Electrostatics problem

Feb 2020
5
0
England
I am confused regarding a question about Gauss' law. My confusion arises from a non-uniform charge density within a non-conducting sphere. The question is to find the electric field inside and outside the sphere.

I know that for a uniform charge density that it's a straight forward application of Gauss' law, however, given a function where the charge density varies means that the electric field is not constant and so can't be brought outside of the integral.

Also, outside of the sphere, the enclosed charge is constant and hence I think that this is just the enclosed charge divided by the permittivity of free space.
 

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topsquark

Forum Staff
Apr 2008
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On the dance floor, baby!
Let's talk about inside the sphere.

Do the usual thing with the Gaussian surface... a sphere centered on the center of the charge distribution of radius r < R where R is the radius of trhe whole distribution. Then we get the usual
\(\displaystyle \int _{ \sigma } \vec{E} \cdot \vec{dr} = \dfrac{ Q_{enc} }{ \epsilon _0 }\)

The integral is done the way it always is and \(\displaystyle Q_{enc} \) is the charge inside the sphere is the volume integral of the charge density:
\(\displaystyle Q_{enc} = \int _{ \tau } \rho ~ d^3r = \int_0^r \int _0^{ \pi } \int _0^{ 2 \pi } ( \alpha r) ~ r^2 ~ sin( \theta ) ~d \phi ~ d \theta ~ dr\).

Outside the charge distribution you do it the same way. You just use R = the radius of the charge distribution.

Give that a try and let us know how it works out.

-Dan
 
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Feb 2020
5
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England
Let's talk about inside the sphere.

Do the usual thing with the Gaussian surface... a sphere centered on the center of the charge distribution of radius r < R where R is the radius of trhe whole distribution. Then we get the usual
\(\displaystyle \int _{ \sigma } \vec{E} \cdot \vec{dr} = \dfrac{ Q_{enc} }{ \epsilon _0 }\)

The integral is done the way it always is and \(\displaystyle Q_{enc} \) is the charge inside the sphere is the volume integral of the charge density:
\(\displaystyle Q_{enc} = \int _{ \tau } \rho ~ d^3r = \int_0^r \int _0^{ \pi } \int _0^{ 2 \pi } ( \alpha r) ~ r^2 ~ sin( \theta ) ~d \phi ~ d \theta ~ dr\).

Outside the charge distribution you do it the same way. You just use R = the radius of the charge distribution.

Give that a try and let us know how it works out.

-Dan
The triple integral leads to method 1 and integrating directly with respect to volume leads to method 2. I don't understand the difference between the two methods and why they give different answers? And, for outside the charge distribution, does that mean that the expression is the same just with R instead of r?
 

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topsquark

Forum Staff
Apr 2008
3,055
651
On the dance floor, baby!
When calculating the field outside the sphere we know that all the charge is in a sphere of radius R. So the total charge will be \(\displaystyle Q_{enc} = \int _0^R \rho ~d^3r\), rather than \(\displaystyle \int_0^r \rho ~d^3r\), if you'll permit the bad notation.

As for Method 2, you can't pull \(\displaystyle \rho\) outside the integral... It depends on r.

-Dan
 
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Apr 2015
1,156
303
Somerset, England
Before I pitch in here, please tell my why this (and your other stuff) is posted in High School Physics ?

It is way beyond that.
 
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Feb 2020
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England
Before I pitch in here, please tell my why this (and your other stuff) is posted in High School Physics ?

It is way beyond that.

Sorry, I'm new to this forum and didn't scroll down through the other sections - my fault. But thank you for the help, the mistake in pulling rho out of the integral was a silly one but thanks for pointing it out!
 
Feb 2020
5
0
England
I'm fairly convinced that I've solved the problem now. Here's what I did.
I defined R to be a distance outside of the sphere and r to be an arbitrary distance inside the sphere (a radius of a sphere contained in the larger sphere)

To solve it, I considered an element of volume, dV and considered the charge contained within that, dQ. Then came up with an expression for the charge within this element dV as the charge density multiplied by the area of a given sphere with radius r multiplied by its thickness dr.
The total enclosed charge in this sphere contained within the larger sphere is then the integral of this expression which came out as pi*alpha*r^4.
Then, using Gauss' law, rearranged EA=Q/epsilon for the electric field, E.

That was for the value of E within the sphere, then for outside the sphere, I evaluated the integral of the charge contained within dV between the limits of 0 to R, doing the same process with rearranging Gauss' law giving an expression for the electric field outside the sphere.
When I set R=r, the two equations I have give the same result which would be the value for the electric field at the surface. This is what convinced me that the expressions I have are correct. I think I was going wrong with trying to calculate electric flux when the question was asking for the electric field. I also tried this with the method of the volume integral and ended up with the same solution after carrying out the algebra.

I'm only just getting started with physics and love problems like this! Spent the last couple days trying to figure this one out, thank you for all the help!
 
Apr 2015
1,156
303
Somerset, England
Well you seem prepared to put in a lot of good effort to your study and I'm suee that will pay off.

:)

You and topsquark are having a very productive consversation about the maths so I won't interfere, but would like to add something extra since this is a very good question.

Gauss' theorem is actually an example of a very deep connection between (pure) mathematics and physics.

It is remarkable that just by knowing what is happening for certain functions at the boundary of a region we can calculate everything about them within that region.

Not only is it remarkable but extremely useful and is applied not only to electrostatics, but to fluid mechanics, mechanical stress and strain theory and lots more.
The technique of finite element analysis is often greatly simplified by transferring to a boudary element instead of a body element.

In Maths the theorem is called 'The Fundamental Theorem of Calculus'

In Physics it appears in theorems by Gauss, Stokes, Green and others.

The simplest example is the definite integral, where the end points define the value of the function being integrated.


\(\displaystyle \int\limits_a^b {df} = f\left( b \right) - f\left( a \right)\)

A one dimensional problem reduced two two points.

A 3 dimensional problem can be reduced to a two dimensional boundary surface problem and so on.