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 × 10⁻⁶ 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,

\[F = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r^2}\]

Force of friction on B = μmg

Since B is at equilibrium, 
Coulomb force = Frictional force

⇒ F = μmg

\[ \Rightarrow   \frac{9 \times {10}^9 \times 2 \times {10}^{- 6} \times q_2}{\left( 0 . 1 \right)^2} = 0 . 2 \times 0 . 08 \times 9 . 8\] 

\[ \Rightarrow    q_2  = 8 . 71 \times  {10}^{- 8}   C\]

∴ The range of the charge = ∓ 8.71×10^-8 C