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Answer 1 · 2016-10-28T08:05:55

By coulomb's law the force acting between two point charges q and Q-q lying r distance apart is given by

$F=k(q(Q-q))/r^2.....(1)$

Where Q is the total charge, a constant quantity.
k is the Coulmb's constant

q is the varible charge on which the magnitude of force F depends.

Now

$F=k(qQ-q^2)/r^2$

$=>F=k/r^2(Q^2/4-Q^2/4+2*Q/2*q-q^2)$

$=>F=k/r^2(Q^2/4-((Q/2)^2-2*Q/2*q+q^2))$

$=>F=k/r^2(Q^2/4-(Q/2-q)^2)$

This relation suggests that the magnltude of force F will be maximum only when

$Q/2-q=0$
$=>q=Q/2$

This means the force between the two parts of Q will be maximum only when Q is divided in two equal parts and then

$Q/q=2/1$

Differentiating equation (1) w .r to q we get

$(dF)/(dq)=k/r^2(d/(dq)(Qq-q^2))$

$=>(dF)/(dq)=k/r^2(Qd/(dq)(q)-d/(dq)(q^2))$

$=>(dF)/(dq)=k/r^2(Q-2q)$

For maximum value of F

$(dF)/(dq)=0$

This means $k/r^2(Q-2q)=0$

$=>Q=2q$

$=>Q/q=2/1$