Two identical conducting spheres A and B, carry equal charge. They are separated by a distance much larger than their diameter, and the force between them is F. A third identical conducting sphere, C, is uncharged. Sphere C is first touched to A, then to B, and then removed. As a result, the force between A and B would be equal to

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$\frac{3 F}{4}$

$\frac{F}{2}$

$\frac{3 F}{8}$

answer is D.

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Detailed Solution

Sphere A and B carry equal charge say ‘q’
$∴$ Force between them, $F = \frac{kqq}{r^{2}}$ When A and C are touched, charge on both $q_{A} = q_{C} = \frac{q}{2}$
Then when B and C are touched, charge on ${Bq}_{B} = \frac{\frac{q}{2} + q}{2} = \frac{3 q}{4}$
Now, the force between charge $q_{A} and q_{B}$ $F^{1} = \frac{{kq}_{A} q_{B}}{r^{2}} = \frac{k \times \frac{q}{2} \times \frac{3 q}{4}}{r^{2}} = \frac{3}{8} \frac{{kq}^{2}}{r^{2}} = \frac{3}{8} F$

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