Question

What is the net electric force on charge $A$in FIGURE?

Short answer

The net electric force on charge$A$is$0N$

Step-by-step solution

Formula of Force

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 $F=\frac{k{q}_1{q}_2}{r^2}$. $k=9\times {10}^9\frac{N{m}^2}{C^2}$is an vital note to stay in mind.

Forces on A,B and C

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.

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

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.

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

Find the total force

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.

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