Two particles, each having a mass of 5 g and charge 1.0 × 10−7 C, stay in limiting equilibrium on a horizontal table with a separation of 10 cm between them. The coefficient of friction between each particle and the table is the same. The value of the coefficient of friction is: (Take g = 10 ms−2)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.18
Each particle is pushed away from the other by the Coulomb repulsion. In limiting equilibrium this force is just balanced by the maximum (limiting) friction \( \mu mg \) on that particle.
\( m = 5\,\mathrm{g} = 5\times10^{-3}\,\mathrm{kg} \)
\( q = 1.0\times10^{-7}\,\mathrm{C} \)
\( r = 10\,\mathrm{cm} = 0.1\,\mathrm{m} \)
\( \displaystyle F = \frac{1}{4\pi\varepsilon_0}\frac{q^{2}}{r^{2}} \)
\( \displaystyle F = \frac{9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}} \times (1.0\times10^{-7}\,\mathrm{C})^{2}}{(0.1\,\mathrm{m})^{2}} \)
\( \displaystyle F = \frac{9\times10^{-5}}{10^{-2}}\,\mathrm{N} = 9\times10^{-3}\,\mathrm{N} \)
\( F = \mu mg \)
\( \displaystyle \mu = \frac{F}{mg} = \frac{9\times10^{-3}\,\mathrm{N}}{5\times10^{-3}\,\mathrm{kg} \times 10\,\mathrm{m\,s^{-2}}} \)
\( \mu = 0.18 \)