Two identical charged particles each having a mass 10g and charge $2.0\times {10}^{-7}\mathrm{C}$ 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 $\mathrm{g}=10{ \mathrm{ms}}^{-2}$ ]

Given Data:

• Mass of each particle, m = 10 g = 0.01 kg
• Charge on each particle, q = 2.0 × 10^-7 C
• Coefficient of friction between each particle and the table, μ = 0.25
• Acceleration due to gravity, g = 10 m/s²
• Coulomb's constant, k = 9 × 10⁹ N·m²/C²

Approach:

We are tasked with determining the separation distance L between two identical charged particles that are in electrostatic equilibrium, with the effect of friction also considered.

Step-by-Step Solution:

Step 1: Analyzing the Forces

Each particle experiences two forces in the horizontal direction:

• Electrostatic Force: The repulsive force between two charges is given by Coulomb's law: 
  Felectrostatic = k * (q2) / L2
• Frictional Force: The frictional force that resists the motion of the particles is given by: 
  Ffriction = μ * m * g

Step 2: Equilibrium Condition

For the particles to stay in equilibrium, the electrostatic force must be balanced by the frictional force:

Felectrostatic = Ffriction Substituting the expressions for both forces: 
k * (q2) / L2 = μ * m * g

Step 3: Solving for L

Rearranging the equation to solve for L:

L = sqrt((k * q2) / (μ * m * g))

Step 4: Substituting Known Values

Now, substituting the given values into the equation:

L = sqrt((9 × 109 * (2.0 × 10-7)2) / (0.25 * 0.01 * 10))

Step 5: Final Calculation

Performing the calculations:

L = sqrt((9 × 109 * 4.0 × 10-14) / (0.25 * 0.01 * 10)) 
L = sqrt((3.6 × 10-4) / (0.025)) 
L = sqrt(0.0144) 
L ≈ 0.12 m

Conclusion:

The separation distance L between the two charges is approximately 0.12 meters.