How Does Charge Transfer Affect Electrostatic Force Between Spheres?

AI Thread Summary
The discussion focuses on determining the charge transfer ratio (q/Q) needed for the electrostatic force between two spheres to be one-fourth of its maximum value. Initially, one sphere has charge Q, and after transferring charge q to the second sphere, the first sphere's charge becomes Q - q. The maximum force is calculated as F_max = Q^2 / 4, leading to the equation (q)(Q - q) = Q^2 / 16 for the desired force. Participants clarify that to find the correct ratio, one must solve the equation while conserving charge. The conversation highlights the importance of understanding the relationship between charge transfer and resulting electrostatic forces.
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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? I am rather confused. I thought it would just be 1/4 of the max.

any help would be appreciated. thanks.
 
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If you know the max. value, multiply by 1/4 and then solve for q/Q.
 
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.
 
let the ratio be r:
q=rQ
rQ(Q-rQ)=Q^2/16

solve it...
 
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