Physics - Electrostatics Question with Solution | TestHub

Initial potential energy of the system $$\begin{array}{l}=\frac{1}{4\pi {\epsilon }_0}\left[\frac{{\mathrm{q}}^2}{\mathrm{a}}+\frac{{\mathrm{q}}^2}{\mathrm{a}}+\frac{{\mathrm{q}}^2}{\mathrm{a}}\right]=\frac{1}{4\pi {\epsilon }_0}\left(\frac{3{\mathrm{q}}^2}{\mathrm{a}}\right)\\ =9\times 10^9\left(3\times \frac{(0.1{)}^2}{1}\right)=27\times 10^7\mathrm{J}\end{array}$$ Let charge at $\mathrm{A}$ is moved to mid-point $\mathrm{O}$ Then final potential energy of thhe system $$\begin{array}{rl}{\mathrm{U}}_{\mathrm{f}}& =\frac{1}{4\pi {\epsilon }_0}\left[\frac{2{\mathrm{q}}^2}{(\mathrm{a}/2)}+\frac{{\mathrm{q}}^2}{\mathrm{a}}\right]\\ & =5\times \frac{1}{4\pi {\epsilon }_0}\left(\frac{{\mathrm{q}}^2}{{\mathrm{a}}^2}\right)=45\times 10^7\mathrm{J}\end{array}$$ 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 $$\mathrm{t}=\frac{18\times 10^7}{1000}=18\times 10^4\mathrm{s}=50\mathrm{h}$$