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JenniferSmart1

  • 3 years ago

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  1. goformit100
    • 3 years ago
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    Wher is it ?

  2. JenniferSmart1
    • 3 years ago
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    |dw:1361637181191:dw| Given \[q_1=-q_2\] I have to find the potential difference. Here is what I got so far..... \[V_1=\frac{kq_1}{r_1}\rightarrow q_1=\frac{V_1r_1}{k}\] \[V_2=\frac{kq_2}{r_2} \rightarrow q_2=\frac{V_2 r_2}{k}\] \[V_1 r_1=-V_2r_2\] \[V_1r_1+V_2r_2=0\] but I have to get delta V

  3. JenniferSmart1
    • 3 years ago
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    I probably shouldn't cancel out the k's

  4. JenniferSmart1
    • 3 years ago
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    the k's do cancel out. I tried it another way \[\frac1k(V_1r_1-V_2r_2)=0\]

  5. TuringTest
    • 3 years ago
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    q1 is in the inside?

  6. JenniferSmart1
    • 3 years ago
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    q_1 is on the inside and it's positive q_2 is on the outside and it's negative

  7. JenniferSmart1
    • 3 years ago
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    I can set q_1 equal to q_2 correct? since they are equal but opposite in charge?

  8. TuringTest
    • 3 years ago
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    yes, so you can factor it out by superposition\[\sum_i\vec E_i=\frac q{4\pi\epsilon_0r_i}\]

  9. JenniferSmart1
    • 3 years ago
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    elaborate.... how did we add the electric field into the equation?

  10. TuringTest
    • 3 years ago
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    I'm trying to do this from scratch, but I guess I'm a little rusty, hold on please

  11. JenniferSmart1
    • 3 years ago
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    yeah no prob :)

  12. TuringTest
    • 3 years ago
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    \[\Delta V=-\int\limits_C\vec E\cdot d\vec\ell=-\frac q{4\pi\epsilon_0}\int_a^b\frac{dr}{r^2}\]or is it 2q... I should find something to double-check, but the rest is right

  13. JenniferSmart1
    • 3 years ago
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  14. JenniferSmart1
    • 3 years ago
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    oh I see....

  15. TuringTest
    • 3 years ago
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    it's just q, but you are using a different formula

  16. TuringTest
    • 3 years ago
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    your formula is the same, of course. You are probably just highlighting the connection to potential energy and voltage.

  17. JenniferSmart1
    • 3 years ago
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    Does that mean that \(q=k\Delta Vr\) ?

  18. TuringTest
    • 3 years ago
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    \[q=k\Delta Vr=\frac1{4\pi\epsilon_0}\cdot\frac q{4\pi\epsilon_0 r^2}\cdot r\]so no :p

  19. JenniferSmart1
    • 3 years ago
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    how about this: \[\Delta V=-\frac Er\] \[\frac{kq}{r}=-\frac E r\] \[q=\frac{-E} k\]

  20. JenniferSmart1
    • 3 years ago
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    LOL I don't know ...haha

  21. TuringTest
    • 3 years ago
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    for the other, I eant\[q\neq\Delta Vr=\frac1{4\pi\epsilon_0}\cdot\frac q{4\pi\epsilon_0 r}\cdot r=\frac q{(4\pi\epsilon_0)^2}\]about your next one\[\Delta V=-\frac{Er}q\]\(if\) E and r are in the same direction whe whole time

  22. JenniferSmart1
    • 3 years ago
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    I think they are int he same direction since |dw:1361640455529:dw| correct?

  23. TuringTest
    • 3 years ago
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    in this case, yes

  24. JenniferSmart1
    • 3 years ago
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    oh yay! So what would we do with that relationship? \[\Delta V=-\frac{Er}q\]

  25. TuringTest
    • 3 years ago
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    yes, as long as the path we are checking voltages between is parallel to the electric field lines

  26. JenniferSmart1
    • 3 years ago
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    do we assume that that is the path?

  27. JenniferSmart1
    • 3 years ago
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    It probably wouldn't be parallel

  28. TuringTest
    • 3 years ago
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    yeah, I mean the displacement really, we could go in the path|dw:1361640896393:dw|but the *displacement* is all that matters for calculating voltage if the field lines are parallel to it

  29. JenniferSmart1
    • 3 years ago
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    Oh I see...

  30. TuringTest
    • 3 years ago
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    |dw:1361641026090:dw|

  31. TuringTest
    • 3 years ago
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    we need the more general line integral for for things where the E-field is not parallel to the displacement at all times, like|dw:1361641180439:dw|where a charge is going like...

  32. TuringTest
    • 3 years ago
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    |dw:1361641482262:dw|better prediction of the path of the charge in such a field^

  33. JenniferSmart1
    • 3 years ago
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    since we have a sphere, it shouldn't have a potential difference on it's own. But it has a constant potential.....so: no potential difference on a sphere? just between spheres?

  34. TuringTest
    • 3 years ago
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    if there were a potential difference between one part of a conducting sphere and another, current would flow across it until the potential difference was neutralized

  35. TuringTest
    • 3 years ago
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    what's the question?

  36. JenniferSmart1
    • 3 years ago
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    \[\frac{kq}{r}=E\cdot r\] \[q=\frac{E\cdot r}{k}\] I'm soo stuck....LOL! But it seems sooo simple!

  37. TuringTest
    • 3 years ago
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    in that case, all it's basically asking is for you to find\[\Delta V=\frac1q\Delta U=k(E_{outside}-E_{insode})\]and since |Q1|=|Q2| in this case, we can factor it out to give\[\Delta V=-kE\Delta r=-\frac Q{4\pi\epsilon_0}\left(\frac1{r_2^2}-\frac1{r_1^2}\right)\]

  38. JenniferSmart1
    • 3 years ago
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    How did you get the \(-kE\Delta r\) part?

  39. TuringTest
    • 3 years ago
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    so many typos, my connection is terrible getting a new computer soon last line should be\[\Delta V=-\frac1QE\Delta r\] but really this is better seen as an integral..\[\Delta V=-\int_a^b Edr=1\frac Q{4\pi\epsilon_0}\int_a^b\frac{dr}{r^2}\]I don't know what else to say right now... go wat and think of a more definitive question to ask!

  40. JenniferSmart1
    • 3 years ago
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    I'll go eat....I wanna walk away from this problem for a little bit....

  41. JenniferSmart1
    • 3 years ago
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    It will probably make more sense then LOL.

  42. TuringTest
    • 3 years ago
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    yes, it is quite simple and you are over-thining it see you later, I am learning a little statistics

  43. JenniferSmart1
    • 3 years ago
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    see ya Turing!

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