A thin wire is bent into a semicircle of radius 9 cm. It carries a charge of \(3\ \mathrm{nC}\) spread uniformly along it. The work needed to bring a point charge of \(2\ \mu\mathrm{C}\) slowly from infinity to the centre of the semicircle is (Take \(\frac{1}{4\pi\varepsilon_0}=9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}\).)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(6\times10^{-4}\ \mathrm{J}\)
Divide the wire into small bits. Every bit is at the same distance \( R \) from the centre, and potential is a scalar. So the potentials of all the bits add up as numbers, and the shape of the wire does not matter. The work needed is the charge brought in times the potential at the centre.
\( \begin{aligned} V &= \frac{1}{4\pi\varepsilon_0}\,\frac{Q}{R} \\ &= \frac{9\times10^{9}\times3\times10^{-9}}{0.09} \\ &= 300\ \mathrm{V} \end{aligned} \)
\( \begin{aligned} W &= qV \\ &= 2\times10^{-6}\times300 \\ &= 6\times10^{-4}\ \mathrm{J} \end{aligned} \)
NCERT (2026-27):
Potential Due to a System of Charges – “If we have a continuous charge distribution characterised by a charge density ρ (r), we divide it, as before, into small volume elements each of size Δv and carrying a charge ρΔv. We then determine the potential due to each volume element and sum (strictly speaking , integrate) over all such contributions, and thus determine the potential due to the entire distribution.”
Summary – “The potential energy of a charge q in an external potential V(r) is qV(r).”