## anonymous 3 years ago I'm adding my vectors incorrectly. Please give me a min to draw it.

1. anonymous

|dw:1360530853167:dw|

2. anonymous

$\vec{F}_{net}=(\vec F_{41}-\vec F_{31}\cos(45))\hat i +(\vec F_{21}-\vec F_{31}\sin(45)) \hat j$

3. anonymous

|dw:1360531227825:dw|

4. anonymous

$F_{net}=\left( \frac{kq_4q_1}{r_{41}^2}-\frac{kq_3q_1}{r_{31}^2}\sin(45)\right) \hat i \left(\frac{kq_2q_1}{r_{21}^2}-\frac{kq_3q_1}{r_{31}^2}\cos(45)\right) \hat j$

5. anonymous

$q_1=3\times10^{-9}$ $q_2=9\times10^{-9}$ $q_3=-9\times10^{-9}$ $q_4=9\times10^{-9}$

6. anonymous

|dw:1360531777377:dw|

7. anonymous

@phi

8. anonymous

I get $(72974.5 N)\hat i$

9. anonymous

which is wrong...

10. anonymous

11. anonymous

|dw:1360532089439:dw|

12. phi

8 orders of magnitude is quite an error

13. anonymous

it sure is ...haha

14. anonymous

the charges were given in nano coulombs. I converted them to coulombs. $q_1=+3nC=3\times 10^{-9}$

15. anonymous

k=8.99E9

16. phi

First, what units is the answer given in ?

17. anonymous

Newtons \hat{i}

18. anonymous

maybe I read the question wrong

19. phi

your approach is ok. I would write down the components of each vector

20. phi

the k 9e9 times 9e-9 gives 81. that still leaves a 3e-9 are you double counting the k ?

21. anonymous

|dw:1360532909843:dw| oh my gosh ...yes I think soooo

22. anonymous

I'll do it on wolfram... one sec

23. anonymous
24. anonymous
25. anonymous

I found another error...this looks better http://www.wolframalpha.com/input/?i=%28coulombs+constant+*9nC*3nC%29%2F%282cm^2%29++-+++%28coulombs+constant+*-9nC*3nC%29%2F%282.8cm^2%29

26. anonymous
27. anonymous

silly question... $F_{31}=F_{31x}+F_{31y}$ is this true?

28. anonymous

or is it $F_{31}=\sqrt{F_{31x}^2+F_{31y}^2}$

29. phi

ignoring the units (we can figure those out later) the force from the top left corner is <9*3/4 i 0 j> the force from the bottom right is < 0 i 9*3/4 j> the force from the bottom left is < -9*3/8 * 1/sqrt(2) i - 9*3/8 *1/sqrt(2) j> add up the components to get 27/4( 1- 1/2sqrt(2)) for both i and j components that number is 4.3635 now to get the units 1e-18 C^2 * 9e-9 N-M^2 /C^2 * 1e4 M^2 (to fix the cm) that factor is 9e-5 and our result is 4.3635*9e-5 = 0.0003927 N for both the i and j components

30. phi

as for your "silly" question. if by F31x you mean < F31x 0> and by F31y < 0 F31y> then yes, F31= F31x+F31y the magnitude of F31 is as you wrote it.

31. anonymous

why is "the force from the bottom left is < -9*3/8 * 1/sqrt(2) i" Why did you divide by 8

32. phi

the distance (diagonal) is 2 sqrt(2). squared you get 4*2= 8 the 1/sqrt(2) is cos(45) (often written as sqrt(2)/2 but we don't need to write it that way)

33. phi

your 2.8 was just an approximation

34. phi

I think the resultant force vector is < 0.0003927 i 0.0003927 j> with a magnitude of 0.000555 N pointing in the opposite direction of the diagonal

35. anonymous

Yep that's correct...I'm still trying to make sense of it...one second...

36. anonymous

$(9\cdot 3 \cdot\frac 14+(-9)3\cdot \frac 18 \cdot \frac 1{\sqrt2})\hat i$ and for \hat j as well Yep that makes sense

37. anonymous

what happened to k

38. phi

I was ignoring the constants and units, just to get a rough idea of what is going on. I see that both the i and j components are positive and equal to each other. but obviously you have to multiply every term by k, and by 1e-9 (twice! to change nC to C) and by 1e4 cm^2 per meter^2 to get the units correct.

39. anonymous

Oh I see

40. phi

if we use 8.99e9 for k, we match the book's answer 0.000392

41. anonymous

shoot me....I continue to get .0000082137 $\frac{(8.99\times10^9N\cdot m^2/C^2)(9\times10^{-9}C)(3\times10^{-9}C)}{0.04m^2}$$-\frac{(8.99\times10^9N\cdot m^2/C^2)(-9\times10^{-9}C)(3\times10^{-9}C)(\frac 1{\sqrt{2}})}{0.08m^2}$

42. anonymous

$\frac{(8.99\times10^9N\cdot m^2/C^2)(9\times10^{-9}C)(3\times10^{-9}C)}{0.04m^2}$-$-\frac{(8.99\times10^9N\cdot m^2/C^2)(-9\times10^{-9}C)(3\times10^{-9}C)(\frac 1{\sqrt{2}})}{\sqrt0.08m^2}$

43. phi

First, I would not write down that expression. I would write down the 3 individual vectors (with components i and j even if 0) Second, your distance squared should be written $$(2e{-2}\ m)^2$$= $$4e{-4}\ m^2$$

44. phi

and $$2 \sqrt{2}$$ cm squared is written as 8e-4 m^2

45. anonymous

finally!!!1

46. phi

Good.