Three point charges of 2 mC each are kept at the vertices of an equilateral triangle of side 50   cm .…

The figure below represents the three point charges forming
equilateral triangle.

The energy system of three charges,

$U=3\left(\frac{k{q}_1{q}_2}{r}\right)$

$U=3\times \frac{k\left(2\times {10}^{-3}\times 2\times {10}^{-3}\right)}{{\displaystyle \frac{1}{2}}}$

$U=24\mathrm{k}\times {10}^{-6} \mathrm{J}$

When one charge moves to the midpoint of the line joining the
other two charges,

The energy of the system,

$U^{\text{'}}=\frac{k\left(2\times 2\times {10}^{-6}\right)}{(1/4)}\times 2+\frac{k\left(2\times 2\times {10}^{-6}\right)}{(1/2)}$

$U^{\text{'}}=40\mathrm{k}\times {10}^{-6} \mathrm{J}$

The net energy,

$\mathrm{\Delta }U=U^{\text{'}}-U$

$\mathrm{\Delta }U=(40-24)k\times {10}^{-6}$

$\mathrm{\Delta }U=16\times 9\times {10}^3 \mathrm{J}$

The energy supplied to the system at the rate of $2 \mathrm{kW}$,

$2 \mathrm{kW}\times t=16\times 9\times {10}^3$

$2\times {10}^3\times t=16\times 9\times {10}^3$

$t=72 \mathrm{s}$