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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
 one year ago
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

This Question is Closed

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
dw:1361480351472:dw
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
I know at this point that V_1 is different from V_2
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
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

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
so \[\Delta V=0\]
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
Let's talk about the electric potential for a moment...
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
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

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
I still have a hard time picturing electric potential .....
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
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

experimentX Group TitleBest ResponseYou've already chosen the best response.1
think of two spheres as two vessels ..dw:1361481079204:dw charges as water ... and wire as tunnel
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
dw:1361481129492:dw
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
oh so they want to come to some kind of equilibrium
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
yes ...
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
dw:1361481183183:dw
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
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

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
Not the problem per se...I get what Potential part :P
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
I can't type ...sorry bout the grammar
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
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

experimentX Group TitleBest ResponseYou've already chosen the best response.1
nvm ... the grammar ,,, not charge difference exactly, charge density ... dw:1361481414886:dw
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
Oh the charge densities are different. got it.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
what's charge density?? in case of conductors ... metals??
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
\[\lambda=\frac{Q}{surface \; area}\]
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
\[E=kQ/r^2\]
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
yep .. that's correct, use \( \sigma \) for for surface charge density .. \( \lambda \) for linear charge density.
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
ooops! \[SA=4\pi r^2\] \[\sigma=Q/SA\] \[E=\frac{k(\sigma4\pi R^2)}{R^2}\]
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
\[\sigma=\frac{E}{k4\pi}\] REally?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
hmm ... what are you trying to do?? calculate surface charge density from given Electric Field?
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
Yep, I have to find the surface charge density on each sphere.
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
since they're connect by a long thin wire, the charge density should be equal for the spheres
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
what are given so far??
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
I have 2 spheres. Little sphere with radius r Big sphere with radius R We know E at the large sphere.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
at surface?? or ...
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
@TuringTest
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
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

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
yep, looks good so far
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
I'm given the electric field at the surface of the large sphere E. so \[E=\frac{kq}{r^2}\]
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{k\sigma_1(SA)}{r_1}=\frac{k\sigma_2(SA)}{r_2}\] \[E=\frac{kq}{r^2}\] would that make.... \[\frac{k\sigma_1(SA)}{r_1}=\frac{kEr^2}{kr_2}\] ?
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{k\sigma_1(SA)}{r_1}=\frac{\cancel{k}Er^\cancel{2}}{\cancel{k}\cancel{r_2}}\]
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
remember that the surace area of each sphere is different, and depends on the radius of that sphere
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{k\sigma_1(SA)}{r_1}=\frac{\cancel{k}Er_2^\cancel{2}}{\cancel{k}\cancel{r_2}}\] \[\sigma_1=\frac{Er_2r_1}{k(SA)}\] 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

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
oh I should probably differentiate between \(SA_1\) and \(SA_2\)
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
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

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
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

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
oh wow! that's AWESOME!!!!
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
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

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
... 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

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
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

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
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

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
nevermind I see it now.
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
We should have \[\sigma=E\epsilon_0\]
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
and I have \[\sigma=\frac{E\epsilon_0}{r}\]
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
LOL I see it again....sorry :S
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
:P Well, it's great that you really make sure your work makes sense to you!
 one year ago
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