Flux from a given charge distribution

In summary, the problem involves three concentric spherical shells with charge distributions of 2, -4, and 5 μC/m2 and asks to calculate the flux through r=1.5m and r=2.5m. The solution involves finding the total enclosed charge (Qenc) by multiplying the charge distribution by the surface area of each shell and then using the formula flux = Qenc/ε0. After solving, the flux is found to be 2.84x106C, which is not out of scale. However, the mistaken assumption that ε0 was unitless caused some confusion in the solution process.
  • #1
ryukyu
20
0

Homework Statement


Three concentric spherical shells ,r=1m ,r=2m and r=3m, respectively, have charge
distributions 2, −4, and 5 μC/m2.

Calculate the flux through r=1.5m and r=2.5m



Homework Equations



Since we are talking about Gaussian surfaces I assume that the flux at 1.5m is just the flux contained in the 1m shell.

I'm particularly having trouble figuring out what to do with the data. Given that the charge distribution is given in uC/m^2, I assume we need to multiply this by the surface area of each sphere to find the total enclosed charge (Qenc).

Then I have that flux = Qenc0. But these answers seem way too high.


The Attempt at a Solution



SA1=4(pi)r2=4(pi)m2
Qenc1=4(pi)m2*2uC/m2=8(pi)uC
flux=8(pi)uC/ε0=2.84x106C
where ε0=(10-9/(36*pi))
 
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  • #2
Your answer is correct.
 
  • #3
Thanks for the quick response. The whole mega-Coulomb just had me worried.
 
  • #4
ryukyu said:
Thanks for the quick response. The whole mega-Coulomb just had me worried.
It is not mega -Coulomb. Flux is not measured in C, but in Weber.
 
  • #5
I've got it straightened out now. I was assuming eps0 was unitless. But even given that enlightenment, are these really mega-Webers that seems even more out of scale?
 
  • #6
No. It is not out of scale.
 
  • #7
Thanks again. I have been working on the homework for this class for at least 30 hours on and off and it's only 7 problems. The professor's method of instruction and my method of learning just don't blend well and that's not his fault, I just need to learn a new way to learn.
 

1. What is flux from a given charge distribution?

Flux from a given charge distribution is the measure of the flow of electric field lines through a given surface, caused by the presence of electric charges. It is a crucial concept in understanding the behavior of electric fields.

2. How is flux calculated from a given charge distribution?

Flux is calculated by taking the dot product of the electric field vector and the normal vector of the surface, multiplied by the surface area. Mathematically, it can be represented as Φ = ∫∫E⋅dA, where E is the electric field and dA is the differential area element of the surface.

3. What factors affect the flux from a given charge distribution?

The flux from a given charge distribution is affected by the magnitude of the electric field, the distance between the charges and the surface, and the orientation of the surface with respect to the electric field lines. It is also affected by the presence of any other charges in the vicinity.

4. Can flux be negative from a given charge distribution?

Yes, flux can be negative from a given charge distribution. This occurs when the electric field and the normal vector of the surface are in opposite directions, resulting in a negative dot product. Negative flux indicates that the electric field lines are entering the surface instead of leaving it.

5. What are the units of flux from a given charge distribution?

The SI unit of flux is coulomb (C), which is equivalent to joules per volt (J/V). In terms of base units, it is represented as kg⋅m^3⋅s^-3⋅A^-1. However, it is often expressed in terms of electric flux density (D), which has units of C/m^2 or V⋅m^-2.

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