Charges of \(+3\ \mu\mathrm{C}\) and \(-1\ \mu\mathrm{C}\) are held fixed 20 cm apart. Point P is between the two charges, 5 cm from the \(+3\ \mu\mathrm{C}\) charge. The work needed to bring a charge of \(+2\ \mu\mathrm{C}\) from infinity to P 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) → \(0.96\ \mathrm{J}\)
The work needed to bring a charge \( q_3 \) from infinity to a point is \( q_3 \) times the potential there. This potential is due to the charges already in place. P is 5 cm (0.05 m) from the \( +3\ \mu\mathrm{C} \) charge and 15 cm (0.15 m) from the \( -1\ \mu\mathrm{C} \) charge.
\( \begin{aligned} V_P &= \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}\right) \\ &= 9\times10^{9}\left(\frac{3\times10^{-6}}{0.05}-\frac{1\times10^{-6}}{0.15}\right) \\ &= 5.4\times10^{5}-0.6\times10^{5} \\ &= 4.8\times10^{5}\ \mathrm{V} \end{aligned} \)
\( \begin{aligned} W &= q_3V_P \\ &= 2\times10^{-6}\times4.8\times10^{5} \\ &= 0.96\ \mathrm{J} \end{aligned} \)
NCERT (2026-27):
Potential Energy of a System of Charges – “The charges q1 and q2 produce a potential, which at any point P is given by … Work done next in bringing q3 from infinity to the point r3 is q3 times V1, 2 at r3”
\( q_3V_{1,2}(\mathbf{r}_3)=\frac{1}{4\pi\varepsilon_0}\left(\frac{q_1q_3}{r_{13}}+\frac{q_2q_3}{r_{23}}\right) \qquad (2.25) \)