Question 23 Solution
Step 1: Motion before collision (Attraction)
The balls start at separation $l$ with charges $Q$ and $-q$. They attract each other until they touch (distance $2r$).
- Initial Potential Energy: $U_i = -\frac{1}{4\pi\epsilon_0}\frac{Qq}{l}$
- Potential Energy just before contact: $U_{contact} = -\frac{1}{4\pi\epsilon_0}\frac{Qq}{2r}$
- Kinetic Energy gained ($K_1$) = Loss in Potential Energy: $$K_1 = U_i – U_{contact} = \frac{1}{4\pi\epsilon_0}Qq \left( \frac{1}{2r} – \frac{1}{l} \right)$$
Step 2: The Collision and Charge Redistribution
During contact, charges redistribute. Since the balls are identical, the total charge splits equally.
- Total Charge: $Q_{net} = Q + (-q) = Q – q$
- Charge on each ball after contact: $q’ = \frac{Q-q}{2}$
Note: The hint states that energy loss during redistribution is heat (electrical), but the balls exchange velocities mechanically. The mechanical kinetic energy $K_1$ is conserved after the collision moment.
Step 3: Motion after collision (Repulsion)
The balls now have charge $q’$ and repel each other. This repulsion converts electrostatic potential energy into additional kinetic energy.
- Potential Energy at contact (new charges): $$U’_{contact} = \frac{1}{4\pi\epsilon_0} \frac{q’ \cdot q’}{2r} = \frac{1}{4\pi\epsilon_0} \frac{(Q-q)^2}{8r}$$
- Potential Energy at infinity: $0$
- Additional Kinetic Energy ($K_2$) = $U’_{contact} – 0$
Step 4: Total Energy Calculation
Total Final KE = $K_1 + K_2$.
Since there are two balls, Total KE = $2 \times \frac{1}{2}mv^2 = mv^2$.
Simplifying the term inside the bracket with common denominator $8r$ (ignoring $1/l$ for a moment):
$$\frac{4Qq}{8r} + \frac{Q^2 – 2Qq + q^2}{8r} = \frac{Q^2 + 2Qq + q^2}{8r} = \frac{(Q+q)^2}{8r}$$Combining everything:
$$mv^2 = \frac{1}{4\pi\epsilon_0} \left[ \frac{(Q+q)^2}{8r} – \frac{Qq}{l} \right]$$