## anonymous 4 years ago just need this answer real quick...timed test..got 13 minutes left: A +48.26 nC point charge is placed on the x axis at 3.007 m, and a -21.49 nC point charge is placed on the y axis at y= -6.779 m. What is the direction of the electric field at the origin?

1. anonymous

counterclockwise from the positive x axis

2. anonymous

plz tell me u can do this?

3. ash2326

yeah let me try

4. ash2326

|dw:1328301704314:dw|

5. anonymous

i got 8 minutes to submit answer

6. anonymous

i hate timed tests

7. anonymous

Just think about it yourself. First hint: You can regard the resulting electrical field as the sum of the two electrical fields of each charge.

8. anonymous

chris i don't have the time...i got 7 minutes

9. anonymous

Enough time to think ...

10. anonymous

afterwards i'll worry about how's and why's...right now i just need the correct answer

11. anonymous

26.77 is net electric charge

12. ash2326

Electric field at origin due to positive charge=9*10^9*48.26*10^-9/(3.007)^2( -i) = 48.0355 (-i) (toward left ) Electric field at origin due to negative charge=9*10^9*(-21.49)*10^-9/(6.779)^2 (j) =4.208(-j) towards downward direction I think clockwise from x direction

13. anonymous

Second hint: Use coulomb's law to calculate the contribution of each charge at the origin.

14. anonymous

plz for the love of god just gimme the answer and i'll work through it after i submit it

15. anonymous

plz

16. ash2326

got it?

17. anonymous

I'm not going to support you to cheat, sorry.

18. anonymous

then leave

19. anonymous

ashe no i don't

20. ash2326

are there options

21. anonymous

22. anonymous

what numbers?

23. anonymous

i got 2 minutes left?

24. anonymous

u wnana throw an airball up?

25. ash2326

net electric field= $\sqrt{{48.0355}^2+{4.028}^2}$=48.204 Volt/meter in clockwise direction from =ve x-axis

26. ash2326

it'd be +ve x -axis

27. anonymous

i submitted 128...i think it's 132 though

28. anonymous

29. ash2326

but from where you got 128

30. anonymous

i appreciate the effort

31. anonymous

i added the two numbers u had earlier....as per what chris said

32. anonymous

then just subtracted taht from 180

33. anonymous

to get the angle counterclockwise

34. ash2326

angle is -175 degrees

35. ash2326

that was magnitude

36. anonymous

If ash's numbers are correct it's just $\vec E = ({-48.0},{-4.2} )^T \frac{\textrm{N}}{\textrm{C}}$, i.e. 180.5° (counter-clockwise).

37. anonymous

yea thx....coulda used that 15 minutes ago

38. anonymous

good lookin out

39. anonymous

In the future you might consider: a.) Not wait until the last minute to do a test like this. b.) Being respectful to those who helped you. They aren't required to do so.

40. anonymous

to xishem....i did not wait until the last minute to do this test. It's a timed 20 minute 1 question test. I'm very respectful of those who help me. I help many many people on here in the math section. I was very respectful and appreciative of ash for taking the time to help me. Chris was not helping, therefore at the time, was not appreciative