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

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(C) Let the point $P$ be at a distance $x$ from the origin $(x=0)$.The electric field due to $+8q$ at $P$ is $E_1 = \frac{k(8q)}{x^2}$ (directed away

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$2\,L$

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(C) Let the point $P$ be at a distance $x$ from the origin $(x=0)$.The electric field due to $+8q$ at $P$ is $E_1 = \frac{k(8q)}{x^2}$ (directed away from the origin).The electric field due to $-2q$ at $P$ is $E_2 = \frac{k(2q)}{(x-L)^2}$ (directed towards the origin).For the net electric field to be zero at $P$,the magnitudes must be equal: $E_1 = E_2$.$\frac{k(8q)}{x^2} = \frac{k(2q)}{(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 = 2L$Thus,the point is located at $x = 2L$.

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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