply by 1/4 and then solve for q/Q.
     
  4. Aug 31, 2006 #3

    lightgrav

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    Homework Helper

    gee, that was ambiguous ...

    make sur eyou CONSERVE charge ...

    They want the Force to be 1/4 of the maximum Force.
    Ignoring r^2 and k, F_max = (Q/2)(Q/2) = Q^2 /4 .
    . . . You want : (q)(Q-q) = Q^2 /16.
    F = (Q/8)(7Q/8) = Q^2 7/64 is almost 2x too strong.
     
  5. Aug 31, 2006 #4
    let the ratio be r:
    q=rQ
    rQ(Q-rQ)=Q^2/16

    solve it...
     
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