anonymous 4 years ago Three 20 cm-long rods form an equilateral triangle. Two of the rods are charged to + 14 nC, the third to - 14nC.What is the electric field strength at the center of the triangle?

1. anonymous

All i need is help on how to start it.

2. TuringTest

Well it seems we're gonna want to start with Gauss' law to find the field due to each rod. I'm assuming that is applicable here, but that is a simplification.

3. TuringTest

$\oint\vec E\cdot d\vec A=\frac q{\epsilon_0}$$E(2\pi r\ell)=\frac{\lambda\ell}{\epsilon_0}\to\vec E=\frac{\lambda}{2\pi r}\vec r$so now that we have our field formula we can get our three vectors at the point and superimpose them.

4. TuringTest

by symmetry the situation will be this way:|dw:1329356841736:dw|

5. TuringTest

|dw:1329356974346:dw|the angles should be 60/2=30degrees

6. anonymous

ok, im not sure if i know that equation yet. so how do you use it?

7. TuringTest

don't know Gauss' law? I wonder how else it can be solved...

8. TuringTest

The way it would go from here is that we have the formula for the electric field a distance r from the rods use geometry to find this point, then add the three vectorially: define straight down as zero: First note that the point is equidistant from each side of the triangle, so the magnitude of the fields there will be the same for each rod. Also note that by symmetry the horizontal components cancel, and the vertical components sum will be$\sum\vec E=\vec E_1+\vec E_2+\vec E_3=|E|\cos\theta_1+|E|\cos\theta_2+|E|\cos\theta_3$$=|E|(\cos\theta_1+\cos\theta_2+\cos\theta_3)=\frac{\lambda}{2\pi r}[\cos(30)+\cos(0)+\cos(-30)]$

9. TuringTest

lambda is the charge per unit length by the way...

10. anonymous

ok i understand this so far. Thanks

11. TuringTest

welcome, let me know if you get stuck

12. anonymous

Real Quick, how do i find the radius in this situation?

13. TuringTest

the distance from each rod to the center is found through geometry if you forgot how, break the triangle in half and use the pythagorean theorem to find the height. then divide the height by 2 that will be the midpoint of an equilateral triangle

14. anonymous

got it

15. anonymous

I hate to say it but i did everything and i still got it wrong. In the end i got 136.1.

16. TuringTest

let me give it a try...

17. TuringTest

sorry, I get$1.76\times10^{-8}\frac Vm$

18. anonymous

nope still wrong. :/

19. TuringTest

ok, closer investigation it seems strange to me that you got such a high number though, there isn't than much charge...

20. anonymous

I have no clue.

21. TuringTest

ok I'm really hoping it's about$3.51\times10^{-7}$

22. TuringTest

wait no!

23. TuringTest

I forgot e0 all the way through!

24. anonymous

sadly nope. i only have one try left. I'm not sure whats happening

25. TuringTest

Oh I didn't realize you had limited tries, sorry :(

26. TuringTest

ok I will just write out all my work and you can look for a mistake... I stand by my formula, but I omitted epsilon zero

27. anonymous

ok sounds good. And thank you soooooo much for doing that

28. TuringTest

(everything in meters) linear charge density:$\lambda=\frac{14\times10^{-9}}{0.2}=7\times10^{-8}$finding the distance to the midpoint r:$a^2+b^2=c^2\to0.2^2=0.1^2+a^2$$a=\sqrt{0.03}=0.1\sqrt3$$r=a/2=0.05\sqrt3$and then all that business I derived earlier$E=\frac{\lambda}{2\pi\epsilon_0r}[\cos(30)+\cos0+\cos(-30)]$$=\frac{7\times10^{-8}}{2\pi(8.85\times10^{-12})(0.05)\sqrt3}(\sqrt3+1)$let me know if you see something strange...

29. anonymous

Well, the correct answer was 3.8*10^4. i didnt get it right but i just thought i would let you know

30. anonymous

31. TuringTest

the above comes to 3.97*10^4 ...

32. TuringTest

sorry, but I stand by the formulation, and would have to use the same techniques to solve your latest question above

33. TuringTest

sucks I forgot epsilon0 for so long though, otherwise might have figured it out :/

34. anonymous

I'll ask me teacher when i go to school