Three particles, each having a charge of 10 \( \mu\mathrm{C} \), are placed at the vertices of an equilateral triangle of 10 mm. Find the work done by a person in pulling them apart to infinite separation.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( -270\,\mathrm{J} \)
The electrostatic potential energy of the three charges is the sum over the three pairs. Each pair has charges \( q \) a distance \( a \) apart.
\( q = 10\,\mu\mathrm{C} = 10^{-5}\,\mathrm{C} \)
\( a = 10\,\mathrm{mm} = 10^{-2}\,\mathrm{m} \)
\( \displaystyle U_{i} = 3 \times \frac{k q^{2}}{a} = \frac{3 \times 9\times10^{9} \times (10^{-5})^{2}}{10^{-2}} = +270\,\mathrm{J} \)
When the charges are at infinite separation the energy is zero, \( U_{f} = 0 \). The work done by an external agent (the person) is the change in the potential energy.
\( \displaystyle W_{\mathrm{person}} = U_{f} - U_{i} = 0 - 270 = -270\,\mathrm{J} \)
The sign is negative because the charges repel each other: the repulsion itself pushes them apart, so the person only has to hold them back.
Note: The sign of the answer depends on who is said to do the work. The repulsion between the charges does +270 J of work as they move apart, and the person, who holds them back, does −270 J, which is the value given here. If the question is read as the work done by the electric force, or as the magnitude of the energy involved, the value is +270 J, which is Option (1).