Two point charges +8q and -2q are located at | vedclass

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(C) The net electric field will be zero at a point outside the charges and closer to the charge which is smaller in magnitude.Let the point $P$ be at

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$2\,L$

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(C) The net electric field will be zero at a point outside the charges and closer to the charge which is smaller in magnitude.Let the point $P$ be at a distance $l$ from the charge $-2q$ at $x = L$. The distance of $P$ from the charge $+8q$ at $x = 0$ is $(L + l)$.At point $P$,the magnitude of the electric field due to $+8q$ must be equal to the magnitude of the electric field due to $-2q$:$\frac{k \cdot |8q|}{(L + l)^2} = \frac{k \cdot |2q|}{l^2}$Simplifying the equation:$\frac{8}{(L + l)^2} = \frac{2}{l^2}$$\frac{4}{(L + l)^2} = \frac{1}{l^2}$Taking the square root on both sides:$\frac{2}{L + l} = \frac{1}{l}$$2l = L + l$$l = L$The position of point $P$ on the $x$-axis is $x = L + l = L + L = 2L$.

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

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