ELECTROSTATICS O27

Given Data:

• Mass $m = 100 \text{ g} = 0.1 \text{ kg}$
• Charge $q = 10 \mu\text{C} = 10^{-5} \text{ C}$
• Initial separation $r_1 = 1.0 \text{ m}$
• Coefficient of friction $\mu = 0.1$
• $g = 10 \text{ m/s}^2$

Step 1: Determine Condition for Maximum Speed

The particles accelerate as long as the repulsive electrostatic force ($F_e$) is greater than the kinetic friction force ($f_k$). The maximum speed is attained when the acceleration becomes zero, i.e., when $F_e = f_k$.

Friction force on one particle: $f_k = \mu N = \mu mg = 0.1 \times 0.1 \times 10 = 0.1 \text{ N}$.

Electrostatic force: $F_e = \frac{kq^2}{r^2}$.

Equating them to find the separation $r_2$ at max speed:

$$ \frac{9 \times 10^9 \times (10^{-5})^2}{r_2^2} = 0.1 $$ $$ \frac{0.9}{r_2^2} = 0.1 \Rightarrow r_2^2 = 9 \Rightarrow r_2 = 3 \text{ m} $$

Step 2: Apply Work-Energy Theorem

We apply the Work-Energy theorem from the initial state ($r_1=1$) to the state of maximum speed ($r_2=3$).

$$ W_{electric} + W_{friction} = \Delta K $$

Work done by Electric Force (Conservative):

$$ W_{elec} = -\Delta U = k q^2 \left( \frac{1}{r_1} – \frac{1}{r_2} \right) $$ $$ W_{elec} = 0.9 \left( \frac{1}{1} – \frac{1}{3} \right) = 0.9 \times \frac{2}{3} = 0.6 \text{ J} $$

Work done by Friction (Non-Conservative):

Since both particles move, the total displacement is the change in separation $(r_2 – r_1) = 2 \text{ m}$. Wait, each particle moves half this distance ($1 \text{ m}$ each). Total work by friction is the sum for both particles.

$$ W_{fric} = -f_k \times (\text{Total relative displacement}) = -0.1 \times (3 – 1) = -0.2 \text{ J} $$

Alternatively: Each particle moves $1\text{m}$. Work on one = $-0.1 \times 1$. Total = $-0.2$.