Two electric charges of $9 μ C$ and $- 3 μ C$ are placed $0.16 m$ apart in air. There will be a point P at which electric potential is $z e r o$ on the line joining the two charges and in between them. The distance of P from $9 μ C$ charge is:

0.14 m

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0.12 m

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0.08 m

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0.06 m

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Solution

The correct option is B $0.12 m$
let point p be at at distance $x$ form $1 μ c$ ,
we know potential $v = \frac{k q}{r}$
now potential at $p = 0$
$\Rightarrow \frac{k (9)}{x} + \frac{k (- 3)}{(0.16 - x)} = 0$

$\Rightarrow \frac{9}{x} = \frac{3}{0.16 - x}$

$\Rightarrow 1.44 - 9 x = 3 x$
$\Rightarrow 1.44 = 12 x$
$\Rightarrow x = 0.12 m$

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