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

In summary: Thanks for trying.In summary, the electric field inside a spherical shell is outward pointing and of uniform constant magnitude. The charge on the inner surface of the shell is ρ and the charge per unit volume is μ.
  • #1
MichelCarroll
3
0

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?
 
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  • #2
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
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
vanhees71 said:
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 learned that form yet though.
 

What is Gauss's Law and how does it apply to a spherical shell with non-uniform charge?

Gauss's Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the net electric charge enclosed within that surface. In the case of a spherical shell with non-uniform charge, Gauss's Law tells us that the electric flux through any closed surface surrounding the shell is equal to the net charge inside the surface divided by the permittivity of free space. This is true regardless of the distribution of charge on the shell.

How do you calculate the electric field for a point outside the spherical shell with non-uniform charge?

To calculate the electric field at a point outside the spherical shell with non-uniform charge, we use the formula E = Q/4πε₀r², where Q is the net charge inside the shell, ε₀ is the permittivity of free space, and r is the distance from the point to the center of the shell. This is due to the fact that the electric field outside the shell is the same as if all the charge were concentrated at the center of the shell.

How do you calculate the electric field for a point inside the spherical shell with non-uniform charge?

Inside the spherical shell with non-uniform charge, the electric field is zero. This is because the electric field from each element of charge on the shell cancels out at the point due to the symmetry of the shell. Therefore, the net electric field inside the shell is zero.

What happens to the electric field if the charge on the spherical shell is concentrated on one side?

If the charge on the spherical shell is concentrated on one side, the electric field will be stronger on that side and weaker on the opposite side. This is because the electric field is directly proportional to the amount of charge present, and the closer the charge is to a point, the stronger the electric field will be at that point. Therefore, the side with more charge will have a stronger electric field.

How does the electric field change if the radius of the spherical shell is increased?

If the radius of the spherical shell is increased, the electric field at a point outside the shell will decrease. This is because the electric field is inversely proportional to the square of the distance from the point to the center of the shell. Therefore, as the distance increases, the electric field decreases. However, the electric field inside the shell will remain zero regardless of the change in radius.

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