Gauss's Law for Spherical Symmetry

In summary, the electric field for a non-conducting sphere of radius R = 1 meter that is surrounded by air in the region r > 1 is k(pi)/r^2. However, when calculating the enclosed charge, you must first break the sum into two separate integrals, one for when the charge density is nonzero (within the non-conducting sphere) and the other for when the charge density is zero (within the surrounding air). When doing so, the enclosed charge is ##\pi## Coulombs.
  • #1
Vectorspace17
5
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Find the electric field for a non-conducting sphere of radius R = 1 meter that is surrounded by air in the region r > 1. The interior of the sphere has a charge density of ρ(r) = r.

The answer is k(pi)/r^2, but I can't seem to get that. My problem is with finding the enclosed charge. I've tried getting the answer many ways, but I keep getting it wrong. Generally, I know you have to set up an integral of the from ∫ρ(r)4(pi)r^2 dr to get the enclosed charge and then plug it into E(r) = kQ/r^2, but something keeps going wrong somewhere. I would greatly appreciate any help/advice anyone could offer. Thanks.
 
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  • #2
Post your calculation so we can see where you're going awry.
 
  • #3
So from the center to the edge of the sphere (R = 1), I know the enclosed charge should be:

from 0 to 1, ∫ρ(r)4(pi)r^2 dr = 4pi[(1/4)r^4] from 0 to 1
= pi

When put into E(r), this does give k(pi)/r^2, but I'm confused about this result because this is supposed to be the final answer to the region r > 1, not 0 to R = 1.

From 1 to r > 1, its surrounded by air, which means the charge density should be constant, so from 1 to r, the integral to find the enclosed charge is:

∫4(pi)r^2 dr = 4pi[(1/3)r^3] from 1 to pi
= (4/3)(pi)[r^3 - 1]

This is the integral that I though would give the charge enclosed for r > 1, but it yields a really ugly solution. The only thing I can think of (and I'm not sure if it makes any sense) is that my first result would be the charge on the surface of the sphere, and since its surrounded be air, that charge would remain constant as you move away from the surface of the sphere.
 
  • #4
Vectorspace17 said:
So from the center to the edge of the sphere (R = 1), I know the enclosed charge should be:

from 0 to 1, ∫ρ(r)4(pi)r^2 dr = 4pi[(1/4)r^4] from 0 to 1
= pi

When put into E(r), this does give k(pi)/r^2, but I'm confused about this result because this is supposed to be the final answer to the region r > 1, not 0 to R = 1.
You've calculated the total charge on the sphere to be ##\pi## Coulombs. This is correct. Gauss' law tells us we can treat a spherically symmetric charge distribution as a point charge when we're outside of it, so the external field for r > 1 obeys Coulombs law, which is what you've arrived at.
From 1 to r > 1, its surrounded by air, which means the charge density should be constant, so from 1 to r, the integral to find the enclosed charge is:

∫4(pi)r^2 dr = 4pi[(1/3)r^3] from 1 to pi
= (4/3)(pi)[r^3 - 1]

This is the integral that I though would give the charge enclosed for r > 1, but it yields a really ugly solution. The only thing I can think of (and I'm not sure if it makes any sense) is that my first result would be the charge on the surface of the sphere, and since its surrounded be air, that charge would remain constant as you move away from the surface of the sphere.

Clearly there are two distinct regions to deal with, one where the charge density is nonzero (within the non-conducting sphere), and the other where the charge density is zero (within the surrounding air). So your charge density function is discontinuous at r = 1.
$$
\rho (r) = \left \{
\begin{array}{l}
r & : 0 \leq r \leq 1 \\
0 & : r > 1
\end{array}
\right .
$$
mathematically you deal with this by breaking the sum into two separate integrals. Since the charge density function is uniformly zero for the second region, that integral must vanish.
 
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  • #5


I would suggest double checking your calculations and making sure you are using the correct formula for Gauss's Law. The correct formula for Gauss's Law for spherical symmetry is:

Φ(E) = ∫E * dA = Q/ε0

Where Φ(E) is the electric flux, Q is the enclosed charge, and ε0 is the permittivity of free space.

In this case, the electric field for a non-conducting sphere with a charge density of ρ(r) = r can be found by setting up the integral:

∫E * dA = Q/ε0

Since the sphere is surrounded by air, the electric field outside the sphere will be zero (since air is an insulator). Therefore, the only contribution to the electric flux will come from the interior of the sphere.

Using the given charge density, ρ(r) = r, we can rewrite the integral as:

∫E * dA = ∫r * 4(pi)r^2 dr

Integrating this, we get:

∫E * dA = 4(pi)r^4/4

Simplifying, we get:

∫E * dA = (pi)r^4

Since we know that the electric flux must equal Q/ε0, we can set these two expressions equal to each other:

(pie)r^4 = Q/ε0

Solving for Q, we get:

Q = (pi)r^4 * ε0

Substituting this value for Q into the formula for electric field, we get:

E = kQ/r^2 = k(pi)r^4 * ε0/ r^2 = k(pi)/r^2

Therefore, the electric field for a non-conducting sphere of radius R = 1 meter with a charge density of ρ(r) = r, surrounded by air in the region r > 1, is k(pi)/r^2. I would suggest checking your calculations and making sure you are using the correct formula for Gauss's Law.
 

What is Gauss's Law for Spherical Symmetry?

Gauss's Law for Spherical Symmetry is a fundamental law in electrostatics that relates the electric flux through a closed surface to the charge enclosed within that surface. It states that the flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space (ε0). This law is derived from Coulomb's Law and is used to simplify calculations of electric fields for spherically symmetric charge distributions.

What is the formula for Gauss's Law for Spherical Symmetry?

The formula for Gauss's Law for Spherical Symmetry is ΦE = Qenc0, where ΦE is the electric flux through a closed surface, Qenc is the charge enclosed within that surface, and ε0 is the permittivity of free space.

How do you use Gauss's Law for Spherical Symmetry to calculate electric fields?

To use Gauss's Law for Spherical Symmetry to calculate electric fields, you must first determine the charge enclosed within the closed surface. Then, you can use the formula ΦE = Qenc0 and solve for the electric field (E). This involves taking the derivative of the flux with respect to the radial distance (r) and simplifying the equation to solve for E.

What are the limitations of Gauss's Law for Spherical Symmetry?

Gauss's Law for Spherical Symmetry is only applicable to spherically symmetric charge distributions. This means that the charge must be evenly distributed on a spherical surface or at the center of a sphere. It also assumes that the electric field is constant on the surface of the sphere and that the surface is a perfect conductor.

How is Gauss's Law for Spherical Symmetry related to Gauss's Law for general cases?

Gauss's Law for Spherical Symmetry is a special case of Gauss's Law for general cases. Gauss's Law for general cases states that the electric flux through a closed surface is equal to the net charge enclosed by that surface divided by the permittivity of free space. This law applies to all types of charge distributions, not just spherically symmetric ones.

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