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JenniferSmart1 Group Title

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.

  • one year ago
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  1. JenniferSmart1 Group Title
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    |dw:1361480351472:dw|

    • one year ago
  2. JenniferSmart1 Group Title
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    I know at this point that V_1 is different from V_2

    • one year ago
  3. experimentX Group Title
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    why??

    • one year ago
  4. JenniferSmart1 Group Title
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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

    • one year ago
  5. JenniferSmart1 Group Title
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    so \[\Delta V=0\]

    • one year ago
  6. JenniferSmart1 Group Title
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    Let's talk about the electric potential for a moment...

    • one year ago
  7. experimentX Group Title
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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.

    • one year ago
  8. JenniferSmart1 Group Title
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    I still have a hard time picturing electric potential .....

    • one year ago
  9. JenniferSmart1 Group Title
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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?

    • one year ago
  10. experimentX Group Title
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    think of two spheres as two vessels ..|dw:1361481079204:dw| charges as water ... and wire as tunnel

    • one year ago
  11. JenniferSmart1 Group Title
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    |dw:1361481129492:dw|

    • one year ago
  12. JenniferSmart1 Group Title
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    oh so they want to come to some kind of equilibrium

    • one year ago
  13. experimentX Group Title
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    yes ...

    • one year ago
  14. JenniferSmart1 Group Title
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    |dw:1361481183183:dw|

    • one year ago
  15. JenniferSmart1 Group Title
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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

    • one year ago
  16. JenniferSmart1 Group Title
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    Not the problem per se...I get what Potential part :P

    • one year ago
  17. JenniferSmart1 Group Title
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    I can't type ...sorry bout the grammar

    • one year ago
  18. JenniferSmart1 Group Title
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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...

    • one year ago
  19. experimentX Group Title
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    nvm ... the grammar ,,, not charge difference exactly, charge density ... |dw:1361481414886:dw|

    • one year ago
  20. JenniferSmart1 Group Title
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    Oh the charge densities are different. got it.

    • one year ago
  21. experimentX Group Title
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    what's charge density?? in case of conductors ... metals??

    • one year ago
  22. JenniferSmart1 Group Title
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    \[\lambda=\frac{Q}{surface \; area}\]

    • one year ago
  23. JenniferSmart1 Group Title
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    \[E=kQ/r^2\]

    • one year ago
  24. experimentX Group Title
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    yep .. that's correct, use \( \sigma \) for for surface charge density .. \( \lambda \) for linear charge density.

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

    • one year ago
  26. JenniferSmart1 Group Title
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    \[\sigma=\frac{E}{k4\pi}\] REally?

    • one year ago
  27. experimentX Group Title
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    hmm ... what are you trying to do?? calculate surface charge density from given Electric Field?

    • one year ago
  28. JenniferSmart1 Group Title
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    Yep, I have to find the surface charge density on each sphere.

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

    • one year ago
  30. experimentX Group Title
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    what are given so far??

    • one year ago
  31. JenniferSmart1 Group Title
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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.

    • one year ago
  32. experimentX Group Title
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    at surface?? or ...

    • one year ago
  33. JenniferSmart1 Group Title
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    @TuringTest

    • one year ago
  34. JenniferSmart1 Group Title
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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?

    • one year ago
  35. TuringTest Group Title
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    yep, looks good so far

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

    • one year ago
  37. JenniferSmart1 Group Title
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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.... \[\frac{k\sigma_1(SA)}{r_1}=\frac{kEr^2}{kr_2}\] ?

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

    • one year ago
  39. TuringTest Group Title
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    remember that the surace area of each sphere is different, and depends on the radius of that sphere

    • one year ago
  40. JenniferSmart1 Group Title
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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 \[\frac{k\sigma_1(SA)}{r_1}=V_2\]

    • one year ago
  41. JenniferSmart1 Group Title
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    oh I should probably differentiate between \(SA_1\) and \(SA_2\)

    • one year ago
  42. TuringTest Group Title
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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.

    • one year ago
  43. JenniferSmart1 Group Title
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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.

    • one year ago
  44. TuringTest Group Title
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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

    • one year ago
  45. JenniferSmart1 Group Title
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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

    • one year ago
  46. TuringTest Group Title
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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 :)

    • one year ago
  47. JenniferSmart1 Group Title
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    oh wow! that's AWESOME!!!!

    • one year ago
  48. JenniferSmart1 Group Title
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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?)

    • one year ago
  49. TuringTest Group Title
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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...

    • one year ago
  50. TuringTest Group Title
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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 :)

    • one year ago
  51. JenniferSmart1 Group Title
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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."

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

    • one year ago
  53. JenniferSmart1 Group Title
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    nevermind I see it now.

    • one year ago
  54. TuringTest Group Title
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    :)

    • one year ago
  55. JenniferSmart1 Group Title
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    We should have \[\sigma=E\epsilon_0\]

    • one year ago
  56. JenniferSmart1 Group Title
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    and I have \[\sigma=\frac{E\epsilon_0}{r}\]

    • one year ago
  57. JenniferSmart1 Group Title
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    LOL I see it again....sorry :S

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

    • one year ago
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