Charge Distribution.

1. Aug 31, 2006

mb85

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. Aug 31, 2006

e(ho0n3

If you know the max. value, multiply by 1/4 and then solve for q/Q.

3. Aug 31, 2006

lightgrav

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.

4. Aug 31, 2006

tim_lou

let the ratio be r:
q=rQ
rQ(Q-rQ)=Q^2/16

solve it...