Two point charges, each \(+q\), are fixed at points A and B, a distance \(r\) apart. The potential energy of this pair is \(U\). A third charge \(+q\) is brought from infinity to a point P on the line AB extended beyond B, at a distance \(r/2\) from B. The work needed is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(8U/3\)
The work needed to bring a charge from infinity to a point is the charge times the potential at that point. For the pair, \( U=q^{2}/4\pi\varepsilon_0 r \). Point P is \( r/2 \) from B and \( r+r/2=3r/2 \) from A.
\( \begin{aligned} V_P &= \frac{q}{4\pi\varepsilon_0(r/2)}+\frac{q}{4\pi\varepsilon_0(3r/2)} \\ &= \frac{2q}{4\pi\varepsilon_0 r}+\frac{2q}{3\times4\pi\varepsilon_0 r} \\ &= \frac{8}{3}\,\frac{q}{4\pi\varepsilon_0 r} \end{aligned} \)
\( \begin{aligned} W &= q\,V_P \\ &= \frac{8}{3}\,\frac{q^{2}}{4\pi\varepsilon_0 r} \\ &= \frac{8U}{3} \end{aligned} \)
NCERT (2026-27):
Potential Energy of a System of Charges – “… Thus, the potential energy of a system of two charges q1 and q2 is”
\( U=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r_{12}} \qquad (2.22) \)
Potential Energy of a System of Charges – “… Work done next in bringing q3 from infinity to the point r3 is q3 times V1, 2 at r3”
\( q_3V_{1,2}(\mathbf{r}_3)=\frac{1}{4\pi\varepsilon_0}\left(\frac{q_1q_3}{r_{13}}+\frac{q_2q_3}{r_{23}}\right) \qquad (2.25) \)