Equivalence of Integral and Differential Forms of Gauss's Law?

In summary: The correct result is that the electric field is E=\frac{k\cdot r^2}{4\epsilon} anywhere inside the sphere, but has a half-value at the surface due to the charge distribution being discontinuous there.
  • #1
BucketOfFish
60
1
A sphere has charge density [itex]\rho=k\cdot r[/itex]. Using the integral form of Gauss's Law, one easily finds that the electric field is [itex]E=\frac{k\cdot r^2}{4\epsilon}[/itex] anywhere inside the sphere. However, [itex]\nabla\cdot E=\frac{k\cdot r}{2\epsilon}[/itex], which is half of what should be expected from the differential form of Gauss's Law, since [itex]\frac{\rho}{\epsilon}=\frac{k\cdot r}{\epsilon}[/itex]. Why is this?
 
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  • #2
In Heaviside-Lorentz units, the equation for the electrostatic potential reads

[tex]\Delta \phi(\vec{x})=-\rho(\vec{x}).[/tex]

In spherical coordinates, for a radially symmetric charge distribution, this equation simplifies to the ode,

[tex]\frac{1}{r} \mathrm{d}_r^2 (r \phi)=:\frac{1}{r} (r \phi)''=-\rho(r).[/tex]

Here, [itex]r=|\vec{x}|[/itex]. In your case we have

[tex](r \phi)''=-k r^2.[/tex]

This you can integrate up successively:

[tex](r \phi)'=-\frac{k}{3} r^3+C_1, \quad r \phi=-\frac{k}{12} r^4 + C_1 r+C_2[/tex]

or, finally,

[tex]\phi(r)=-\frac{k}{12} r^3 + C_1 + \frac{C_2}{r}.[/tex]

The integration constants, [itex]C_1[/itex] and [itex]C_2[/itex], are determined by appropriate boundary conditions, where [itex]C_1[/itex] is physically irrelevant since the electric field is the observable quantity, which is given by the gradient of the potential. So I set [itex]C_1=0[/itex] Further, since the charge distribution is continuous at the origin, we must have [itex]C_2=0[/itex]. So finally we get

[tex]\phi(r)=-\frac{k}{12} r^3, \quad \vec{E}=-\vec{\nabla} \phi = \frac{k}{4} r^2 \vec{e}_r = \frac{k}{4} r \vec{x}.[/tex]

To check this result, we take the divergence, which for our case of [itex]\vec{E}=E_r(r) \vec{e}_r [/itex] simplifies to

[tex]\vec{\nabla} \cdot \vec{E}=\frac{1}{r^2} (r^2 E_r)'=\frac{1}{r^2} \left (\frac{k r^4}{4} \right )'=k r=\rho[/tex]

as it should be since Maxwell's equations for the electrostatic case read

[tex]\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{E}=\rho.[/tex]

I guess, your confusion comes from forgetting that the differential operators look different in curvilinear coordinates compared to the expressions in cartesian ones.

Of course, you can calculate the divergence also in Cartesian coordinates. Although this is a bit more inconvenient in your case of spherical symmetry, you must get the same result. Indeed using the Cartesian form

[tex]\vec{E}=\frac{k}{4} \sqrt{x^2+y^2+z^2} \vec{x}[/tex]

you get again

[tex]\vec{\nabla} \cdot \vec{E}=\partial_x E_x+\partial_y E_y+\partial_z E_z=\frac{k}{4} \left (\frac{2 x^2+y^2+z^2}{r}+\frac{x^2+2y^2+z^2}{r}+\frac{x^2+y^2+2 z^2}{r} \right) =\frac{k}{4} \frac{4(x^2+y^2+z^2)}{r}=k r.[/tex]
 
  • #3
Thank you. I did take the divergence incorrectly.
 

1. What is Gauss's Law and why is it important in physics?

Gauss's Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the charge enclosed within that surface. It is important in physics because it helps us understand the behavior of electric fields and how they are affected by charges.

2. What is the integral form of Gauss's Law and how is it derived?

The integral form of Gauss's Law 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 (ε0). It is derived from the differential form of Gauss's Law by applying the divergence theorem.

3. What is the differential form of Gauss's Law and how is it related to the integral form?

The differential form of Gauss's Law states that the divergence of the electric field at a point is equal to the charge density at that point divided by the permittivity of free space (ε0). It is related to the integral form by the divergence theorem, which allows us to convert between surface integrals and volume integrals.

4. Why are there two different forms of Gauss's Law?

The two forms of Gauss's Law (integral and differential) are mathematically equivalent and can be used interchangeably. However, the differential form is more useful for solving problems involving continuous charge distributions, while the integral form is better suited for calculating the electric field at a specific point or over a closed surface.

5. How can we use Gauss's Law to solve practical problems in electromagnetism?

Gauss's Law is a powerful tool for solving problems involving electric fields and charges. It allows us to calculate the electric field at a point or over a closed surface, and can also be used to determine the charge distribution given a known electric field. It is a fundamental law in electromagnetism and is used extensively in many practical applications, such as designing electrical circuits and understanding the behavior of electromagnetic waves.

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