Two point charges +8q and -2q are located at x = 0 and x = L respectively. location of a point on x-

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1. Electric Charges and Fields

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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 net electric field due to these two point charges is zero, is

A

$8\,L$

B

$4\,L$

C

$2\,L$

D

$L/4$

Solution

The net electric field at $P$ is

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

$\Rightarrow \frac{4}{x^{2}}=\frac{1}{(x-L)^{2}}$

$\Rightarrow \frac{2}{x}=\frac{1}{(x-L)}$

or $x=2 x-2 L \Rightarrow x=2 L$

Standard 12

Physics

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