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.

- anonymous

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- anonymous

|dw:1361480351472:dw|

- anonymous

I know at this point that V_1 is different from V_2

- experimentX

why??

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## More answers

- 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

so \[\Delta V=0\]

- anonymous

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

- 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

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

- 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

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

- anonymous

|dw:1361481129492:dw|

- anonymous

oh so they want to come to some kind of equilibrium

- experimentX

yes ...

- anonymous

|dw:1361481183183:dw|

- 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

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

- anonymous

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

- 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

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

- anonymous

Oh the charge densities are different. got it.

- experimentX

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

- anonymous

\[\lambda=\frac{Q}{surface \; area}\]

- anonymous

\[E=kQ/r^2\]

- experimentX

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

- anonymous

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

- anonymous

\[\sigma=\frac{E}{k4\pi}\]
REally?

- experimentX

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

- anonymous

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

- anonymous

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

- experimentX

what are given so far??

- anonymous

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

- experimentX

at surface?? or ...

- anonymous

@TuringTest

- 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

yep, looks good so far

- anonymous

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

- 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

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

- TuringTest

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

- 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

oh I should probably differentiate between \(SA_1\) and \(SA_2\)

- 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

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

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

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

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

oh wow! that's AWESOME!!!!

- 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

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

... and as always, happy to help! As I said, I'm Taking E&M too (again) , so I need the refreshers like this :)

- 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

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

nevermind I see it now.

- TuringTest

:)

- anonymous

We should have
\[\sigma=E\epsilon_0\]

- anonymous

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

- anonymous

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

- TuringTest

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

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