Gauss's Law Problem - Spherical Shell with Non-uniform Charge

  • #1

Homework Statement



Consider a spherical shell with inner radius r1=0.30 m and outer radius r2=1.00 m. The hollow inside the shell contains no charge; and charge is distributed on the inside surface of the shell and within the shell itself, such that the electrical field inside the shell itself is everywhere outward pointing and of uniform constant magnitude 28 N/C.

a) What is the charge per unit area on the inner surface at r=r1?

b) What is the charge per unit volume at radius r=0.65 m (within the material of the shell)?


Homework Equations



[itex]\phi[/itex]ₑ = [itex]\oint[/itex] E · dA = Q/[itex]\epsilon[/itex]ₒ
area of sphere = 4πr^2

The Attempt at a Solution



I managed to get the a) part of the question using Gauss's Law. I made the Gaussian surface directly on the inner surface of the sphere (an infinitesimally small thickness, you could say). Since the electric field is constant everywhere on the surface, I could rewrite the equation as:

E4πr^2[itex]\epsilon[/itex]ₒ = Q

Since I'm looking for the charge per unit area, I rewrote it like this, and got the right answer:

[itex]\mu[/itex] = Q/A = q/(4πr^2)
[itex]\mu[/itex] = E[itex]\epsilon[/itex]ₒ

For the b) part of the question, I'm really struggling.

What I tried to do this: Since the volumetric charge density is ρ = Q/V, we could say that for an infinitesimally small volume, ρ = dQ/dV.

From this logic, I rewrote my equation for charge in terms of volume instead of radius. Then, I wanted to take the derivative of this function in terms of volume. Then, I reformulated the function in terms of radius. Here's my steps:

Q = E4πr^2[itex]\epsilon[/itex]ₒ
V = (4/3)πr^3
r = [itex]\sqrt[3]{3V/4π}[/itex]
Q = E4π(3V/4π)^(2/3)[itex]\epsilon[/itex]ₒ
dQ/dV = 2πr^2E[itex]\epsilon[/itex]ₒ/r

It's possible that my calculus got rusty, but this function is not giving me the right answer.

Can somebody lead me in the right direction please?
 
Last edited:

Answers and Replies

  • #2
49
0
Your first equation is correct and not only does it tell you the total charge on the inner surface of the shell, it also tells you the total charge inside any concentric sphere of any radius inside the shell. (Can you see why?)

Where you have gone wrong is to introduce the equation for the volume of a sphere. This is not relevant as the charge density within the shell is not uniform. What you must do is differentiate the first equation to find the total charge within a thin shell of thickness δr; then divide by the volume of the shell to get the charge per unit volume.
 
  • #3
Thanks for your help. I think I understand. Here's my steps:

Differentiating the function for charge to find the charge inside a thin layer of the shell of thickness dr:

Q = E4πr²[itex]\epsilon[/itex]ₒ
dQ/dr = 8πrE[itex]\epsilon[/itex]ₒ

Finding the volume of a thin spherical shell of thickness dr: (between "r" and "r+dr")

dV = (4/3)π( (r + dr)³ - r³ )
dV = (4/3)π( 3r²dr + 3rdr² + dr³ ) (ignoring 3rdr² and dr³ because they are very small)
dV = 4πr² dr

Swapping dr for dV in the first equation:

dQ/dV = 2E[itex]\epsilon[/itex]ₒ/r = ρ

Is this correct?
 
  • #4
vanhees71
Science Advisor
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Gold Member
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If you are allowed to use the local form of Gauß's Law,
[tex]\vec{\nabla} \cdot \vec{E}=\frac{\rho}{\epsilon}[/tex]
the whole thing becomes much easier!
 
  • #5
If you are allowed to use the local form of Gauß's Law,
[tex]\vec{\nabla} \cdot \vec{E}=\frac{\rho}{\epsilon}[/tex]
the whole thing becomes much easier!

Unfortunately haven't learnt that form yet though.
 

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