Point charge q₁ = 2 μC and q₂ = -1 μC are kept at points x = 0 and x = 6, respectively. Electrical potential will be zero at points

Potential will be zero at two points.

At internal point (M),

$\frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \left[\frac{2\times {10}^{-6}}{(6-\mathrm{l})}+\frac{\left(-1\times {10}^{-6}\right)}{\mathrm{l}}\right]=0\Rightarrow \mathrm{l}=2$

So distance of M from origin, x = 6 - 2 = 4 At exterior point (N),

$\frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \left[\frac{2\times {10}^{-6}}{\left(6+{\mathrm{l}}^{\text{'}}\right)}+\frac{\left(-1\times {10}^{-6}\right)}{\mathrm{l}}\right]=0\Rightarrow {\mathrm{l}}^{\text{'}}=6$

So distance of N from origin, x = 6 + 6 = 12