Two particles A and B having charges 20 $\mu C$  and $5\mu C$ - respectively are held fixed with a separation of 5cm. At what position a third charged particle should be placed so that it does not experience a net electric force?

 

Correct Option - 4


Detailed Solution:

From observation, we can say that right of $-5\mu c$ charge, net electric force (or electric field) can be zero.

$So,\frac{k\left(5\mu c\right)q}{x^2}=\frac{k\left(20\mu c\right)q}{{\left(5+x\right)}^2}$

$\Rightarrow \frac{1}{x^2}=\frac{4}{{\left(5+x\right)}^2}$

$\Rightarrow \frac{5+x}{x}=2$

x = 5 cm

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