## 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 one year ago

1. JenniferSmart1 Group Title

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2. JenniferSmart1 Group Title

I know at this point that V_1 is different from V_2

3. experimentX Group Title

why??

4. JenniferSmart1 Group Title

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 Group Title

so $\Delta V=0$

6. JenniferSmart1 Group Title

Let's talk about the electric potential for a moment...

7. experimentX Group Title

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 Group Title

I still have a hard time picturing electric potential .....

9. JenniferSmart1 Group Title

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 Group Title

think of two spheres as two vessels ..|dw:1361481079204:dw| charges as water ... and wire as tunnel

11. JenniferSmart1 Group Title

|dw:1361481129492:dw|

12. JenniferSmart1 Group Title

oh so they want to come to some kind of equilibrium

13. experimentX Group Title

yes ...

14. JenniferSmart1 Group Title

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15. JenniferSmart1 Group Title

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 Group Title

Not the problem per se...I get what Potential part :P

17. JenniferSmart1 Group Title

I can't type ...sorry bout the grammar

18. JenniferSmart1 Group Title

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 Group Title

nvm ... the grammar ,,, not charge difference exactly, charge density ... |dw:1361481414886:dw|

20. JenniferSmart1 Group Title

Oh the charge densities are different. got it.

21. experimentX Group Title

what's charge density?? in case of conductors ... metals??

22. JenniferSmart1 Group Title

$\lambda=\frac{Q}{surface \; area}$

23. JenniferSmart1 Group Title

$E=kQ/r^2$

24. experimentX Group Title

yep .. that's correct, use $$\sigma$$ for for surface charge density .. $$\lambda$$ for linear charge density.

25. JenniferSmart1 Group Title

ooops! $SA=4\pi r^2$ $\sigma=Q/SA$ $E=\frac{k(\sigma4\pi R^2)}{R^2}$

26. JenniferSmart1 Group Title

$\sigma=\frac{E}{k4\pi}$ REally?

27. experimentX Group Title

hmm ... what are you trying to do?? calculate surface charge density from given Electric Field?

28. JenniferSmart1 Group Title

Yep, I have to find the surface charge density on each sphere.

29. JenniferSmart1 Group Title

since they're connect by a long thin wire, the charge density should be equal for the spheres

30. experimentX Group Title

what are given so far??

31. JenniferSmart1 Group Title

I have 2 spheres. Little sphere with radius r Big sphere with radius R We know E at the large sphere.

32. experimentX Group Title

at surface?? or ...

33. JenniferSmart1 Group Title

@TuringTest

34. JenniferSmart1 Group Title

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 Group Title

yep, looks good so far

36. JenniferSmart1 Group Title

I'm given the electric field at the surface of the large sphere E. so $E=\frac{kq}{r^2}$

37. JenniferSmart1 Group Title

$\frac{k\sigma_1(SA)}{r_1}=\frac{k\sigma_2(SA)}{r_2}$ $E=\frac{kq}{r^2}$ would that make.... $\frac{k\sigma_1(SA)}{r_1}=\frac{kEr^2}{kr_2}$ ?

38. JenniferSmart1 Group Title

$\frac{k\sigma_1(SA)}{r_1}=\frac{\cancel{k}Er^\cancel{2}}{\cancel{k}\cancel{r_2}}$

39. TuringTest Group Title

remember that the surace area of each sphere is different, and depends on the radius of that sphere

40. JenniferSmart1 Group Title

$\frac{k\sigma_1(SA)}{r_1}=\frac{\cancel{k}Er_2^\cancel{2}}{\cancel{k}\cancel{r_2}}$ $\sigma_1=\frac{Er_2r_1}{k(SA)}$ 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 Group Title

oh I should probably differentiate between $$SA_1$$ and $$SA_2$$

42. TuringTest Group Title

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 Group Title

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 Group Title

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 Group Title

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 Group Title

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 Group Title

oh wow! that's AWESOME!!!!

48. JenniferSmart1 Group Title

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 Group Title

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 Group Title

... 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 Group Title

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 Group Title

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 Group Title

nevermind I see it now.

54. TuringTest Group Title

:)

55. JenniferSmart1 Group Title

We should have $\sigma=E\epsilon_0$

56. JenniferSmart1 Group Title

and I have $\sigma=\frac{E\epsilon_0}{r}$

57. JenniferSmart1 Group Title

LOL I see it again....sorry :S

58. TuringTest Group Title

:P Well, it's great that you really make sure your work makes sense to you!