Two charges 4 × 10^–9 C  and  –16 × 10^–9 C are separated by a distance 20 cm in air. The position of the neutral point from the small charge is

Let the neutral point be at a distance x from smaller charge.

$\Rightarrow \frac{\mathrm{k}\left(4\times {10}^{-9}\right)}{(\mathrm{x}{)}^2}-\frac{\left(16\times {10}^{-9}\right)\mathrm{q}}{{\left(\mathrm{d} + \mathrm{x}\right)}^2} = 0$

$\Rightarrow \frac{\mathrm{k}\left(4\times {10}^{-9}\right)}{(\mathrm{x}{)}^2}=\frac{\left(16\times {10}^{-9}\right)}{{\left(\mathrm{d} + \mathrm{x}\right)}^2}$

$\Rightarrow \frac{{\left(\mathrm{d} + \mathrm{x}\right)}^2}{x^2}=\frac{16}{4}$

$\Rightarrow \frac{\left(\mathrm{d} + \mathrm{x}\right)}{x}=2$

$\Rightarrow x =d = 20 \mathrm{cm}$