Two point charges 8q and 2q are located at x=0 and x=L...

Let the point where the net electric field is zero be at a distance $x$ from the origin.

Since both charges are positive, the null point must lie between them. The distance of this point from the charge $8q$ is $x$ and from the charge $2q$ is $L - x$.

Equating the magnitudes of the electric fields produced by both charges at this point:

$\dfrac{k(8q)}{x^2} = \dfrac{k(2q)}{(L - x)^2}$

$\dfrac{4}{x^2} = \dfrac{1}{(L - x)^2}$

Taking the square root on both sides:

$\dfrac{2}{x} = \dfrac{1}{L - x}$

$2(L - x) = x$

$2L - 2x = x$

$3x = 2L$

$x = \dfrac{2L}{3}$

Answer: $\dfrac{2L}{3}$