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

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(A) Let the point where the net electric field is zero be at a distance $r$ from the charge $+8q$ (at $x=0$).Since the charges have opposite signs,the

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$2L$

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(A) Let the point where the net electric field is zero be at a distance $r$ from the charge $+8q$ (at $x=0$).Since the charges have opposite signs,the null point must lie outside the region between the charges,specifically closer to the smaller magnitude charge $(-2q)$.Let the distance from $+8q$ be $r$. Then the distance from $-2q$ is $(r - L)$.The electric field due to $+8q$ is $E_1 = \frac{k(8q)}{r^2}$ and due to $-2q$ is $E_2 = \frac{k(2q)}{(r - L)^2}$.For the net field to be zero,$E_1 = E_2$,so $\frac{8q}{r^2} = \frac{2q}{(r - L)^2}$.Taking the square root on both sides: $\frac{\sqrt{8}}{r} = \frac{\sqrt{2}}{r - L}$.$2\sqrt{2}(r - L) = \sqrt{2}r \Rightarrow 2r - 2L = r \Rightarrow r = 2L$.Thus,the point is at $x = 2L$.

Two point charges $+8q$ and $-2q$ are placed at $x = 0$ and $x = L$ respectively. The net electric field due to these two point charges will be zero at a point on the $x$-axis at:

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