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 and then placed at the mid-point of the line joining A and B. The magnitude of the net electric force on C is

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\frac{3 F}{4}

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\frac{F}{2}

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\frac{F}{4}

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Solution

The correct option is A F
Initially

Finally

Force on C due to A, $F_{A} = \frac{k (Q / 2)^{2}}{(r / 2)^{2}} = \frac{k Q^{2}}{r^{2}}$
Force on C due to B, $F_{B} = \frac{K Q (Q / 2)}{(r / 2)^{2}} = \frac{2 K Q^{2}}{r^{3}}$
$∴$ Net force on C, $F_{n e t} = F_{B} - F_{A} = \frac{k Q^{2}}{r^{2}} = F$

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The correct option is A FInitially Finally Force on C due to A, FA=k(Q/2)2(r/2)2=kQ2r2 Force on C due to B, FB=KQ(Q/2)(r/2)2=2KQ2r3 ∴ Net force on C, Fnet=FB−FA=kQ2r2=F