Point charges +8q and -2q are placed at x = 0 | vedclass

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(D) Let the point where the electric field is zero be $A$,and let the distance between the charge $-2q$ and point $A$ be $x$.The electric field due to

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$2L$

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(D) Let the point where the electric field is zero be $A$,and let the distance between the charge $-2q$ and point $A$ be $x$.The electric field due to $+8q$ at point $A$ is $E_1 = \frac{k(8q)}{(L+x)^2}$ (directed away from the origin).The electric field due to $-2q$ at point $A$ is $E_2 = \frac{k(2q)}{x^2}$ (directed towards the origin).For the resultant electric field to be zero,$E_1 = E_2$.$\frac{k(8q)}{(L+x)^2} = \frac{k(2q)}{x^2}$$\frac{4}{(L+x)^2} = \frac{1}{x^2}$Taking the square root on both sides:$\frac{2}{L+x} = \frac{1}{x}$$2x = L + x$$x = L$The distance of point $A$ from $x = 0$ is $L + x = L + L = 2L$.

Point charges $+8q$ and $-2q$ are placed at $x = 0$ and $x = L$ respectively. At which point on the $X$-axis will the resultant electric field intensity be zero?

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