91Ó°ÊÓ

I propose drawing a three-point charged graphic because we'll be using superposition and symmetry to create this problem a touch bit easier to answer.

The problem requires us to see the scale and direction of $F_{\text{net}}$. The direction of $F_{\text{net}}$may be determined using superposition. $F_{\text{net}}$must be facing upwards since $A$and $B$have the identical amount of charge, are at the identical distance from $C$, and are all positive. Also, we may use the tip-to-tail approach to work out where $F_{\text{net}}$stops. Because the $x$-components cancel one another out, this proves to be incredibly advantageous. As a result, we all know that $F_{\text{net}}$only features a $y$-component. last but not least, some important information The last piece of data we are able to glean from this diagram is that $\mathrm{θ}=60°$. This can be why images and ideas like superposition and therefore the tip-to-tail approach are so valuable.

To get,

$∑F_x:F_{AC}+F_{BC}=0N→F_{AC}=-F_{BC}$

Because we demonstrated before that $F_{AC}=-F_{BC}$, we will simply double it to induce$F_{net}$. It's worth noting that the negative only indicates that the $y$-component is positive.

$$\begin{gathered} F=\frac{k·q_a{q}_c}{r^2} \\ =\frac{9×{10}^{9}N\frac{m^2}{C^2}\left(2×{10}^{-9}C\right)\left(1×{10}^{-9}C\right)}{(0.01m{)}^2} \\ =1.8×{10}^{-4}N \end{gathered}$$

By, ${\left(F_{AC}\right)}_y={\left(F_{BC}\right)}_y$. This section also understand to, but the calculated number will be thought of because the hypotenuse of a triangle.

$2\left(1.8×{10}^{-4}\mathrm{Nsin}60°\right)=3.11×{10}^{-4}\mathrm{N}\hat{j}$