Two identical small conducting spheres, having charges of opposite sign, attract each other with a force of $0.108 N$ when separated by $0.5 m$ . The spheres are connected by a conducting wire, which is then removed, and thereafter, they repel each other with a force of $0.036 N$ . The initial charges of the spheres are :

\pm 5 \times 10^{- 6} C a n d \mp 15 \times 10^{- 6} C

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\pm 1.0 \times 10^{- 6} C a n d \mp 3.0 \times 10^{- 6} C

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\pm 2.0 \times 10^{- 6} C a n d \mp 6.0 \times 10^{- 6} C

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\pm 0.5 \times 10^{- 6} C a n d \mp 1.5 \times 10^{- 6} C

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Solution

The correct option is B $\pm 1.0 \times 10^{- 6} C a n d \mp 3.0 \times 10^{- 6} C$
Let initial charge be Q1 and Q2.
Initial force $= F = \frac{K \times Q 1 \times Q 2}{r^{2}} = - 0.108$
$F = \frac{9 \times 10^{9} \times Q 1 \times Q 2}{0.25} = - 0.108$
$Q 1 \cdot Q 2 = - 3 \times 10^{- 12}$
Now on connecting them, since they are identical so they will have equal charges.
Final charge $= \frac{Q 1 + Q 2}{2}$
Final force $= F = \frac{9 \times 10^{9} \times (Q 1 + Q 2)^{2}}{4 \times 0.25} = 0.036$
$Q 1 + Q 2 = \pm 2 \times 10^{- 6}$
$Q 2 = \pm 1.0 \times 10^{- 6} C, Q 1 = \mp 3.0 \times 10^{- 6} C$

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