Two equally charged, identical metal spheres A and B repel each other with a force $F$. The spheres are kept f

Question

Two equally charged, identical metal spheres A and B repel each other with a force $F$. The spheres are kept fixed with a distance $r$ between them. A third identical, but uncharged sphere C is brought in contact with A, then brought in contact with B, and finally placed at the mid-point of the line joining A and B. The magnitude of the net electric force on C is:

Accepted Answer

Let the initial charge on spheres A and B be $q$. The force $F = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2}$.

1.  **Contact with A:** When uncharged sphere C touches A, the charge $q$ distributes equally. Charge on A becomes $q_A = q/2$, and charge on C becomes $q_C = q/2$.
2.  **Contact with B:** Sphere C (now with charge $q/2$) touches sphere B (charge $q$). Total charge is $q + q/2 = 3q/2$. This splits equally, so $q_B = 3q/4$ and $q_C = 3q/4$.
3.  **Placed at Midpoint:** C is placed at distance $r/2$ from both A and B.
    *   Force from A on C ($F_{CA}$): Repulsive. $F_{CA} = \frac{1}{4\pi\epsilon_0} \frac{(q/2)(3q/4)}{(r/2)^2} = \frac{1}{4\pi\epsilon_0} \frac{3q^2/8}{r^2/4} = 1.5 F$.
    *   Force from B on C ($F_{CB}$): Repulsive. $F_{CB} = \frac{1}{4\pi\epsilon_0} \frac{(3q/4)(3q/4)}{(r/2)^2} = \frac{1}{4\pi\epsilon_0} \frac{9q^2/16}{r^2/4} = 2.25 F$.
4.  **Net Force:** Since forces are opposite, $F_{net} = F_{CB} - F_{CA} = 2.25F - 1.5F = 0.75F = \frac{3F}{4}$.

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