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(D) Let the point where the net electric field is zero be at a distance $x$ from the origin $(x=0)$.Since the charges have opposite signs,the null poi

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$2 L$

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(D) Let the point where the net electric field is zero be at a distance $x$ from the origin $(x=0)$.Since the charges have opposite signs,the null point must lie outside the region between the charges,specifically on the side of the smaller magnitude charge $(-2 q)$.Let the point be at $x > L$. The distance from $+8 q$ is $x$ and the distance from $-2 q$ is $(x - L)$.For the net electric field to be zero,the magnitudes of the electric fields must be equal:$E_1 = E_2$$\frac{K(8 q)}{x^2} = \frac{K(2 q)}{(x - L)^2}$$\frac{4}{x^2} = \frac{1}{(x - L)^2}$Taking the square root on both sides:$\frac{2}{x} = \frac{1}{x - L}$$2(x - L) = x$$2x - 2L = x$$x = 2 L$Thus,the point is at a distance of $2 L$ from the origin.

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

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