Question

Two identical charged particles each having a mass 10 g and charge \(2.0 × 10^{–7}\) C are placed on a horizontal table with a separation of L between them such that they stay in limited equilibrium. If the coefficient of friction between each particle and the table is \(0.25\), find the value of L. [Use g = \(10\) \(ms^{–2}\)]

• A. 12 cm
• B. 10 cm
• C. 8 cm
• D. 5 cm

Answer 1

The Correct Option is A

 To find the separation distance \(L\) at which the two charged particles remain in equilibrium, we can analyze the forces acting on the particles.

Each particle experiences two forces: the electrostatic force of repulsion and the frictional force. These should balance each other for equilibrium.

1. Electrostatic Force: The electrostatic force \(F_{\text{e}}\) between two like charges is given by Coulomb's law: \(F_{\text{e}} = \dfrac{k \cdot q^2}{L^2}\), where \(k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2\) is the Coulomb's constant, and \(q = 2.0 \times 10^{-7} \, \text{C}\) is the charge on each particle.
2. Maximum Frictional Force: The maximum frictional force \(F_{\text{fric}}\) is given by: \(F_{\text{fric}} = \mu \cdot m \cdot g\), where \(\mu = 0.25\) is the coefficient of friction, \(m = 10 \, \text{g} = 0.01 \, \text{kg}\) is the mass of each particle, and \(g = 10 \, \text{m/s}^2\) is the acceleration due to gravity.

For equilibrium, these forces must be equal: \(F_{\text{e}} = F_{\text{fric}}\)

Substituting the expressions for these forces, we get: \(\dfrac{k \cdot q^2}{L^2} = \mu \cdot m \cdot g\)

Re-arranging for \(L\), we have: \(L^2 = \dfrac{k \cdot q^2}{\mu \cdot m \cdot g}\)

Plugging in the given values: \(L^2 = \dfrac{9 \times 10^9 \cdot (2.0 \times 10^{-7})^2}{0.25 \cdot 0.01 \cdot 10}\)

Calculating further: \(L^2 = \dfrac{9 \times 10^9 \cdot 4 \times 10^{-14}}{0.025}\)

\(L^2 = \dfrac{36 \times 10^{-5}}{0.025}\)

 

\(L^2 = 1.44 \times 10^{-2} = 0.0144\)

 

Solving for \(L\), we find: \(L = \sqrt{0.0144} = 0.12 \, \text{m} = 12 \, \text{cm}\)

Thus, the value of \(L\) is 12 cm, ensuring the particles remain in limited equilibrium.