Question

A particle A with a charge of 2.0 × 10⁻⁶ C is held fixed on a horizontal table. A second charged particle of mass 80 g stays in equilibrium on the table at a distance of 10 cm from the first charge. The coefficient of friction between the table and this second particle is μ = 0.2. Find the range within which the charge of this second particle may lie.

Answer 1

Given:
Magnitude of charge of particle A, q₁ = 2.0 × 10⁻⁵ C
Separation between the charges, r = 0.1 m
Mass of particle B, m = 80 g = 0.08 kg
Let the magnitude of charge of particle B be q₂.
By Coulomb's Law, force on B due to A,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_0}\frac{q_1{q}_2}{r^2}$
Force of friction on B = $\mu mg$
Since B is at equilibrium,
Coulomb force = Frictional force
$$\begin{gathered} \Rightarrow F=\mu mg \\ \Rightarrow \frac{9\times {10}^9\times 2\times {10}^{-5}\times q_2}{{\left(0.1\right)}^2}=0.2\times 0.08\times 9.8 \\ \Rightarrow q_2=8.71\times {10}^{-8}\mathrm{C} \end{gathered}$$
∴ The range of the charge = ∓8.71×10^-8 C