Q103 / 125 · BITSAT 2015 · Physics
At the corners of an equilateral triangle of side a (1 metre), three point charges are placed (each of $0.1 \mathrm{C}$ ). If this system is supplied energy at the rate of $1 \mathrm{kw},$ then calculate the time required to move one of the mid-point of the line joining the other two.
- Correct answerA$50 \mathrm{~h}$
- B$60 \mathrm{~h}$
- C$48 \mathrm{~h}$
- D$54 \mathrm{~h}$
Solution
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
$$
\mathrm{t}=\frac{18 \times 10^{7}}{1000}=18 \times 10^{4} \mathrm{~s}=50 \mathrm{~h}
$$
Hence, the correct answer is option: 'A'.