Two particles A and B having charges 8 × 10^-6 C and –2 × 10^-6 C respectively are held fixed with a separation of 20 cm. Where should a third charged particle be placed so that it does not experience a net electric force ?

20 cm

As the net electric force on C should be equal to zero, the force due to A and B must be opposite in direction. Hence, the particle should be placed on the line AB. As A and B have charges of opposite signs, C cannot be between A and B

Also A has larger magnitude of charge than B. Hence, C should be placed closer to B than A. The situation is shown in figure. Suppose BC = x and the charge on C is Q

$\vec{F}_{C A}=\frac{1}{4 \pi \epsilon_0} \frac{\left(8.0 \times 10^{-6}\right) Q}{(0.2+x)^2} \hat{i}$

and $\vec{F}_{C B}=\frac{-1}{4 \pi \epsilon_0} \frac{\left(2.0 \times 10^{-6}\right) Q}{x^2} \hat{i}$

$\vec{F}_C=\vec{F}_{C A}+\vec{F}_{C B}=\frac{1}{4 \pi \epsilon_0}\left[\frac{\left(8.0 \times 10^{-6}\right) Q}{(0.2+x)^2}-\frac{\left(2.0 \times 10^{-6}\right) Q}{x^2}\right] i$

But $\left|\vec{F}_C\right|=0$

Hence $\frac{1}{4 \pi \in_0}\left[\frac{\left(8.0 \times 10^{-6}\right) Q}{(0.2+x)^2}-\frac{\left(2.0 \times 10^{-6}\right) Q}{x^2}\right]=0$

Which gives x = 0.2 m = 20 cm