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Q.

A point charge of $10 μ C$ is placed at the origin. At what location on the x-axis should a point charge of $40 μ C$ be placed so that net electric field is zero at $x = 2 c m$ on the x-axis

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a

$x = 8 c m$

b

$x = 6 c m$

c

$x = 4 c m$

d

$x = - 4 c m$

answer is B.

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Detailed Solution

$x = \frac{d}{\sqrt{\frac{q_{2}}{q_{1}}} + 1} \Rightarrow 2 = \frac{d}{\sqrt{\frac{40}{10}} + 1}$
$\Rightarrow d = 6 c m$

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