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anonymous

  • 4 years ago

Three 1.0 nC charges are placed as shown in the figure. Each of these charges creates an electric field vector E at a point 3.0 cm in front of the middle charge.What are the three fields vector E1, vector E2, and vector E3 created by the three charges? Write your answer for each as a vector in component form.

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  1. anonymous
    • 4 years ago
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    I mainly need help on starting it.

  2. TuringTest
    • 4 years ago
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    Got a picture?

  3. anonymous
    • 4 years ago
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    sorry yes there is just hold on

  4. anonymous
    • 4 years ago
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    |dw:1327883171291:dw|

  5. anonymous
    • 4 years ago
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    q1,q2q3 all equal 1nC

  6. TuringTest
    • 4 years ago
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    so you need the field vector from each charge at that point on the right?

  7. anonymous
    • 4 years ago
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    Yes

  8. TuringTest
    • 4 years ago
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    |dw:1327883385609:dw|The easiest one to find is the field due the the second charge\[\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\overrightarrow{r}=<0,\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}>\]You can just plug the numbers in, and you know the angle will be 0 because the vector points in the +x direction. Got it so far?

  9. anonymous
    • 4 years ago
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    I do understand that so far but it says to put it in vector component form. How would i do that? Like (Ex, Ey). I know Ey should be zero but how would you find Ex?

  10. TuringTest
    • 4 years ago
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    this is component form:\[<0,\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}>\]^^^ ^^^ Ex=0 Ey=that^

  11. anonymous
    • 4 years ago
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    ohhh ok

  12. TuringTest
    • 4 years ago
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    sorry Ex and Ey are backwards above...

  13. TuringTest
    • 4 years ago
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    \[<\frac{1}{4\pi\epsilon_0}\frac{q}{r^2},0>\] Ex^^ ^^Ey=0 I just noted that it says that the point is directly to the right of q2, so since the sine of 0=0 the y-component disappears.

  14. anonymous
    • 4 years ago
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    now how do you find the components q1?

  15. TuringTest
    • 4 years ago
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    let's put a grid on our point so we can do a little trigonometry:|dw:1327884328041:dw|The next one I want to find is the vector due to q3. You will see why soon.

  16. anonymous
    • 4 years ago
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    ok

  17. TuringTest
    • 4 years ago
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    notice all the geometry we can do here, and all the places this angle theta turns up|dw:1327884443364:dw|we can find r with the pythagorean theorem, and the angle with the inverse tangent.

  18. TuringTest
    • 4 years ago
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    \[r=\sqrt{3^2+1}\]\[\theta=\tan^{-1}(\frac13)\]so we have\[\overrightarrow{E_3}=\frac{1}{4\pi\epsilon_0}\frac{q_3}{r^2}\overrightarrow{r}=<\frac{1}{4\pi\epsilon_0}\frac{q_3}{r^2}\cos\theta,\frac{1}{4\pi\epsilon_0}\frac{q_3}{r^2}\sin\theta>\]so this can be solved quickly as well now

  19. TuringTest
    • 4 years ago
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    ^I suppose you have to put the centimeters in terms of meters above actually

  20. TuringTest
    • 4 years ago
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    \[r=\sqrt{0.03^2+0.01^2}\]\[\theta=\tan^{-1}(\frac13)\]

  21. anonymous
    • 4 years ago
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    Ok, i understand all of that so far.

  22. TuringTest
    • 4 years ago
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    well if you have the radius and angle you can solve that guy, right? now E1 will be simple, because by symmetry it will have the same vector as E3, only the y-component will be negative\[\overrightarrow{E_1}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\overrightarrow{r}=<\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\cos\theta,-\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\sin\theta>\]where theta is the angle we used earlier. Same r as well.

  23. TuringTest
    • 4 years ago
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    |dw:1327885322988:dw|

  24. anonymous
    • 4 years ago
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    Thanks for helping this the set up. I'll let you know if it doesn't work on the homework. (it can be finicky sometimes)

  25. TuringTest
    • 4 years ago
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    please do, good luck!

  26. anonymous
    • 4 years ago
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    For some reason it keeps saying I'm wrong for my q2 answer. I wrote that the answer was (1,0). I'm pretty sure it's right but just want to check with you (the answer is in kiloNewtons by the way).

  27. TuringTest
    • 4 years ago
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    let me bust out the calculator...

  28. anonymous
    • 4 years ago
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    thanks for doing that :).

  29. TuringTest
    • 4 years ago
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    wait, the answer is in V/m or N/c or something, not just N that's force

  30. anonymous
    • 4 years ago
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    kN/c

  31. TuringTest
    • 4 years ago
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    I got about 9.9kN/c

  32. anonymous
    • 4 years ago
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    Yep that worked. thanks, i now see what i did wrong.

  33. TuringTest
    • 4 years ago
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    Have you tried the others yet? if not be careful of the units is all I can say...

  34. anonymous
    • 4 years ago
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    yes, they all worked. Thank you very much

  35. TuringTest
    • 4 years ago
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    awesome! glad to help :)

  36. anonymous
    • 4 years ago
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    Your saving the world by helping one physics student at a time. lol

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