91Ó°ÊÓ

I suggest drawing and labelling what's stated within the problem, then calculating the force on $A$from $B$($F_{BA}$) and therefore the force on $A$from $C$($F_{CA}$).

To compute the forces on $A$from each of the purpose charges, we use $F=\frac{k{q}_1{q}_2}{r^2}$. $k=9×{10}^9\frac{N{m}^2}{C^2}$is an vital note to stay in mind.

We start by calculating the forces that $B$has on a. Because $A$is positive and $B$is negative, they're going to be interested to one other, and also the net force on $A$are going to be the proper.

$$\begin{gathered} F_{BA}=\frac{k\left(1×{10}^{-9}C\right)\left(1×{10}^{-9}C\right)}{(0.01\mathrm{m}{)}^2} \\ =9×{10}^{-5}\mathrm{N} \end{gathered}$$

Since both $A$and $C$are positive, they repel one other, we must now determine the force $C$exerts on $A$. This means that $C$'s net force is to the left.

$$\begin{gathered} F_{CA}=\frac{k\left(4×{10}^{-9}C\right)\left(1×{10}^{-9}C\right)}{(0.02\mathrm{m}{)}^2} \\ =9×{10}^{-5}\mathrm{N} \end{gathered}$$

Now that we all know the magnitudes of both point charges' forces, we are able to simply add them together to see the overall force. It seems to be $0$as a matter of fact.

$F_{net}=F_{CA}-F_{BA}=0\mathrm{N}$