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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.
Solution
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
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