Two small insulating spheres are rubbed against each other and placed 1 cm apart. If they attract each other with a force, F = 0.1 N, the number of electrons that were transferred from one sphere to another during rubbing are nearly
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2 × 1011
Rubbing transfers electrons from one sphere to the other, so the spheres get equal and opposite charges ±q, with q = ne for n transferred electrons. The small spheres are treated as point charges in Coulomb’s law, with r = 1 cm = 0.01 m.
\( \displaystyle F = \frac{1}{4\pi\varepsilon_0}\,\frac{q^{2}}{r^{2}} \Rightarrow q = r\sqrt{\frac{F}{9\times10^{9}}} \)
\( \displaystyle q = 0.01\,\mathrm{m} \times \sqrt{\frac{0.1\,\mathrm{N}}{9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}}}} \)
\( \displaystyle q = 0.01 \times 3.33\times10^{-6} = 3.33\times10^{-8}\,\mathrm{C} \)
\( \displaystyle n = \frac{q}{e} = \frac{3.33\times10^{-8}}{1.6\times10^{-19}} \approx 2.1\times10^{11} \)
So about 2 × 1011 electrons were transferred.