## anonymous 2 years ago The force on the -1.0nC charge is as shown in the figure. What is the magnitude of this force?

1. anonymous

figure attached:

2. theEric

Hi! So, do you have any idea where to start?

3. anonymous

no clue, not sure if q has anything to do with it.

4. anonymous

is the q just to help calculate lengths or does it impact the force?

5. theEric

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?

6. theEric

Problems don't usually include unnecessary information.

7. anonymous

I know why the charge goes down conceptually, but do I have to calculate q and if so, how?

8. theEric

So, conceptually, why does the force go down? Your other questions will be answered momentarily.

9. theEric

(And they might be answered by you, in a moment!)

10. anonymous

the 10nC is pulling down and to the right, and q is pulling down and to the right.

11. theEric

"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....

12. anonymous

...must be equal. I understand, but do i specifically need to find q?

13. theEric

It says I'm typing but I'm not... You finish my earlier sentence!

14. theEric

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!

15. theEric

10nc*, not -10nC.

16. theEric

C*

17. theEric

nC*

18. theEric

Ah...

19. theEric

So the horizontal component of forces from the 10nC and q charges add to zero. Time for some math?

20. theEric

Solve for q, since it will be the only unknown.

21. anonymous

I dont get what you mean...

22. anonymous

What?

23. anonymous

sovling for q, How do I go about doing that? Unsure where to start...

24. theEric

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?

25. theEric

And then you can spit it into horizontal and vertical components?

26. anonymous

yep, can do that. Its just finding q and then F

27. theEric

$F_{10nC}+F_{q}=0\Rightarrow F_q=F_{10nC}$ $F_{10nC}=\frac{\left( 10\ [nC]\right)\ \left( -1\ [nC]\right)}{r_{10nC\rightarrow -1nC}}$ $F_{q}=\frac{\left(q\right)\ \left( -1\ [nC]\right)}{r_{q\rightarrow -1nC}}$

28. theEric

No! Wait! I did that wrong! One change.

29. theEric

$F_{10nC,\ x}+F_{q,\ x}=0\Rightarrow F_{q,\ x}=-F_{10nC,\ x}$

30. anonymous

still lost... do I use the entire bottom length?

31. anonymous

when you say$F _{10nC}=\frac{ (10nC)(1nC) }{r10nC \rightarrow-1nC }$ Is that the length along the diagonal or the x-component?

32. theEric

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.

33. anonymous

yes, so i use the diagonals for both the 10nC, 1nC and q,1nC when finding value of q?

34. theEric

Yep!

35. anonymous

ok

36. theEric

The horizontal forces cancel each other out. $$F_q cos(30^\circ) = -F_{10nC}cos(60^\circ)$$

37. anonymous

ok, starting to get clearer.

38. theEric

$F_q = -F_{10nC}\frac{cos(60^\circ)}{cos(30^\circ) }$ Still good?

39. anonymous

yes

40. anonymous

For q, I got 3e-8C.

41. theEric

$F_{q}=\frac{\left(q\right)\ \left( -1\ [nC]\right)}{r_{q\rightarrow -1nC}}$ 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?

42. theEric

If so, you can solve for $$q$$ after you find the distances.

43. anonymous

ahh, now i'm lost again...

44. theEric

I used substitution.

45. anonymous

Could I just use trig to find the distances?

46. theEric

Yup! You'll need to. I don't see how to get around it, at least.

47. theEric

Do you see what you need to do?

48. theEric

5cm is the total distance between the bottom charges, right?...

49. anonymous

I know how to find the distances. Its just figuring out which distance goes into r for$F=\frac{ Kq1q2 }{ r^2}$

50. theEric

The $$r$$ you use to calculate a force between two charges is the distance between those charges.

51. anonymous

Along the diagonal? or (since x-components cancel) the y-component

52. theEric

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.

53. anonymous

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?

54. anonymous

To find q, I have Fq(cos(30) = F10nC(cos(60)

55. theEric

Yep! And yep.

56. theEric

Except they're not quite equal.

57. theEric

Equal in magnitude, opposite in direction. So they cancel.

58. anonymous

yes,F10nC - Fq = 0. I just moved them.

59. anonymous

q=1.7e-8C

60. theEric

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.

61. anonymous

I got 1.7e-4 N for total force

62. theEric

I still didn't find the distances.. And then I did more math that became jumbled.