Suppose you have a total charge\({{\rm{q}}_{{\rm{tot}}}}\)that you can split in any manner. Once split, the separation distance is fixed. How do you split the charge to achieve the greatest force?

When two-point charges are separated by some distance, the electrostatic force between them is,

\(F = \frac{{K{q_1}{q_2}}}{{{r^2}}}\)

Here, K is the electrostatic force constant, \({q_1}\) and \({q_2}\) are the two-point charges separated by distance r.

The total charge is,

\({q_{tot}} = {q_1} + {q_2}\)

Therefore, the charge \({q_2}\) is,

\({q_2} = {q_1} - {q_{tot}}\)

From equation (1.1),

\(\begin{aligned} F &= \frac{{K{q_1}\left( {{q_{tot}} - {q_1}} \right)}}{{{r^2}}}\\ &= \frac{{K\left( {{q_1}{q_{tot}} - q_1^2} \right)}}{{{r^2}}}\end{aligned}\)

Differencating equation (1.3) with respect to \({q_1}\),

\(\frac{{dF}}{{d{q_1}}} = \frac{K}{{{r^2}}}\left( {{q_{tot}} - 2{q_1}} \right)\)

Equating \(\frac{{dF}}{{d{q_1}}}\) in order to get maxima and minima points,

\(\begin{aligned} 0 = \frac{K}{{{r^2}}}\left( {{q_{tot}} - 2{q_1}} \right)\\{q_1} = \frac{{{q_{tot}}}}{2}\end{aligned}\)

Now, calculating second derivative of force F at \({q_1} = \frac{{{q_{tot}}}}{2}\).

Again, differentiating equation (1.4) with respect to \({q_1}\),

\({\left. {\frac{{{d^2}F}}{{dq_1^2}}} \right|_{{q_1} = \frac{{{q_{tot}}}}{2}}} = \frac{K}{{{r^2}}}\left( { - 2} \right)\)

Since, \({\left. {\frac{{{d^2}F}}{{dq_1^2}}} \right|_{{q_1} = \frac{{{q_{tot}}}}{2}}} < 0\). Therefore, the force would be maximum at\({q_1} = \frac{{{q_{tot}}}}{2}\).

From equation (1.3), the charge \({q_2}\) is,

\(\begin{aligned} {q_2} &= {q_{tot}} - \frac{{{q_{tot}}}}{2}\\ &= \frac{{{q_{tot}}}}{2}\end{aligned}\)

Hence, the force will be maximum when both charges are equal to \(\frac{{{q_{tot}}}}{2}\).