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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?

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  1. anonymous
    • 4 years ago
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    All i need is help on how to start it.

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

  4. TuringTest
    • 4 years ago
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    by symmetry the situation will be this way:|dw:1329356841736:dw|

  5. TuringTest
    • 4 years ago
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    |dw:1329356974346:dw|the angles should be 60/2=30degrees

  6. anonymous
    • 4 years ago
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    ok, im not sure if i know that equation yet. so how do you use it?

  7. TuringTest
    • 4 years ago
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    don't know Gauss' law? I wonder how else it can be solved...

  8. TuringTest
    • 4 years ago
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    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
    • 4 years ago
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    lambda is the charge per unit length by the way...

  10. anonymous
    • 4 years ago
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    ok i understand this so far. Thanks

  11. TuringTest
    • 4 years ago
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    welcome, let me know if you get stuck

  12. anonymous
    • 4 years ago
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    Real Quick, how do i find the radius in this situation?

  13. TuringTest
    • 4 years ago
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    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
    • 4 years ago
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    got it

  15. anonymous
    • 4 years ago
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    I hate to say it but i did everything and i still got it wrong. In the end i got 136.1.

  16. TuringTest
    • 4 years ago
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    let me give it a try...

  17. TuringTest
    • 4 years ago
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    sorry, I get\[1.76\times10^{-8}\frac Vm\]

  18. anonymous
    • 4 years ago
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    nope still wrong. :/

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

  20. anonymous
    • 4 years ago
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    I have no clue.

  21. TuringTest
    • 4 years ago
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    ok I'm really hoping it's about\[3.51\times10^{-7}\]

  22. TuringTest
    • 4 years ago
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    wait no!

  23. TuringTest
    • 4 years ago
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    I forgot e0 all the way through!

  24. anonymous
    • 4 years ago
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    sadly nope. i only have one try left. I'm not sure whats happening

  25. TuringTest
    • 4 years ago
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    Oh I didn't realize you had limited tries, sorry :(

  26. TuringTest
    • 4 years ago
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    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
    • 4 years ago
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    ok sounds good. And thank you soooooo much for doing that

  28. TuringTest
    • 4 years ago
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    (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
    • 4 years ago
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    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
    • 4 years ago
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    Thanks for your help anyways

  31. TuringTest
    • 4 years ago
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    the above comes to 3.97*10^4 ...

  32. TuringTest
    • 4 years ago
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    sorry, but I stand by the formulation, and would have to use the same techniques to solve your latest question above

  33. TuringTest
    • 4 years ago
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    sucks I forgot epsilon0 for so long though, otherwise might have figured it out :/

  34. anonymous
    • 4 years ago
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    I'll ask me teacher when i go to school

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