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Two identical 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 on the spheres are
$\pm 5 \times 10^{-6}\ C $ and $ \mp 15 \times 10^{-6}\ C$
$\pm 1.0 \times 10^{-6}\ C $ and $\mp 3.0 \times 10^{-6}\ C$
$\pm 2.0 \times 10^{-6}\ C $ and $\mp 6.0 \times 10^{-6}\ C$
$\pm 0.5 \times 10^{-6}\ C$ and $\mp 1.5 \times 10^{-6}\ C$
Solution
Let initial charge be $Q 1$ and $Q 2$.
Initial force $=F=\frac{K \times Q 1 \times Q 2}{r^{2}}=-0.108$
$F=\frac{9 \times 10^{9} \times Q 1 \times Q 2^{r^{2}}}{0.25}=-0.108$
$Q 1 . Q 2=-3 \times 10^{-12}$
Now on connecting them, since they are identical so they will have equal charges.
Final charge $=\frac{Q 1+Q 2}{2}$
Final force $=F=\frac{9 \times 10^{9} \times(Q 1+Q 2)^{2}}{4 \times 0.25}=0.036$
$Q 1+Q 2=\pm 2 \times 10^{-6}$
$Q 2=\pm 1.0 \times 10^{-6} \mathrm{C} \quad, Q 1=\mp 3.0 \times 10^{-6} \mathrm{C}$
