Two equally charged small balls placed at a fixed distance experience a force 'F'. A similar uncharged ball after touching one of them is placed at the middle point between the two balls. The force experienced by this ball is ______.
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When identical conducting spheres touch, they behave as a single system. The total charge is summed up and then divided equally by 2 upon separation.
Step 1: Understanding the Concept:
When a charged conductor touches an identical uncharged one, the charge is shared equally. Coulomb's Law states $F = k \frac{q_1 q_2}{r^2}$. Step 2: Formula Application:
Initial state: Two balls with charge $Q$ at distance $r$. $F = k \frac{Q^2}{r^2}$.
After touching: One ball (A) and the uncharged ball (C) share charge $Q$, so each has $Q/2$. The other original ball (B) still has $Q$. Step 3: Explanation:
Ball C is placed at $r/2$.
Force from A on C: $F_{AC} = k \frac{(Q/2)(Q/2)}{(r/2)^2} = k \frac{Q^2/4}{r^2/4} = k \frac{Q^2}{r^2} = F$ (repulsive).
Force from B on C: $F_{BC} = k \frac{(Q)(Q/2)}{(r/2)^2} = k \frac{Q^2/2}{r^2/4} = 2k \frac{Q^2}{r^2} = 2F$ (repulsive).
Net Force on C: $2F - F = F$. Step 4: Final Answer:
The force experienced by the ball is $F$.