anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1361480351472:dw|
anonymous
  • anonymous
I know at this point that V_1 is different from V_2
experimentX
  • experimentX
why??

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anonymous
  • anonymous
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
anonymous
  • anonymous
so \[\Delta V=0\]
anonymous
  • anonymous
Let's talk about the electric potential for a moment...
experimentX
  • experimentX
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.
anonymous
  • anonymous
I still have a hard time picturing electric potential .....
anonymous
  • anonymous
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?
experimentX
  • experimentX
think of two spheres as two vessels ..|dw:1361481079204:dw| charges as water ... and wire as tunnel
anonymous
  • anonymous
|dw:1361481129492:dw|
anonymous
  • anonymous
oh so they want to come to some kind of equilibrium
experimentX
  • experimentX
yes ...
anonymous
  • anonymous
|dw:1361481183183:dw|
anonymous
  • anonymous
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
anonymous
  • anonymous
Not the problem per se...I get what Potential part :P
anonymous
  • anonymous
I can't type ...sorry bout the grammar
anonymous
  • anonymous
ok let's see here....I have to relate the electric field of the big sphere to the potential on the big sphere...
experimentX
  • experimentX
nvm ... the grammar ,,, not charge difference exactly, charge density ... |dw:1361481414886:dw|
anonymous
  • anonymous
Oh the charge densities are different. got it.
experimentX
  • experimentX
what's charge density?? in case of conductors ... metals??
anonymous
  • anonymous
\[\lambda=\frac{Q}{surface \; area}\]
anonymous
  • anonymous
\[E=kQ/r^2\]
experimentX
  • experimentX
yep .. that's correct, use \( \sigma \) for for surface charge density .. \( \lambda \) for linear charge density.
anonymous
  • anonymous
ooops! \[SA=4\pi r^2\] \[\sigma=Q/SA\] \[E=\frac{k(\sigma4\pi R^2)}{R^2}\]
anonymous
  • anonymous
\[\sigma=\frac{E}{k4\pi}\] REally?
experimentX
  • experimentX
hmm ... what are you trying to do?? calculate surface charge density from given Electric Field?
anonymous
  • anonymous
Yep, I have to find the surface charge density on each sphere.
anonymous
  • anonymous
since they're connect by a long thin wire, the charge density should be equal for the spheres
experimentX
  • experimentX
what are given so far??
anonymous
  • anonymous
I have 2 spheres. Little sphere with radius r Big sphere with radius R We know E at the large sphere.
experimentX
  • experimentX
at surface?? or ...
anonymous
  • anonymous
@TuringTest
anonymous
  • anonymous
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?
TuringTest
  • TuringTest
yep, looks good so far
anonymous
  • anonymous
I'm given the electric field at the surface of the large sphere E. so \[E=\frac{kq}{r^2}\]
anonymous
  • anonymous
\[\frac{k\sigma_1(SA)}{r_1}=\frac{k\sigma_2(SA)}{r_2}\] \[E=\frac{kq}{r^2}\] would that make.... \[\frac{k\sigma_1(SA)}{r_1}=\frac{kEr^2}{kr_2}\] ?
anonymous
  • anonymous
\[\frac{k\sigma_1(SA)}{r_1}=\frac{\cancel{k}Er^\cancel{2}}{\cancel{k}\cancel{r_2}}\]
TuringTest
  • TuringTest
remember that the surace area of each sphere is different, and depends on the radius of that sphere
anonymous
  • anonymous
\[\frac{k\sigma_1(SA)}{r_1}=\frac{\cancel{k}Er_2^\cancel{2}}{\cancel{k}\cancel{r_2}}\] \[\sigma_1=\frac{Er_2r_1}{k(SA)}\] 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 \[\frac{k\sigma_1(SA)}{r_1}=V_2\]
anonymous
  • anonymous
oh I should probably differentiate between \(SA_1\) and \(SA_2\)
TuringTest
  • TuringTest
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.
anonymous
  • anonymous
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.
TuringTest
  • TuringTest
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
anonymous
  • anonymous
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
TuringTest
  • TuringTest
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 :)
anonymous
  • anonymous
oh wow! that's AWESOME!!!!
anonymous
  • anonymous
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?)
TuringTest
  • TuringTest
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...
TuringTest
  • TuringTest
... and as always, happy to help! As I said, I'm Taking E&M too (again) , so I need the refreshers like this :)
anonymous
  • anonymous
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."
anonymous
  • anonymous
is that legal? \[\frac{1}{\cancel{4\pi}\cancel\epsilon_0}\frac{\sigma_1\cancel{4\pi} r_1^2}{r_1}=\frac{\sigma_2\cancel{4\pi} r_2^2}{r_2}\frac{1}{\cancel{4\pi} \cancel\epsilon_0}\]
anonymous
  • anonymous
nevermind I see it now.
TuringTest
  • TuringTest
:)
anonymous
  • anonymous
We should have \[\sigma=E\epsilon_0\]
anonymous
  • anonymous
and I have \[\sigma=\frac{E\epsilon_0}{r}\]
anonymous
  • anonymous
LOL I see it again....sorry :S
TuringTest
  • TuringTest
:P Well, it's great that you really make sure your work makes sense to you!

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