## anonymous 4 years ago THREE CHARGES ARE AT THE corners of equilateral triangle.calculate electric feild at left corner charge?

1. anonymous

|dw:1349626330386:dw|

2. anonymous

|dw:1349626433674:dw|

3. anonymous

How would you actually do this? Well, define E2 to be on the x-axis, so $E_{2(x)}=E_2$$E_{2(y)}=0$ And$E_{1(x)}=E_1\cos(60)$$E_{1(y)}=E_1\sin(60)$Add the x and y components

4. anonymous

$E=\frac{F}{q}=\frac{Qk}{r^2}$

5. anonymous

i dnt understand:/

6. anonymous

http://www.youtube.com/watch?v=OsWDUqJQcpk Watch the first 9 minutes, he explains something very similar to your problem on the 7th-9th minutes.

7. anonymous

utube iz bloked now:/

8. anonymous

theree charges are at the corners of an equilateral triangle as shown in fig below , calculte the electric feild at the position of tha 2.00 micro colum charge due to the 7.00 micro columb and -4.00 micro columb charges.(b) use ur answer to detrmine the force on the 2.00 m|dw:1349627210442:dw|icro columb chare.

9. anonymous

|dw:1349627494021:dw| Are you fine this far?

10. anonymous

calculationx?

11. anonymous
12. anonymous

+ve direction=right E1x=$\frac{-4* (9*10^9)}{r^2}$ E2x=$\frac{7* (9*10^9)}{r^2} cos(60)$ E2y=$\frac{7* (9*10^9)}{r^2} sin(60)$

13. anonymous

force?

14. anonymous

$E=F/q$Multiply by the charge Sorry! In my earlier calculations it should have been MICRO coulombs- multiply each by 10^-6!

15. anonymous

*Multiply E field by the charge to get force

16. anonymous

|dw:1349627940207:dw|

17. anonymous

r sq is o.5 sq. na?