Maxwell's equations

**Maxwell's equations** are a set of four partial differential equations that describe the electric and magnetic fields in a region of space and how they interact. The four Maxwells' equations are

∬

∂
Ω

E
→

⋅

d

S
→

=

1

ε
0

∭
Ω

ρ

d

V

{\displaystyle \iint\_{\partial \Omega} \vec{E}\cdot\mathrm{d}\vec{S} = \frac{1}{\varepsilon\_0} \iiint\_\Omega \rho \,\mathrm{d}V}

∇
⋅

E
→

=

ρ

ε
0

{\displaystyle \nabla \cdot \vec{E} = \frac{\rho}{\varepsilon\_0}}
 Gauss' law for magnetism

∬

∂
Ω

B
→

⋅

d

S
→

=
0

{\displaystyle \iint\_{\partial \Omega} \vec{B}\cdot\mathrm{d}\vec{S} = 0}

∇
⋅

B
→

=
0

{\displaystyle \nabla \cdot \vec{B} = 0}
 Ampere's law

∮

∂
Σ

B
→

⋅

d

ℓ
→

=

μ
0

∬

Σ

J
→

⋅

d

S
→

μ
0

ε
0

d

d

t

∬

Σ

E
→

⋅

d

S
→

{\displaystyle \oint\_{\partial \Sigma} \vec{B} \cdot \mathrm{d}\vec{\ell} = \mu\_0 \iint\_{\Sigma} \vec{J} \cdot \mathrm{d}\vec{S} + \mu\_0 \varepsilon\_0 \frac{\mathrm{d}}{\mathrm{d}t} \iint\_{\Sigma} \vec{E} \cdot \mathrm{d}\vec{S} }

∇
×

B
→

=

μ
0

(

J
→

ε
0

∂

E
→

∂
t

)

{\displaystyle \nabla \times \vec{B} = \mu\_0\left(\vec{J} + \varepsilon\_0 \frac{\partial \vec{E}} {\partial t} \right) }
 Faraday's law

∮

∂
Σ

E
→

⋅

d

ℓ
→

=
−

d

d

t

∬

Σ

B
→

⋅

d

S
→

{\displaystyle \oint\_{\partial \Sigma} \vec{E} \cdot \mathrm{d}\vec{\ell} = - \frac{\mathrm{d}}{\mathrm{d}t} \iint\_{\Sigma} \vec{B} \cdot \mathrm{d}\vec{S}}

∇
×

E
→

=
−

∂

B
→

∂
t

{\displaystyle \nabla \times \vec{E} = -\frac{\partial \vec{B}} {\partial t}}

Here,

* **E** and **B** are the electric and magnetic fields
* ε0 and μ0 are the electric and magnetic constants
* ρ and **J** are the charge and current densities
* Ω (with volume element dV) is a volume with boundary ∂Ω (with surface element d**S**)
* Σ (with surface element d**S**) is a surface with boundary ∂Σ (with line element d**ℓ**)
* ∇× and ∇· are the curl and divergence operators

## Electrostatics

In the electrostatic approximation, the charge density is fixed and there are no currents. This means that the time derivative of the electric field is always zero, and the magnetic field is always zero. In electrostatics, all information about the system can be derived using only Gauss' law.

Given a charge density ρ(**r**), Gauss' law states that the flux of the electric field over a surface is proportional to the charge enclosed by that surface; charges outside of the surface do not influence the net flux.

### Point Charge

The simplest example is the electric field of a point charge at the origin with charge q. This can be found using Gauss' law:

∬

∂
Ω

E
→

⋅

d

S
→

=

1

ε
0

∭
Ω

ρ

d

V

{\displaystyle \iint\_{\partial \Omega} \vec{E}\cdot\mathrm{d}\vec{S} = \frac{1}{\varepsilon\_0} \iiint\_\Omega \rho \,\mathrm{d}V}

A spherical Gaussian surface centered on this point charge has the same flux at every point of its surface, so the surface integral of the flux is equal to the area of the sphere multiplied by the magnitude of the electric field. The charge enclosed is equal to q, because the point charge has charge q. Hence,

E
# Σ

q

ε
0

{\displaystyle E \Sigma = \frac{q}{\varepsilon\_0}}

The area of the spherical surface Σ is equal to 4πr2, so

4
π

r
2

# E

q

ε
0

{\displaystyle 4\pi r^2 E = \frac{q}{\varepsilon\_0}}

This gives Coulomb's law,

E
→

=

q

4
π

ε
0

r
2

r
^

{\displaystyle \vec{E} = \frac{q}{4\pi \varepsilon\_0 r^2} \hat{r}}

Because of the principle of superposition, an arbitrary arrangement of point charges will have each individual point charge follow Coulomb's law, and hence a distribution of n charges will have an electric field given by

E
→

(

r
→

# i
1

n

q
i

4
π

ε
0

r
→

i
′

−

r
→

2

(

r
→

i
′

−

r
→

)

r
→

i
′

−

r
→

This also works for a continuous distribution of charges. For a charge density distribution ρ(r), Coulomb's law can be made continuous by changing the sum into an integral and changing the charge into a charge density, giving

E
→

(

r
→

# )

∭
Ω

ρ
(

r
→

′

)

4
π

ε
0

r
→

′

−

r
→

2

(

r
→

′

−

r
→

)

r
→

′

−

r
→

d
Ω

### Line Charge

Another example is a line charge, with linear charge density λ. Using Gauss' law,

∬

∂
Ω

E
→

⋅

d

S
→

=

1

ε
0

∭
Ω

ρ

d

V

{\displaystyle \iint\_{\partial \Omega} \vec{E}\cdot\mathrm{d}\vec{S} = \frac{1}{\varepsilon\_0} \iiint\_\Omega \rho \,\mathrm{d}V}

The surface chosen will be a cylinder of radius r and length z enclosing the line charge. The flux across both of the end caps of the cylinder is 0 (because the electric field is perpendicular to the wire, and hence parallel with the ends of the cylinder), and the flux across the curved surface of the cylinder is equal at all points (due to the cylindrical symmetry of the situation) and is equal to the magnitude of the electric field multiplied by the area of the cylinder. The charge enclosed by this cylinder is λz.

E
# Σ

λ
z

ε
0

{\displaystyle E \Sigma = \frac{\lambda z}{\varepsilon\_0}}

The area of the curved surface of a cylinder with radius r is 2πrz, so

E
2
π
r
# z

λ
z

ε
0

{\displaystyle E 2 \pi r z = \frac{\lambda z}{\varepsilon\_0}}

And hence the electric field from a line charge is

E
→

=

λ

2
π

ε
0

r

r
^

{\displaystyle \vec{E} = \frac{\lambda}{2 \pi \varepsilon\_0 r} \hat{r}}

### Plane Charge

For a plane charge, the charge is evenly distributed over an infinite plane with surface charge density σ. Using Gauss' law,

∬

∂
Ω

E
→

⋅

d

S
→

=

1

ε
0

∭
Ω

ρ

d

V

{\displaystyle \iint\_{\partial \Omega} \vec{E}\cdot\mathrm{d}\vec{S} = \frac{1}{\varepsilon\_0} \iiint\_\Omega \rho \,\mathrm{d}V}

The surface chosen will be a prism, with area Σ, enclosing a section of the plane charge. The charge enclosed by this is Σσ, and the total flux of the electric field is even is equal to the area of the two ends of the prism multiplied by the magnitude of the field (the sides of the prism, being parallel to the electric field, have no flux through them).

2
E
# Σ

Σ
σ

ε
0

{\displaystyle 2 E \Sigma = \frac{\Sigma\sigma}{\varepsilon\_0}}

And hence the electric field is

E
→

=

σ

2

ε
0

E
^

{\displaystyle \vec{E} = \frac{\sigma}{2 \varepsilon\_0} \hat{E}}

The direction of this is worth commenting on; the electric field points upwards above the plane of charge, and downwards below the plane of charge.

### Spherical Shell

The electric field surrounding a spherical shell of charge can also be found using Gauss' law. The shell has a total charge q (equivalent to a surface charge density of

# σ

q

4
π

R
2

{\displaystyle \sigma = \frac{q}{4 \pi R^2}}

), and a radius R.

Taking a Gaussian sphere with a radius r, and having r > R, the flux of the electric field is the same over the entire Gaussian surface and the charge enclosed is equal to q, so

4
π

r
2

# E

q

ε
0

{\displaystyle 4\pi r^2 E = \frac{q}{\varepsilon\_0}}

And hence the electric field outside of the spherical shell is

E
→

=

q

4
π

ε
0

r
2

r
^

{\displaystyle \vec{E} = \frac{q}{4\pi \varepsilon\_0 r^2} \hat{r}}

 (for r > R)

When the radius of the Gaussian sphere is less than R (r < R), the charge enclosed is zero, so

E
→

=

0
→

{\displaystyle \vec{E} = \vec{0}}

 (for r < R)

This result is called the shell theorem; outside of a spherical shell, it appears electrically like all of the charge is concentrated at a single point in the center, and inside of a spherical shell, it has no electric field. This occurs regardless of the position at which the electric field is being measured.

### Uniform Charged Sphere

A uniform charged sphere is the simplest result that has a full, three-dimensional charge distribution through space. For a sphere with a uniform volume charge density ρ and a radius R, the electric field at a given radius can be found using Gauss' law.

Outside of the sphere, with the Gaussian surface a sphere with radius r > R, the charge enclosed is

4
3

ρ

R
3

{\displaystyle \frac{4}{3} \rho R^3}

 and the total flux across the Gaussian surface is the magnitude of the electric field multiplied by the area.

E
4
π

r
2

=

4
3

ρ

R
3

ε
0

{\displaystyle E 4 \pi r^2 = \frac{\frac{4}{3} \rho R^3}{\varepsilon\_0}}

And hence the electric field outside the sphere is

E
→

=

ρ

R
3

3
π

ε
0

r
2

r
^

{\displaystyle \vec{E} = \frac{\rho R^3}{3 \pi \varepsilon\_0 r^2} \hat{r}}

 (for r > R)

This is identical to the charge all being concentrated as a single point in the center of the sphere, as stated by the shell theorem.

When the Gaussian surface is inside of the sphere (r < R), only some of the sphere's charge is enclosed, equal to

4
3

ρ

r
3

{\displaystyle \frac{4}{3} \rho r^3}

.

E
4
π

r
2

=

4
3

ρ

r
3

ε
0

{\displaystyle E 4 \pi r^2 = \frac{\frac{4}{3} \rho r^3}{\varepsilon\_0}}

Hence, the electric field inside of the uniformly charged sphere is

E
→

=

ρ
r

3
π

ε
0

r
^

{\displaystyle \vec{E} = \frac{\rho r}{3 \pi \varepsilon\_0} \hat{r}}

 (for r < R)

This increases proportionally to the distance from the center of the sphere. At r = R, these two expressions for the electric field are the same, meaning that unlike for the other examples the electric field is continuous everywhere.

## Magnetostatics

In the magnetostatic approximation, the current density is fixed and there are no charges. This means that the time derivative of the magnetic field is always zero, and the electric field is always zero. In magnetostatics, all information about the system can be derived using only Ampere's law.

Given a current density **J**(**r**), Ampere's law states that the line integral of the magnetic field over a loop is proportional to the current flowing through the loop, plus an additional factor called the displacement current. The displacement current represents the magnetic field induced by the current moving charges from one location to another, and is therefore not applicable in magnetostatics.

Because of the displacement current term, however, the magnetostatic approximation does not work for all arrangements of currents. For example, a wire of finite length with a constant current flowing through it can not be solved with magnetostatics because the flow of current results in positive and negative charges building up at either end of the wire.

### Infinite Wire

For an infinite (or very long) wire with a current I flowing through it, the problem of buildup of charge on either end is avoided. This means that the magnetic field can be found using the magnetostatic version of Ampere's law without involving the displacement current term.

∮

∂
Σ

B
→

⋅

d

ℓ
→

=

μ
0

∬

Σ

J
→

⋅

d

S
→

{\displaystyle \oint\_{\partial \Sigma} \vec{B} \cdot \mathrm{d}\vec{\ell} = \mu\_0 \iint\_{\Sigma} \vec{J} \cdot \mathrm{d}\vec{S} }

For a circular Amperian loop of radius r surrounding the wire, the circulation of the magnetic field is constant over the loop so the line integral is equal to the magnitude of the magnetic field multiplied by the length of the loop, and the current flowing through the loop is equal to I. This means that

B
# ℓ

μ
0

I

{\displaystyle B \ell = \mu\_0 I }

The length of the loop ℓ is 2πr, so

2
π
r
# B

μ
0

I

{\displaystyle 2 \pi r B = \mu\_0 I }

And hence the magnetic field surround the wire (with the direction found using the right-hand rule) is

B
→

=

μ
0

I

2
π
r

θ
^

{\displaystyle \vec{B} = \frac{\mu\_0 I}{2 \pi r} \hat{\theta}}