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Shhot
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The force on the 1.0nC charge is as shown in the figure. What is the magnitude of this force?
 one year ago
 one year ago
Shhot Group Title
The force on the 1.0nC charge is as shown in the figure. What is the magnitude of this force?
 one year ago
 one year ago

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Shhot Group TitleBest ResponseYou've already chosen the best response.0
figure attached:
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
Hi! So, do you have any idea where to start?
 one year ago

Shhot Group TitleBest ResponseYou've already chosen the best response.0
no clue, not sure if q has anything to do with it.
 one year ago

Shhot Group TitleBest ResponseYou've already chosen the best response.0
is the q just to help calculate lengths or does it impact the force?
 one year ago

theEric Group TitleBest 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?
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
Problems don't usually include unnecessary information.
 one year ago

Shhot Group TitleBest 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?
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
So, conceptually, why does the force go down? Your other questions will be answered momentarily.
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
(And they might be answered by you, in a moment!)
 one year ago

Shhot Group TitleBest 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.
 one year ago

theEric Group TitleBest 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....
 one year ago

Shhot Group TitleBest ResponseYou've already chosen the best response.0
...must be equal. I understand, but do i specifically need to find q?
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
It says I'm typing but I'm not... You finish my earlier sentence!
 one year ago

theEric Group TitleBest 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!
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
10nc*, not 10nC.
 one year ago

theEric Group TitleBest 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?
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
Solve for q, since it will be the only unknown.
 one year ago

Shhot Group TitleBest ResponseYou've already chosen the best response.0
I dont get what you mean...
 one year ago

Shhot Group TitleBest ResponseYou've already chosen the best response.0
sovling for q, How do I go about doing that? Unsure where to start...
 one year ago

theEric Group TitleBest 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?
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
And then you can spit it into horizontal and vertical components?
 one year ago

Shhot Group TitleBest ResponseYou've already chosen the best response.0
yep, can do that. Its just finding q and then F
 one year ago

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

theEric Group TitleBest ResponseYou've already chosen the best response.0
No! Wait! I did that wrong! One change.
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
\[F_{10nC,\ x}+F_{q,\ x}=0\Rightarrow F_{q,\ x}=F_{10nC,\ x}\]
 one year ago

Shhot Group TitleBest ResponseYou've already chosen the best response.0
still lost... do I use the entire bottom length?
 one year ago

Shhot Group TitleBest 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?
 one year ago

theEric Group TitleBest 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.
 one year ago

Shhot Group TitleBest 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?
 one year ago

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

Shhot Group TitleBest ResponseYou've already chosen the best response.0
ok, starting to get clearer.
 one year ago

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

Shhot Group TitleBest ResponseYou've already chosen the best response.0
For q, I got 3e8C.
 one year ago

theEric Group TitleBest 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?
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
If so, you can solve for \(q\) after you find the distances.
 one year ago

Shhot Group TitleBest ResponseYou've already chosen the best response.0
ahh, now i'm lost again...
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
I used substitution.
 one year ago

Shhot Group TitleBest ResponseYou've already chosen the best response.0
Could I just use trig to find the distances?
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
Yup! You'll need to. I don't see how to get around it, at least.
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
Do you see what you need to do?
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
5cm is the total distance between the bottom charges, right?...
 one year ago

Shhot Group TitleBest 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\[F=\frac{ Kq1q2 }{ r^2}\]
 one year ago

theEric Group TitleBest 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.
 one year ago

Shhot Group TitleBest ResponseYou've already chosen the best response.0
Along the diagonal? or (since xcomponents cancel) the ycomponent
 one year ago

theEric Group TitleBest 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.
 one year ago

Shhot Group TitleBest 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?
 one year ago

Shhot Group TitleBest ResponseYou've already chosen the best response.0
To find q, I have Fq(cos(30) = F10nC(cos(60)
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
Yep! And yep.
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
Except they're not quite equal.
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
Equal in magnitude, opposite in direction. So they cancel.
 one year ago

Shhot Group TitleBest ResponseYou've already chosen the best response.0
yes,F10nC  Fq = 0. I just moved them.
 one year ago

theEric Group TitleBest 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.
 one year ago

Shhot Group TitleBest ResponseYou've already chosen the best response.0
I got 1.7e4 N for total force
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
I still didn't find the distances.. And then I did more math that became jumbled.
 one year ago
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