Gauss's Law

- anonymous

Gauss's Law

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- schrodinger

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- anonymous

What about it, can you please copy and past the question...please, or i can't help.

- anonymous

call me when you are ready..

- anonymous

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## More answers

- anonymous

What is the question

- 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

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

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can you show me where exactly? I'm am a visual learner.

- 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

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?

- anonymous

Just a moment...

- 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

Since I made my shell outside the sphere....My charge inside is just Q, correct?

- anonymous

nope, I need to consider both, right?

- 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

@JamesJ The specifically said "a uniformly charged solid sphere"

- anonymous

*they

- JamesJ

ok. So the sphere is not a conductor.

- anonymous

I doubt it. What do you think @Jemurray3 ?

- anonymous

No, it's not a conductor

- 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

I would still watch the lecture. He does still give you a lot of intuition for this sort of problem.

- 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

Qinside means the charge inside the shell that you've drawn.

- 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

Yep

- anonymous

2 hours later.....I finally get it.

- anonymous

awesome! Thanks guys

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