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(B) Let the point $P$ be at a distance $x$ from the origin. Since the charges have opposite signs,the point where the net electric field is zero must

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$2 L$

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(B) Let the point $P$ be at a distance $x$ from the origin. Since the charges have opposite signs,the point where the net electric field is zero must lie outside the region between the charges,specifically on the side of the smaller magnitude charge $(-2 q)$.Let the distance of point $P$ from the origin be $x$. Then the distance of $P$ from the charge $-2 q$ at $X=L$ is $(x-L)$.The electric field due to $+8 q$ at $P$ is $E_1 = \frac{1}{4 \pi \varepsilon_0} \frac{8 q}{x^2}$.The electric field due to $-2 q$ at $P$ is $E_2 = \frac{1}{4 \pi \varepsilon_0} \frac{2 q}{(x-L)^2}$.For the net field to be zero,$E_1 = E_2$:$\frac{8 q}{x^2} = \frac{2 q}{(x-L)^2}$$\frac{4}{x^2} = \frac{1}{(x-L)^2}$Taking the square root on both sides: $\frac{2}{x} = \frac{1}{x-L}$ (taking the positive root as $x > L$)$2(x-L) = x$$2x - 2L = x$$x = 2L$.Thus,the location of point $P$ from the origin is $2 L$.

Two point charges $+8 q$ and $-2 q$ are located at $X=0$ (origin) and $X=L$ respectively. The net electric field due to these two charges is zero at point $P$ on the $X$-axis. The location of point $P$ from the origin is:

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