Two identical copper spheres carrying charge $+ q$ and $- 9 q$ separated by a certain distance d has an attractive force of F. If the spheres are allowed to touch each other and moved to distance of separation x so that force between them becomes $\frac{4 F}{9}$ then x is equal to

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\frac{d}{2}

answer is B.

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Detailed Solution

\frac{F = \frac{1}{4 π ε_{0}} \frac{q \times 9 q}{d^{2}}}{\frac{4 F}{9} = \frac{1}{4 π ε_{0}} \times \frac{4 q \times 4 q}{x^{2}}}

$\frac{9}{4} = \frac{{9/d}^{2}}{\frac{16}{x^{2}}}$

$1 = \frac{x^{2}}{4 d^{2}}$

$x^{2} = 4 d^{2}$

$x = 2 d$

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