A point charge of \( 10\,\mu\mathrm{C} \) is placed at the origin O (the lower-left point in the figure). A charge of \( 2\,\mu\mathrm{C} \) is brought from B to C along the path shown by the arrow in the figure. The work done is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.075 J
The work done in moving a charge \( q \) from B to C is \( q\,(V_C-V_B) \). In an electrostatic field this work does not depend on the path. It depends only on the potentials at the end points. The potential of a point charge \( Q \) at distance \( r \) is \( kQ/r \).
From the figure, B is 80 cm and C is 60 cm from O. In SI units:
\( Q=10\,\mu\mathrm{C}=10\times10^{-6}\,\mathrm{C} \)
\( q=2\,\mu\mathrm{C}=2\times10^{-6}\,\mathrm{C} \)
\( r_B=0.8\,\mathrm{m} \)
\( r_C=0.6\,\mathrm{m} \)
\( k=9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}} \)
The potentials at B and C are:
\( \begin{aligned} V_B &= \frac{kQ}{r_B} \\ &= \frac{9\times10^{9}\times10\times10^{-6}}{0.8}\,\mathrm{V} \\ &= 1.125\times10^{5}\,\mathrm{V} \end{aligned} \)
\( \begin{aligned} V_C &= \frac{kQ}{r_C} \\ &= \frac{9\times10^{9}\times10\times10^{-6}}{0.6}\,\mathrm{V} \\ &= 1.5\times10^{5}\,\mathrm{V} \end{aligned} \)
The work done is:
\( \begin{aligned} W &= q\,(V_C-V_B) \\ &= 2\times10^{-6}\,\mathrm{C}\times(1.5\times10^{5}-1.125\times10^{5})\,\mathrm{V} \\ &= 2\times10^{-6}\,\mathrm{C}\times3.75\times10^{4}\,\mathrm{V} \\ &= 0.075\,\mathrm{J} \end{aligned} \)
The work is positive because C is nearer to the positive charge, so the charge is moved against the repulsion.