Question

A total charge q is broken in two parts q1 and q2 and they are placed at a distance r from each other the maximum force of repulsion between them will occur when

Answer 1

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Answer 2

There is a charge q which is broken into 2 charges as $q_1$ and $q_2$ with their placement at a distance r from each other.

Now, the maximum force of repulsion will occur when the force per unit total charge becomes zero. Therefore, putting the values into the formula of electrostatic force, we get,

                  $- F = q_1 q_2 / 4 \pi \epsilon_o r^2$

=> since, $q_2 = \theta -q_1,$

we get,

             => $F = q_1 ( \theta -q_1) / 4\pi\epsilon_o r^2$

=> for repulsion maximum force, dF / dq = 0,

Therefore,

               $=> dF / dq = k / r^2 d / dq (q_1 ) ( \theta -q_1)$

               $=> dF / dq = k / r^2 dq ( \theta q_1 -q_1 ^2) = k / r^2 [ \theta - 2q_1]$

               $=> dF / dq = 0 = k / r^2 [ \theta - 2q_1]$

               $=> \theta - 2q_1 = 0$

               $=> q_1 = \theta / 2$

               $=> q_2 = \thata - q_1 = \theta - \theta/ 2 = \theta/ 2.$