Two point charges +8q  and -2q are located at x = 0 and x = L respectively. The location of a point on the x-axis at which the net electric field due to these two point charges is zero is

The net field will be zero at a point outside the charges and near the charge which is smaller in magnitude.

Suppose E.F. is zero at P as shown. 

Hence at P; $\mathrm{k}\cdot \frac{8\mathrm{q}}{(\mathrm{L}+\mathrm{l}{)}^2}=\frac{\mathrm{k}\cdot \left(2\mathrm{q}\right)}{{\mathrm{l}}^2}\Rightarrow \mathrm{I}=\mathrm{L}$

So distance of P from origin is L + L = 2L.