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Two point charges $+ 8 q$ and $- 2 q$ 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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$2 L$

$\frac{L}{4}$

$8 L$

$4 L$

answer is A.

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Detailed Solution

$- \frac{k 2 q}{{(x - L)}^{2}} + \frac{k 8 q}{x^{2}} = 0$

$\Rightarrow x = 2 L$

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