Two charges 2 μC and 1 μC are placed at a distance of 10 cm. The position of third charge from 2 μC between them so that it does not experience any force is:

Let the charge is placed at distance x from 2 μC.

Now the net force acting on charge q should be zero.

$\Rightarrow \frac{\mathrm{k}\left(1\times {10}^{-6}\right)\mathrm{q}}{(10-\mathrm{x}{)}^2}=\frac{\left(2\times {10}^{-6}\right)\mathrm{q}}{{\mathrm{x}}^2}$

$\Rightarrow \frac{1}{10-x}=\frac{\sqrt{2}}{x}$

$\Rightarrow x=10\sqrt{2}-x\sqrt{2}$

$\Rightarrow x(1+\sqrt{2})=10\sqrt{2}$

$\Rightarrow x=\frac{10\sqrt{2}}{1+\sqrt{2}}=\frac{14.14}{2.414}=5.857 \mathrm{cm}$