Question

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

Answer 1

As the net electric force $C$ 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 changes 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 $B$ that $A$. The situation is shown in figure. Suppose $BC=x$ and the charge on $C$ is $Q$
${\overrightarrow{F}}_{CA}=\frac{1}{4\pi {\in }_0}\frac{(8.0\times {10}^{-6})Q}{(0.2+x{)}^2}\hat{i}$
and ${\overrightarrow{F}}_{CB}=\frac{-1}{4\pi {\in }_0}\frac{(2.0\times {10}^{-6})Q}{x^2}\hat{i}$
${\overrightarrow{F}}_C={\overrightarrow{F}}_{CA}+{\overrightarrow{F}}_{CB}=\frac{1}{4\pi {\in }_0}[\frac{(8.0\times {10}^{-6})Q}{(0.2\times {10}^{-6})}-\frac{(2.0\times {10}^{-6})Q}{x^2}]\hat{i}$
But $|{\overrightarrow{F}}_C|=0$
Hence $\frac{1}{4\pi {\in }_0}[\frac{(8.0\times {10}^{-6})Q}{(0.2\times x{)}^2}-\frac{(2.0\times {10}^{-6})Q}{x^2}]=0$
Which gives $x=0.2m$