Gauss's Law and positive charge

In summary: Can you help me out?Electric flux is a measure of the rate of flow of electric charge. It's usually measured in coulombs per second (C/s). In your problem, you're looking for the flux of electric charge through a spherical surface. You can think of this as the rate of flow of electric charge through the surface, in terms of r and E and the constants involved. To find this flux, you need to integrate the charge density between two radii - the radius at which the charge density is largest (R/2) and the radius where the charge density is smallest (R).The flux of electric charge through a surface is proportional to the charge contained inside the surface. So, including
  • #1
hitek131
14
0
A region in space contains a total positive charge Q that is distributed spherically such that the volume charge density rho (r) is given by:

rho (r)=3 * alpha * r /(2R) for r ≤ R / 2

rho (r)= alpha * [1-(r/R)^(2) ] for (R/2) ≤ r ≤ R

rho (r)=0 for r ≥ R
Here alpha is a positive constant having units of C/m^3What fraction of the total charge is contained within the region (R/2) ≤ r ≤ R?

If anyone can help me with this problem it would be greatly appreciated. I am getting lost in all the integration with everything. Thanks
 
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  • #2
Please show us what you've done so far...
 
  • #3
This is what I got so far. It aint pretty.

The integral from (r/2) to r of:

Q/ (3R^3*pi/32)+(47R^3*pi/120) all this * (1 -(r/R)^2) * (4 * pi*r^2 dr)
all divided by 4 * pi *r^2 * eplison not.
 
  • #4
Please back up a few steps and explain how you got to that point.
 
  • #5
I am not really sure what I did. I used some integration formulas's, but I don't think they are right. Do you have a better starting way then what I did?
 
  • #6
For some reason, i don't know where to start off on this problem. There are so many specifications and I don't know if I'm doing the problem the right way.
 
  • #7
OK, start with Gauss's Law. What does it say? How can you relate it to the conditions of your problem?
 
  • #8
I know that Gauss's Law is the integral of vector E * vector dA
 
  • #9
I could use Gauss's Law and use the second condition from the specs in the problem. I could take the integral from (R/2) to R of the second condition.
 
  • #11
I found the integral. It is ( a (1- (r^2/R^2) * R)/2
Now I don't know how to use the Law to compute anything?
 
  • #12
hitek131 said:
I know that Gauss's Law is the integral of vector E * vector dA

That's the definition of the flux of [itex]\vec E[/itex] through a surface.

Look up Gauss's Law in your textbook. It says that the total flux of [itex]\vec E[/itex] through a closed surface equals the total charge enclosed inside that surface divided by [itex]\epsilon_0[/itex]. Both of these can be written as integrals:

[tex]\int{\vec E \cdot d \vec A} = \frac{1}{\epsilon_0} \int {\rho dV}[/tex]

where [itex]\rho[/itex] is of course the charge density.
 
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  • #13
Bleah, the server must be overloaded. My equations haven't shown up in my preceding posting yet. [oops, I had to fix a bug... now they're OK]

Anyway, to use Gauss's Law, imagine a spherical surface of radius r (the radius that you want to find E at). You need to find (a) the flux of E through that suface, in terms of r and E and constants, and (b) the total charge enclosed inside the surface, using the densities that you've been given. Then substitute into Gauss's Law and solve for E.

Note that integrating the charge density between R/2 and R (as you proposed) gives you the total charge contained inside that spherical shell. The r that you want to find E at, is < R, so your imaginary spherical surface doesn't contain all of that charge.

Also, don't forget to include the charge that lies between radius 0 and R/2!
 
  • #14
I don't know how to relate electric flux to this problem.
 

Related to Gauss's Law and positive charge

What is Gauss's Law?

Gauss's Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the net electric charge enclosed by that surface.

How does Gauss's Law apply to positive charge?

Gauss's Law states that the net electric flux through a closed surface is equal to the net charge enclosed by that surface divided by the permittivity of free space (ε0). This means that for a positive charge, the electric flux will radiate outward from the charge, resulting in a positive net flux through a closed surface surrounding the charge.

What is the formula for Gauss's Law?

The mathematical expression for Gauss's Law is Φe = Qenc / ε0, where Φe represents the electric flux through a closed surface, Qenc is the net charge enclosed by that surface, and ε0 is the permittivity of free space.

How is Gauss's Law used in practical applications?

Gauss's Law is used in various practical applications, such as calculating the electric field strength around a charged object, determining the capacitance of a capacitor, and analyzing the behavior of electric fields in conductors and insulators.

Does Gauss's Law apply to all types of charges?

Yes, Gauss's Law applies to both positive and negative charges. However, the direction of the electric field and the sign of the net electric flux will depend on the type of charge enclosed by the surface.

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