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.

#### Solution

Given:

Magnitude of charge of particle A, q_{1} = 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 q_{2}.

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

\[ \Rightarrow \frac{9 \times {10}^9 \times 2 \times {10}^{- 5} \times q_2}{\left( 0 . 1 \right)^2} = 0 . 2 \times 0 . 08 \times 9 . 8\]

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

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