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JenniferSmart1

  • 2 years ago

I have 2 spheres. Little sphere with radius r Big sphere with radius R We know E at the large sphere. Find the charge density. Allow me to draw it..... Oh...and they're connected by a long thin wire.

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  1. JenniferSmart1
    • 2 years ago
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    |dw:1361480351472:dw|

  2. JenniferSmart1
    • 2 years ago
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    I know at this point that V_1 is different from V_2

  3. experimentX
    • 2 years ago
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    why??

  4. JenniferSmart1
    • 2 years ago
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    THe potentials are possibly different....or no? Well they didn't mention that. It must be the same since they're attached by a long thin wire

  5. JenniferSmart1
    • 2 years ago
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    so \[\Delta V=0\]

  6. JenniferSmart1
    • 2 years ago
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    Let's talk about the electric potential for a moment...

  7. experimentX
    • 2 years ago
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    since they are connected with wire ,,, they are at same potential. If they are not at same potential ... charges will flow to make at same potential.

  8. JenniferSmart1
    • 2 years ago
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    I still have a hard time picturing electric potential .....

  9. JenniferSmart1
    • 2 years ago
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    I can manipulate the equation relating potential to electric field and so on, but I don't quite understand it. How and why does a sphere have electric potential?

  10. experimentX
    • 2 years ago
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    think of two spheres as two vessels ..|dw:1361481079204:dw| charges as water ... and wire as tunnel

  11. JenniferSmart1
    • 2 years ago
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    |dw:1361481129492:dw|

  12. JenniferSmart1
    • 2 years ago
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    oh so they want to come to some kind of equilibrium

  13. experimentX
    • 2 years ago
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    yes ...

  14. JenniferSmart1
    • 2 years ago
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    |dw:1361481183183:dw|

  15. JenniferSmart1
    • 2 years ago
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    So we go from high potential to low potential and vice versa till we find equilibrium. Uhm.....so there is a charge difference which caused the potential difference....Yeah I get it now

  16. JenniferSmart1
    • 2 years ago
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    Not the problem per se...I get what Potential part :P

  17. JenniferSmart1
    • 2 years ago
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    I can't type ...sorry bout the grammar

  18. JenniferSmart1
    • 2 years ago
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    ok let's see here....I have to relate the electric field of the big sphere to the potential on the big sphere...

  19. experimentX
    • 2 years ago
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    nvm ... the grammar ,,, not charge difference exactly, charge density ... |dw:1361481414886:dw|

  20. JenniferSmart1
    • 2 years ago
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    Oh the charge densities are different. got it.

  21. experimentX
    • 2 years ago
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    what's charge density?? in case of conductors ... metals??

  22. JenniferSmart1
    • 2 years ago
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    \[\lambda=\frac{Q}{surface \; area}\]

  23. JenniferSmart1
    • 2 years ago
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    \[E=kQ/r^2\]

  24. experimentX
    • 2 years ago
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    yep .. that's correct, use \( \sigma \) for for surface charge density .. \( \lambda \) for linear charge density.

  25. JenniferSmart1
    • 2 years ago
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    ooops! \[SA=4\pi r^2\] \[\sigma=Q/SA\] \[E=\frac{k(\sigma4\pi R^2)}{R^2}\]

  26. JenniferSmart1
    • 2 years ago
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    \[\sigma=\frac{E}{k4\pi}\] REally?

  27. experimentX
    • 2 years ago
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    hmm ... what are you trying to do?? calculate surface charge density from given Electric Field?

  28. JenniferSmart1
    • 2 years ago
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    Yep, I have to find the surface charge density on each sphere.

  29. JenniferSmart1
    • 2 years ago
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    since they're connect by a long thin wire, the charge density should be equal for the spheres

  30. experimentX
    • 2 years ago
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    what are given so far??

  31. JenniferSmart1
    • 2 years ago
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    I have 2 spheres. Little sphere with radius r Big sphere with radius R We know E at the large sphere.

  32. experimentX
    • 2 years ago
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    at surface?? or ...

  33. JenniferSmart1
    • 2 years ago
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    @TuringTest

  34. JenniferSmart1
    • 2 years ago
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    Let's see...There is a mistake in here. From the top. 1.) The potential of the two spheres are equal. 2.) The surface charge densities are different. 3.) I can relate the two surface charge densities through this relationship \[V_1=V_2\] \[\frac{kq_1}{r_1}=\frac{kq_2}{r_2}\] \[\sigma_1=\frac{q_1}{SA}\;\;\;\;\; \sigma_2=\frac{q_2}{SA}\] \[\frac{k\sigma_1(SA)}{r_1}=\frac{k\sigma_2(SA)}{r_2}\] am I correct so far?

  35. TuringTest
    • 2 years ago
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    yep, looks good so far

  36. JenniferSmart1
    • 2 years ago
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    I'm given the electric field at the surface of the large sphere E. so \[E=\frac{kq}{r^2}\]

  37. JenniferSmart1
    • 2 years ago
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    \[\frac{k\sigma_1(SA)}{r_1}=\frac{k\sigma_2(SA)}{r_2}\] \[E=\frac{kq}{r^2}\] would that make.... \[\frac{k\sigma_1(SA)}{r_1}=\frac{kEr^2}{kr_2}\] ?

  38. JenniferSmart1
    • 2 years ago
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    \[\frac{k\sigma_1(SA)}{r_1}=\frac{\cancel{k}Er^\cancel{2}}{\cancel{k}\cancel{r_2}}\]

  39. TuringTest
    • 2 years ago
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    remember that the surace area of each sphere is different, and depends on the radius of that sphere

  40. JenniferSmart1
    • 2 years ago
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    \[\frac{k\sigma_1(SA)}{r_1}=\frac{\cancel{k}Er_2^\cancel{2}}{\cancel{k}\cancel{r_2}}\] \[\sigma_1=\frac{Er_2r_1}{k(SA)}\] Should I continue in this manner or .... Should I solve for the potential of the second sphere (the one with the given E) first and then relate it to \[\frac{k\sigma_1(SA)}{r_1}=V_2\]

  41. JenniferSmart1
    • 2 years ago
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    oh I should probably differentiate between \(SA_1\) and \(SA_2\)

  42. TuringTest
    • 2 years ago
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    Yes, you should! I don't see where you got E from... just keep working based on the fact that you can set the voltages equal\[V_1=V_2\]\[\frac{\cancel k\sigma_1(SA_1)}{r_1}=\frac{\cancel k\sigma_2(SA_2)}{r_2}\]\[\frac{\sigma_1(4\pi r_1^2)}{r_1}=\frac{\sigma_1(4\pi r_2^2)}{r_1}2\]\[\vdots\] etc.

  43. JenniferSmart1
    • 2 years ago
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    yeah that makes sense... Did they give me E=(240 kV/m on the big sphere) just to confuse me? It seems useless for this problem.

  44. TuringTest
    • 2 years ago
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    typo on the last line earlier...\[V_1=V_2\]\[\frac{\cancel k\sigma_1(SA_1)}{r_1}=\frac{\cancel k\sigma_2(SA_2)}{r_2}\]\[\frac{\sigma_1(4\pi r_1^2)}{r_1}=\frac{\sigma_1(4\pi r_2^2)}{r_2}\]\[\vdots\]knowing the value of E allowes you to find the value of \(\sima\) on that sphere

  45. JenniferSmart1
    • 2 years ago
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    oh yeah, because the way we have set it up now we two unknowns....yep. I'll solve for sigma of the big sphere and then plug it into that equation to solve for simga of sphere 1

  46. TuringTest
    • 2 years ago
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    yep :) also, since\[k=\frac1{4\pi\epsilon_0 r^2}\]we have\[\frac{\sigma_1(4\pi r_1^2)}{4\pi \epsilon_0r_2r_1}=\frac{\sigma_2(4\pi r_2^2)}{4\pi \epsilon_0r_2}\]\[\sigma_1r_1=\sigma_2r_2\]just to make you calculations easier :)

  47. JenniferSmart1
    • 2 years ago
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    oh wow! that's AWESOME!!!!

  48. JenniferSmart1
    • 2 years ago
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    I guess a rule of thumb is to simplify as much as I can before plugging in numbers....man this makes physics soo much more manageable! Thanks Maxwell! (Maxwell's Silver Hammer?)

  49. TuringTest
    • 2 years ago
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    That's *my* rule of thumb: "as simple os possible, but not simpler" (a quote from Einstein actually) and yeah, I am the maxwell of the Beatles song another typo: I meant \(k=\frac1{4\pi\epsilon_0}\) but you get the point...

  50. TuringTest
    • 2 years ago
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    ... and as always, happy to help! As I said, I'm Taking E&M too (again) , so I need the refreshers like this :)

  51. JenniferSmart1
    • 2 years ago
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    Yeah I get the point. I feel elated seeing the equation simplified to this extent, genuinely happy =D Thanks again!!!!!! btw one of my favorite Einstein quotes is: "If you want to live a happy life, tie it to a goal, not to people or things."

  52. JenniferSmart1
    • 2 years ago
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    is that legal? \[\frac{1}{\cancel{4\pi}\cancel\epsilon_0}\frac{\sigma_1\cancel{4\pi} r_1^2}{r_1}=\frac{\sigma_2\cancel{4\pi} r_2^2}{r_2}\frac{1}{\cancel{4\pi} \cancel\epsilon_0}\]

  53. JenniferSmart1
    • 2 years ago
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    nevermind I see it now.

  54. TuringTest
    • 2 years ago
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    :)

  55. JenniferSmart1
    • 2 years ago
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    We should have \[\sigma=E\epsilon_0\]

  56. JenniferSmart1
    • 2 years ago
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    and I have \[\sigma=\frac{E\epsilon_0}{r}\]

  57. JenniferSmart1
    • 2 years ago
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    LOL I see it again....sorry :S

  58. TuringTest
    • 2 years ago
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    :P Well, it's great that you really make sure your work makes sense to you!

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