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The force on the -1.0nC charge is as shown in the figure. What is the magnitude of this force?

Physics
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figure attached:
Hi! So, do you have any idea where to start?
no clue, not sure if q has anything to do with it.

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Other answers:

is the q just to help calculate lengths or does it impact the force?
Well, it does if it has an effect on the \(-1\ [nC]\) charge! It will definitely have an effect on the force. Here's a question that will get you started: why is the force straight down? And then, what does that mean?
Problems don't usually include unnecessary information.
I know why the charge goes down conceptually, but do I have to calculate q and if so, how?
So, conceptually, why does the force go down? Your other questions will be answered momentarily.
(And they might be answered by you, in a moment!)
the 10nC is pulling down and to the right, and q is pulling down and to the right.
"the 10nC is pulling down and to the" \(left\), "and q is pulling down and to the right." I'm sure you thought that, but mistyped. And for there to be no horizontal component of force on the -1nC charge, the horizontal forces from 10nC and q....
...must be equal. I understand, but do i specifically need to find q?
It says I'm typing but I'm not... You finish my earlier sentence!
Yep, you're right! And to find the net force on the -1nC charge, you do need \(q\)! It obviously has an impact on the -1nC charge, since it cancels the horizontal force of the -10nC charge! And knowing \(q\) will be necessary to calculate the force on the -1nC charge!
10nc*, not -10nC.
C*
nC*
Ah...
So the horizontal component of forces from the 10nC and q charges add to zero. Time for some math?
Solve for q, since it will be the only unknown.
I dont get what you mean...
What?
sovling for q, How do I go about doing that? Unsure where to start...
Okay, well, using the formula\[F=k\frac{q_1\ q_2}{r^2}\]you can "find" the force of the 10nC charge on the -1nC charge. Right?
And then you can spit it into horizontal and vertical components?
yep, can do that. Its just finding q and then F
\[F_{10nC}+F_{q}=0\Rightarrow F_q=F_{10nC}\] \[F_{10nC}=\frac{\left( 10\ [nC]\right)\ \left( -1\ [nC]\right)}{r_{10nC\rightarrow -1nC}}\] \[F_{q}=\frac{\left(q\right)\ \left( -1\ [nC]\right)}{r_{q\rightarrow -1nC}}\]
No! Wait! I did that wrong! One change.
\[F_{10nC,\ x}+F_{q,\ x}=0\Rightarrow F_{q,\ x}=-F_{10nC,\ x}\]
still lost... do I use the entire bottom length?
when you say\[F _{10nC}=\frac{ (10nC)(1nC) }{r10nC \rightarrow-1nC }\] Is that the length along the diagonal or the x-component?
Diagonal, sorry. It must be the distance between the 10nC and -1nC charges if we want to calculate the force between them. Have you tried drawing a force diagram for forces on the -1nC charge? They're very helpful! I'll post mine momentarily.
yes, so i use the diagonals for both the 10nC, 1nC and q,1nC when finding value of q?
Yep!
ok
The horizontal forces cancel each other out. \( F_q cos(30^\circ) = -F_{10nC}cos(60^\circ)\)
ok, starting to get clearer.
\[F_q = -F_{10nC}\frac{cos(60^\circ)}{cos(30^\circ) }\] Still good?
yes
For q, I got 3e-8C.
\[F_{q}=\frac{\left(q\right)\ \left( -1\ [nC]\right)}{r_{q\rightarrow -1nC}}\] So we substitute \(F_q\)'s value in where \(F_q\) is in\[F_q = -F_{10nC}\frac{cos(60^\circ)}{cos(30^\circ) }\]to get\[\frac{\left(q\right)\ \left( -1\ [nC]\right)}{r_{q\rightarrow -1nC}} = -F_{10nC}\frac{cos(60^\circ)}{cos(30^\circ) }\] Still with me?
If so, you can solve for \(q\) after you find the distances.
ahh, now i'm lost again...
I used substitution.
Could I just use trig to find the distances?
Yup! You'll need to. I don't see how to get around it, at least.
Do you see what you need to do?
5cm is the total distance between the bottom charges, right?...
I know how to find the distances. Its just figuring out which distance goes into r for\[F=\frac{ Kq1q2 }{ r^2}\]
The \(r\) you use to calculate a force between two charges is the distance between those charges.
Along the diagonal? or (since x-components cancel) the y-component
Along the diagonal, for sure. To get any one component alone, you have to get the total force and break it up. If you used just the \(x\) distance in your formula, you'd be calculating the force as if the charges themselves were that far away.
oh, so use the diagonal distance for r and then multiply \[\frac{ Kq1q2 }{ r^2 }\] by either sin/cos(theta) to get x or y component?
To find q, I have Fq(cos(30) = F10nC(cos(60)
Yep! And yep.
Except they're not quite equal.
Equal in magnitude, opposite in direction. So they cancel.
yes,F10nC - Fq = 0. I just moved them.
q=1.7e-8C
F10nC + Fq = 0. The net horizontal force, which is the sum of all horizontal forces, must equal zero. That number seems reasonable! I didn't find the distances.
I got 1.7e-4 N for total force
I still didn't find the distances.. And then I did more math that became jumbled.

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