Two identical spheres with charge $Q$ exert a force $F$ on each other. 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 & B. The magnitude of net electric force on C is

F

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\frac{F}{4}

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\frac{F}{2}

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F

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Solution

The correct option is A $F$
$F = \frac{k Q^{2}}{r^{2}} A \cdot Q B \cdot Q$

When, uncharged sphere will touch charged sphere charges will distribute equally as they are identical, so as to maintain same potential. Therefore, charges on A and C $= \frac{Q}{2}$

$A \cdot Q / 2 C \cdot Q / 2 B \cdot Q$

$F_{o n C b y A} = \frac{K \frac{Q}{2} . \frac{Q}{2}}{\frac{r}{2} . \frac{r}{2}} = \frac{K Q^{2}}{r^{2}} = F F_{o n C b y B} = \frac{K Q . \frac{Q}{2}}{\frac{r}{2} . \frac{r}{2}} = 2 \frac{K Q^{2}}{r^{2}} = 2 F F_{n e t} = 2 F - F = F$

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Two identical spheres with charge Q exert a force F on each other. 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 & B. The magnitude of net electric force on C is