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

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

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

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

theEric
 one year ago
Best ResponseYou've already chosen the best response.0Well, 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?

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

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

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

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

Shhot
 one year ago
Best ResponseYou've already chosen the best response.0the 10nC is pulling down and to the right, and q is pulling down and to the right.

theEric
 one year ago
Best 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....

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

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

theEric
 one year ago
Best ResponseYou've already chosen the best response.0Yep, 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!

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

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

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

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

theEric
 one year ago
Best ResponseYou've already chosen the best response.0Okay, 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?

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

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

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

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

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

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

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

theEric
 one year ago
Best ResponseYou've already chosen the best response.0Diagonal, 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.

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

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

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

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

theEric
 one year ago
Best 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?

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

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

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

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

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

theEric
 one year ago
Best ResponseYou've already chosen the best response.05cm is the total distance between the bottom charges, right?...

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

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

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

theEric
 one year ago
Best ResponseYou've already chosen the best response.0Along 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.

Shhot
 one year ago
Best ResponseYou've already chosen the best response.0oh, 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?

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

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

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

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

theEric
 one year ago
Best ResponseYou've already chosen the best response.0F10nC + 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.

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

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