Physics - Elecrostatics Question with Solution | TestHub

Normal reaction applied by table on object will be, $N=mg$.

Both charges are same, hence a repulsive force will act between them.

Value of force acting will be, $\frac{k{Q}^2}{L^2}$.

Now the maximum value of friction force which table can provide is,

$=\mu N=\mu mg$

For equilibrium, $\frac{k{Q}^2}{L^2}=\mu mg$.

$L=\sqrt{\frac{k{Q}^2}{\mu mg}}=\sqrt{\frac{k}{0.25mg}}Q=2\sqrt{\frac{k}{mg}}Q$

$=2\sqrt{\frac{9\times {10}^9}{10\times {10}^{-3}\times 10}}\left(2\times {10}^{-7}\right)=0.12 \mathrm{m}=12 \mathrm{cm}$