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

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theEricBest ResponseYou've already chosen the best response.0
Hi! So, do you have any idea where to start?
 9 months ago

ShhotBest ResponseYou've already chosen the best response.0
no clue, not sure if q has anything to do with it.
 9 months ago

ShhotBest ResponseYou've already chosen the best response.0
is the q just to help calculate lengths or does it impact the force?
 9 months ago

theEricBest ResponseYou've already chosen the best response.0
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?
 9 months ago

theEricBest ResponseYou've already chosen the best response.0
Problems don't usually include unnecessary information.
 9 months ago

ShhotBest ResponseYou've already chosen the best response.0
I know why the charge goes down conceptually, but do I have to calculate q and if so, how?
 9 months ago

theEricBest ResponseYou've already chosen the best response.0
So, conceptually, why does the force go down? Your other questions will be answered momentarily.
 9 months ago

theEricBest ResponseYou've already chosen the best response.0
(And they might be answered by you, in a moment!)
 9 months ago

ShhotBest ResponseYou've already chosen the best response.0
the 10nC is pulling down and to the right, and q is pulling down and to the right.
 9 months ago

theEricBest ResponseYou've already chosen the best response.0
"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....
 9 months ago

ShhotBest ResponseYou've already chosen the best response.0
...must be equal. I understand, but do i specifically need to find q?
 9 months ago

theEricBest ResponseYou've already chosen the best response.0
It says I'm typing but I'm not... You finish my earlier sentence!
 9 months ago

theEricBest ResponseYou've already chosen the best response.0
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!
 9 months ago

theEricBest ResponseYou've already chosen the best response.0
So the horizontal component of forces from the 10nC and q charges add to zero. Time for some math?
 9 months ago

theEricBest ResponseYou've already chosen the best response.0
Solve for q, since it will be the only unknown.
 9 months ago

ShhotBest ResponseYou've already chosen the best response.0
I dont get what you mean...
 9 months ago

ShhotBest ResponseYou've already chosen the best response.0
sovling for q, How do I go about doing that? Unsure where to start...
 9 months ago

theEricBest ResponseYou've already chosen the best response.0
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?
 9 months ago

theEricBest ResponseYou've already chosen the best response.0
And then you can spit it into horizontal and vertical components?
 9 months ago

ShhotBest ResponseYou've already chosen the best response.0
yep, can do that. Its just finding q and then F
 9 months ago

theEricBest ResponseYou've already chosen the best response.0
\[F_{10nC}+F_{q}=0\Rightarrow F_q=F_{10nC}\] \[F_{10nC}=\frac{\left( 10\ [nC]\right)\ \left( 1\ [nC]\right)}{r_{10nC\rightarrow 1nC}}\] \[F_{q}=\frac{\left(q\right)\ \left( 1\ [nC]\right)}{r_{q\rightarrow 1nC}}\]
 9 months ago

theEricBest ResponseYou've already chosen the best response.0
No! Wait! I did that wrong! One change.
 9 months ago

theEricBest ResponseYou've already chosen the best response.0
\[F_{10nC,\ x}+F_{q,\ x}=0\Rightarrow F_{q,\ x}=F_{10nC,\ x}\]
 9 months ago

ShhotBest ResponseYou've already chosen the best response.0
still lost... do I use the entire bottom length?
 9 months ago

ShhotBest ResponseYou've already chosen the best response.0
when you say\[F _{10nC}=\frac{ (10nC)(1nC) }{r10nC \rightarrow1nC }\] Is that the length along the diagonal or the xcomponent?
 9 months ago

theEricBest ResponseYou've already chosen the best response.0
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.
 9 months ago

ShhotBest ResponseYou've already chosen the best response.0
yes, so i use the diagonals for both the 10nC, 1nC and q,1nC when finding value of q?
 9 months ago

theEricBest ResponseYou've already chosen the best response.0
The horizontal forces cancel each other out. \( F_q cos(30^\circ) = F_{10nC}cos(60^\circ)\)
 9 months ago

ShhotBest ResponseYou've already chosen the best response.0
ok, starting to get clearer.
 9 months ago

theEricBest ResponseYou've already chosen the best response.0
\[F_q = F_{10nC}\frac{cos(60^\circ)}{cos(30^\circ) }\] Still good?
 9 months ago

theEricBest ResponseYou've already chosen the best response.0
\[F_{q}=\frac{\left(q\right)\ \left( 1\ [nC]\right)}{r_{q\rightarrow 1nC}}\] 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?
 9 months ago

theEricBest ResponseYou've already chosen the best response.0
If so, you can solve for \(q\) after you find the distances.
 9 months ago

ShhotBest ResponseYou've already chosen the best response.0
ahh, now i'm lost again...
 9 months ago

ShhotBest ResponseYou've already chosen the best response.0
Could I just use trig to find the distances?
 9 months ago

theEricBest ResponseYou've already chosen the best response.0
Yup! You'll need to. I don't see how to get around it, at least.
 9 months ago

theEricBest ResponseYou've already chosen the best response.0
Do you see what you need to do?
 9 months ago

theEricBest ResponseYou've already chosen the best response.0
5cm is the total distance between the bottom charges, right?...
 9 months ago

ShhotBest ResponseYou've already chosen the best response.0
I know how to find the distances. Its just figuring out which distance goes into r for\[F=\frac{ Kq1q2 }{ r^2}\]
 9 months ago

theEricBest ResponseYou've already chosen the best response.0
The \(r\) you use to calculate a force between two charges is the distance between those charges.
 9 months ago

ShhotBest ResponseYou've already chosen the best response.0
Along the diagonal? or (since xcomponents cancel) the ycomponent
 9 months ago

theEricBest ResponseYou've already chosen the best response.0
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.
 9 months ago

ShhotBest ResponseYou've already chosen the best response.0
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?
 9 months ago

ShhotBest ResponseYou've already chosen the best response.0
To find q, I have Fq(cos(30) = F10nC(cos(60)
 9 months ago

theEricBest ResponseYou've already chosen the best response.0
Except they're not quite equal.
 9 months ago

theEricBest ResponseYou've already chosen the best response.0
Equal in magnitude, opposite in direction. So they cancel.
 9 months ago

ShhotBest ResponseYou've already chosen the best response.0
yes,F10nC  Fq = 0. I just moved them.
 9 months ago

theEricBest ResponseYou've already chosen the best response.0
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.
 9 months ago

ShhotBest ResponseYou've already chosen the best response.0
I got 1.7e4 N for total force
 9 months ago

theEricBest ResponseYou've already chosen the best response.0
I still didn't find the distances.. And then I did more math that became jumbled.
 9 months ago
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