At the corners of an equilateral triangle of side a (1 metre),...

Initial potential energy of the system \begin{array}{l} =\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{\mathrm{q}^{2}}{\mathrm{a}}+\frac{\mathrm{q}^{2}}{\mathrm{a}}+\frac{\mathrm{q}^{2}}{\mathrm{a}}\right]=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{3 \mathrm{q}^{2}}{\mathrm{a}}\right) \\ =9 \times 10^{9}\left(3 \times \frac{(0.1)^{2}}{1}\right)=27 \times 10^{7} \mathrm{~J} \end{array} Let charge at $\mathrm{A}$ is moved to mid-point $\mathrm{O}$ Then final potential energy of thhe system \begin{aligned} \mathrm{U}_{\mathrm{f}} &=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{2 \mathrm{q}^{2}}{(\mathrm{a} / 2)}+\frac{\mathrm{q}^{2}}{\mathrm{a}}\right] \\ &=5 \times \frac{1}{4 \pi \varepsilon_{0}}\left(\frac{\mathrm{q}^{2}}{\mathrm{a}^{2}}\right)=45 \times 10^{7} \mathrm{~J} \end{aligned} Work done $=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}=18 \times 10^{7} \mathrm{~J}$ Also, energy supplied per sec $=1000 \mathrm{~J}$ (given) Time required to move one of the mid-point of the line joining the other two $$\mathrm{t}=\frac{18 \times 10^{7}}{1000}=18 \times 10^{4} \mathrm{~s}=50 \mathrm{~h}$$