A point charge of \(-4\times10^{-9}\ \mathrm{C}\) is fixed in air. A charge of \(+3\times10^{-9}\ \mathrm{C}\) is brought from infinity to a point 12 cm from it. The work done by the electrostatic force on the moving charge is (take \(\frac{1}{4\pi\varepsilon_0}=9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}\))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(+9\times10^{-7}\ \mathrm{J}\)
The potential of a point charge is \( V=Q/4\pi\varepsilon_0 r \). For a negative \( Q \), \( V \) is negative. The external force then does negative work in bringing a positive charge from infinity. The electrostatic force does the same work with the opposite sign.
Here \( Q=-4\times10^{-9}\ \mathrm{C} \), \( q=3\times10^{-9}\ \mathrm{C} \) and \( r=12\ \mathrm{cm}=0.12\ \mathrm{m} \).
\( \begin{aligned} V &= \frac{Q}{4\pi\varepsilon_0 r} \\ &= \frac{9\times10^{9}\times(-4\times10^{-9})}{0.12} \\ &= -300\ \mathrm{V} \end{aligned} \)
The work done by the external force is \( W_{\mathrm{ext}}=qV \).
\( \begin{aligned} W_{\mathrm{ext}} &= qV \\ &= 3\times10^{-9}\times(-300) \\ &= -9\times10^{-7}\ \mathrm{J} \end{aligned} \)
\( \begin{aligned} W_{\mathrm{field}} &= -W_{\mathrm{ext}} \\ &= +9\times10^{-7}\ \mathrm{J} \end{aligned} \)
Therefore, the field does positive work. It pulls the positive charge towards the negative charge, along its motion.
NCERT (2026-27):
Potential Due to a Point Charge – “For Q < 0, V < 0, i.e., work done (by the external force) per unit positive test charge in bringing it from infinity to the point is negative. This is equivalent to saying that work done by the electrostatic force in bringing the unit positive charge form infinity to the point P is positive. …”