anonymous
  • anonymous
Gauss's Law
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
What about it, can you please copy and past the question...please, or i can't help.
anonymous
  • anonymous
call me when you are ready..
anonymous
  • anonymous
|dw:1359922050546:dw|

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anonymous
  • anonymous
What is the question
anonymous
  • anonymous
My goodness have some patience...You can obviously see my typing... Find the electric field everywhere for a uniformly charged solid sphere that has a radius R and a total charge Q that is uniformly distributed throughout the volume of the sphere that has a charge density \[\rho=\frac QV\], where \[V=\frac 43 \pi R^3\] is the volume of the sphere. It's an example in my book. So far I understand that I need to make shell around the charged solid, and use that find the \[\phi_{net}\] and then we would find the Charge because \[Q=\phi_{net}\epsilon_0\] From that charge we find the electric field..... Here is my question What do they mean by \[\textrm{For r}\ge R, Q_{inside}=Q\] \[\textrm{For r}\le R, Q_{inside}=\rho V'\]
anonymous
  • anonymous
They mean that once your shell goes beyond the radius of the charged sphere, the amount of charge inside is constant (obviously, just the total charge of the sphere). But when you're inside, the charge inside your shell is less than that.
anonymous
  • anonymous
|dw:1359923152556:dw| can you show me where exactly? I'm am a visual learner.
anonymous
  • anonymous
"....once your shell goes beyond the radius of the charged sphere, the amount of charge inside is constant..." " But when you're inside, the charge inside your shell is less than that."
anonymous
  • anonymous
|dw:1359923417543:dw| ?
anonymous
  • anonymous
Just a moment...
anonymous
  • anonymous
If the shell is INSIDE the sphere, then the charge enclosed is Q times the fraction of the sphere contained in the shell. If the shell is larger than the sphere, though, the total charge inside is just Q.
anonymous
  • anonymous
Since I made my shell outside the sphere....My charge inside is just Q, correct?
anonymous
  • anonymous
nope, I need to consider both, right?
JamesJ
  • JamesJ
Actually, insides the sphere the net charge should be zero, because all of the charge lies on the surface; i.e., \( \rho = 0 \). To see all of this explained, including this case, you might enjoy watching this excellent lecture: http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/video-lectures/lecture-3-electric-flux-and-gausss-law/
anonymous
  • anonymous
@JamesJ The specifically said "a uniformly charged solid sphere"
anonymous
  • anonymous
*they
JamesJ
  • JamesJ
ok. So the sphere is not a conductor.
anonymous
  • anonymous
I doubt it. What do you think @Jemurray3 ?
anonymous
  • anonymous
No, it's not a conductor
JamesJ
  • JamesJ
So now... What do they mean by For r≥R,Qinside=Q For r≤R,Qinside=ρV′ What is meant is that when we are outside the sphere, the charge is the entire charge of the sphere Q. However, once you draw a Gaussian surface inside the sphere, you just want the charge within that surface, which is just the volume of that surface time the charge density. In other words, the charge outside of the surface is irrelevant.
JamesJ
  • JamesJ
I would still watch the lecture. He does still give you a lot of intuition for this sort of problem.
anonymous
  • anonymous
When we say Qinside, we mean the Q inside of the charged sphere, NOT the sphere that I have drawn around it, correct?
anonymous
  • anonymous
Qinside means the charge inside the shell that you've drawn.
anonymous
  • anonymous
I think I get it now: The charge anywhere inside of the charged solid sphere is the density times the newly found volume (which depends on the size of r) The charge anywhere outside of that charged solid sphere, but still within the shell I have drawn is just Q.
anonymous
  • anonymous
Yep
anonymous
  • anonymous
2 hours later.....I finally get it.
anonymous
  • anonymous
awesome! Thanks guys

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