Physics Help?

>>8907768
draw force diagram
plug and chug

draw your fbd

>>8907799
sideways sorry, but yeah idk???

bump, really need help

>>8907768
I got 3.894868419x10^7 at an angle of -11.25223282 degrees. You might want to try putting in 349 degrees as that is with respect to the positive axis? Not sure.

>>8907919
Mind sending me a pic of ur work if possible? Would really appreciate it

>>8907768
>>8907799
OP, my suggestion for solving physics problems is to always solve equations algebraically (with symbols) and only plug in numbers at the end of the day. It's more accurate, it's easier to catch mistakes, things may cancel to give you nice results, etc.

Let's call the bottom 2.0nC charge [math]Q_1[/math] and the top 3.0nC charge [math]Q_2[/math]. The 6.0nC charge is [math]q[/math]. Write out the force on [math]q[/math]:

[eqn]
\begin{align}
\vec{F} &= \vec{F}_1 + \vec{F}_2\\
&= \frac{1}{4\pi\epsilon_0}\frac{qQ_1}{r_1^2}\hat{r}_1 + \frac{1}{4\pi\epsilon_0}\frac{qQ_2}{r_2^2}\hat{r}_2\\
&= \frac{q}{4\pi\epsilon_0} \left ( \frac{Q_1}{r_1^2}\hat{r}_1 + \frac{Q_2}{r_2^2}\hat{r}_2 \right )\\
\end{align}
[/eqn]

so, what are [math]\vec{r}_1[/math] and [math]\vec{r}_2[/math]?

Well, using the coordinate system provided...

[eqn]
\vec{r}_1 = x_1 \hat{x} + y_1 \hat{y} = 0.5 m \hat{x} + 0.5 m \hat{y}\\
\vec{r}_2 = x_2 \hat{x} + y_2 \hat{y} = 0.5 m \hat{x} - 0.5 m \hat{y}\\
[/eqn]

You can now compute the length [math]r^2 = x^2 + y^2[/math] and direction [math]\hat{r} = \vec{r} / r [/math] of both vectors. The angle of the vector is [math]\theta = tan^{-1}\frac{y}{x}[/math].

You should be able to solve this now. Ask if you have trouble.

>>8907974
Shit thanks man appreciate it

>>8908016
glad i could help