## 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

1. Shhot Group Title

figure attached:

2. theEric Group Title

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

3. Shhot Group Title

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

4. Shhot Group Title

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

5. theEric Group Title

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 Group Title

Problems don't usually include unnecessary information.

7. Shhot Group Title

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

8. theEric Group Title

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

9. theEric Group Title

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

10. Shhot Group Title

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

11. theEric Group Title

"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. Shhot Group Title

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

13. theEric Group Title

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

14. theEric Group Title

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 Group Title

10nc*, not -10nC.

16. theEric Group Title

C*

17. theEric Group Title

nC*

18. theEric Group Title

Ah...

19. theEric Group Title

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

20. theEric Group Title

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

21. Shhot Group Title

I dont get what you mean...

22. xxAshxx Group Title

What?

23. Shhot Group Title

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

24. theEric Group Title

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 Group Title

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

26. Shhot Group Title

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

27. theEric Group Title

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

28. theEric Group Title

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

29. theEric Group Title

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

30. Shhot Group Title

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

31. Shhot Group Title

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

32. theEric Group Title

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. Shhot Group Title

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

34. theEric Group Title

Yep!

35. Shhot Group Title

ok

36. theEric Group Title

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

37. Shhot Group Title

ok, starting to get clearer.

38. theEric Group Title

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

39. Shhot Group Title

yes

40. Shhot Group Title

For q, I got 3e-8C.

41. theEric Group Title

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

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

43. Shhot Group Title

ahh, now i'm lost again...

44. theEric Group Title

I used substitution.

45. Shhot Group Title

Could I just use trig to find the distances?

46. theEric Group Title

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

47. theEric Group Title

Do you see what you need to do?

48. theEric Group Title

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

49. Shhot Group Title

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

50. theEric Group Title

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

51. Shhot Group Title

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

52. theEric Group Title

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. Shhot Group Title

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. Shhot Group Title

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

55. theEric Group Title

Yep! And yep.

56. theEric Group Title

Except they're not quite equal.

57. theEric Group Title

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

58. Shhot Group Title

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

59. Shhot Group Title

q=1.7e-8C

60. theEric Group Title

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. Shhot Group Title

I got 1.7e-4 N for total force

62. theEric Group Title

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