Three identical spheres A, B and C of radius r and each carrying charge q are placed on the vertices of an equilateral triangle of side $l (l > > r)$ The resultant force on B is F. Now an uncharged identical sphere is first touched with A and then with B and then removed. Now the resultant force on B is:

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$\frac{3}{8} \sqrt{7} F$

$\frac{2 \sqrt{7}}{4} F$

$\frac{\sqrt{21}}{8} F$

answer is B.

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Detailed Solution

$F = \sqrt{{F_{0}}^{2} + {F_{0}}^{2} + 2 F_{0} F_{0} \cos (60^{0})} = \sqrt{3} F_{0} = \sqrt{3} \frac{K q^{2}}{l^{2}}$

Charges on the sphere after removal of sphere

$F' = \frac{K q^{2}}{l^{2}} \sqrt{{(\frac{3}{8})}^{2} + {(\frac{3}{4})}^{2} + (\frac{3}{8}) (\frac{3}{4})} = \frac{3 \sqrt{7}}{8} \frac{K q^{2}}{l^{2}} = \frac{3 \sqrt{7}}{8} \times \frac{F}{\sqrt{3}} = \frac{\sqrt{21}}{8} F$

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