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 on the x-axis.

Since the charges have opposite signs, the null point must lie outside the region between the charges and closer to the charge with the smaller magnitude. Thus, the point lies at $x > L$.

Equating the magnitudes of the electric fields due to both charges at this point:

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

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

Taking the square root on both sides:

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

$2x - 2L = x$

$x = 2L$

Answer: $2L$