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 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.
 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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JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1361480351472:dw

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.1I know at this point that V_1 is different from V_2

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.1THe 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

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.1so \[\Delta V=0\]

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.1Let's talk about the electric potential for a moment...

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1since 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.

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.1I still have a hard time picturing electric potential .....

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.1I 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
 2 years ago
Best ResponseYou've already chosen the best response.1think of two spheres as two vessels ..dw:1361481079204:dw charges as water ... and wire as tunnel

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1361481129492:dw

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.1oh so they want to come to some kind of equilibrium

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1361481183183:dw

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.1So 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

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.1Not the problem per se...I get what Potential part :P

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.1I can't type ...sorry bout the grammar

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.1ok let's see here....I have to relate the electric field of the big sphere to the potential on the big sphere...

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1nvm ... the grammar ,,, not charge difference exactly, charge density ... dw:1361481414886:dw

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.1Oh the charge densities are different. got it.

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1what's charge density?? in case of conductors ... metals??

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.1\[\lambda=\frac{Q}{surface \; area}\]

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

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

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.1\[\sigma=\frac{E}{k4\pi}\] REally?

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

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.1Yep, I have to find the surface charge density on each sphere.

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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1what are given so far??

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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1at surface?? or ...

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.1Let'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
 2 years ago
Best ResponseYou've already chosen the best response.1yep, looks good so far

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

JenniferSmart1
 2 years ago
Best 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.... \[\frac{k\sigma_1(SA)}{r_1}=\frac{kEr^2}{kr_2}\] ?

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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1remember that the surace area of each sphere is different, and depends on the radius of that sphere

JenniferSmart1
 2 years ago
Best 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 \[\frac{k\sigma_1(SA)}{r_1}=V_2\]

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.1oh I should probably differentiate between \(SA_1\) and \(SA_2\)

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1Yes, 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.

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.1yeah 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
 2 years ago
Best ResponseYou've already chosen the best response.1typo 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

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.1oh 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
 2 years ago
Best ResponseYou've already chosen the best response.1yep :) 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 :)

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.1oh wow! that's AWESOME!!!!

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.1I 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
 2 years ago
Best ResponseYou've already chosen the best response.1That'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
 2 years ago
Best 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 :)

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.1Yeah 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."

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.1is that legal? \[\frac{1}{\cancel{4\pi}\cancel\epsilon_0}\frac{\sigma_1\cancel{4\pi} r_1^2}{r_1}=\frac{\sigma_2\cancel{4\pi} r_2^2}{r_2}\frac{1}{\cancel{4\pi} \cancel\epsilon_0}\]

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.1nevermind I see it now.

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.1We should have \[\sigma=E\epsilon_0\]

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.1and I have \[\sigma=\frac{E\epsilon_0}{r}\]

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.1LOL I see it again....sorry :S

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