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let parts are sepereted at a distance of d
so repulsion force F= k q(Q-q)/d2
here Q and d are constant so for maximum force differentiate the force w.r.t q
dF/dq = k/d2 [Q -2q]
for maximum force dF/dq=0
Q-2q=0
Q/q =2
We get this with the help of some calculus :
F = K*[q1*q2]/r², q1+q2=Q
To maximize F(q1,q2) is to maximize q1*q2 so dF/dq=0
rewriting :
F(q1) = K*[q1*(Q-q1)] / r² => F(q1) = K*[q1Q-q1²)] / r²
dF(q1)/dq1 = [K/r²]* [Q-2q1] = 0 so Q- 2q1 = 0 .: q1 = Q/2 and q2=q1.
Remember and keep this result : the maximum A*B you can get is when A=B.
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