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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$2L$

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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 $-2q$ on the $x$-axis,as shown in the figure.The electric field due to $+8q$ at $P$ is $E_1 = \frac{k(8q)}{(L+l)^2}$ directed away from the origin.The electric field due to $-2q$ at $P$ is $E_2 = \frac{k(2q)}{l^2}$ directed towards the origin.For the net electric field to be zero at $P$,we must have $E_1 = E_2$.$\frac{k(8q)}{(L+l)^2} = \frac{k(2q)}{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 distance of point $P$ from the origin $(x=0)$ is $L + l = L + L = 2L$.

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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