notes/docs/physics/electromagnetism/electrostatics.md

Electrostatics

A Notational remark, let the vector from origin to source point be given by \mathbf{r}' and the vector from origin to field point by \mathbf{r}. The vector from source to field point is given by \bm{\mathfrak{r}} = \mathbf{r} - \mathbf{r}'.

In the electrostatic regime there is no current, no magnetic fields and time dependence is not taken into account

Electrostatics in vacuum

Electrostatics builds entirely on the following axiom:

Axiom 1: In the electrostatic regime the electric field \mathbf{E}: \mathbf{r} \mapsto \mathbf{E}(\mathbf{r}) is described by

\begin{align*} \nabla \cdot \mathbf{E} &= \frac{\rho}{\epsilon_0},\ \nabla \times \mathbf{E} &= \mathbf{0}, \end{align*}

with \rho: \mathbf{r} \mapsto \rho(\mathbf{r}) the volume charge density and \epsilon_0 the vacuum permittivity.

The following definition connects the electric field \mathbf{E}: \mathbf{r} \mapsto \mathbf{E}(\mathbf{r}) to the Newtonian formalism of mechanics:

Definition 1: The electric force \mathbf{F} on a point charge q at a field point \mathbf{r} is defined as

\mathbf{F}(\mathbf{r}) = q \mathbf{E}(\mathbf{r}),

for all \mathbf{E}: \mathbf{r} \mapsto \mathbf{E}(\mathbf{r}) the electric field.

From Axiom 1 we may proof that the following theorem holds (Coulomb law):

Theorem 1: The electric field \mathbf{E}: \mathbf{r} \mapsto \mathbf{E}(\mathbf{r}) may be described as

\mathbf{E}(\mathbf{r}) = \begin{cases} \frac{1}{4\pi\epsilon_0} \cdot \frac{q}{\mathfrak{r}^2} \mathbf{e}\mathfrak{r} \qquad &\mathrm{0D},\ \frac{1}{4\pi\epsilon_0} \int \frac{\lambda}{\mathfrak{r}^2} \mathbf{e}\mathfrak{r} dl' \qquad &\mathrm{1D},\ \frac{1}{4\pi\epsilon_0} \iint \frac{\sigma}{\mathfrak{r}^2} \mathbf{e}\mathfrak{r} da' \qquad &\mathrm{2D},\ \frac{1}{4\pi\epsilon_0} \iiint \frac{\rho}{\mathfrak{r}^2} \mathbf{e}\mathfrak{r} d\tau' \qquad &\mathrm{3D}, \end{cases}

with \lambda and \sigma the line and surface charge density.

??? note "Proof:"

Follows from divergence theorem.

From the divergence theorem it follows as well that

\oint_S \mathbf{E} \cdot d\mathbf{a} = \int_V \nabla \cdot \mathbf{E} d\tau = \int_V \frac{\rho}{\epsilon_0} d\tau = \frac{Q}{\epsilon_0},

with Q the enclosed charge. This result is called Gauß' law.

Since the electric field \mathbf{E} is irrotational in the electrostatic regime we may define the following:

Definition 2: The electric potential V: \mathbf{r} \mapsto V(\mathbf{r}) of the electric field \mathbf{E} is defined as

\mathbf{E} = - \nabla V,

for all \mathbf{r}.

One may simply proof that from Definition 2 it follows that:

Theorem 2: The electric potential V: \mathbf{r} \mapsto V(\mathbf{r}) may be described as

\Big[V(\mathbf{r})\Big]_L = - \int_L \mathbf{E} \cdot d\mathbf{l},

for all \mathbf{r} and a chosen reference.

??? note "Proof:"

Follows from gradient theorem.

Now from (Axiom 1) \nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0} it follows that \nabla^2 V = - \frac{\rho}{\epsilon_0}.

Similarly to Theorem 1 we may state the following about the electric potential V:

Theorem 3: The electric potential V: \mathbf{r} \mapsto V(\mathbf{r}) may be described as

V(\mathbf{r}) = \begin{cases} \frac{1}{4\pi\epsilon_0} \cdot \frac{q}{\mathfrak{r}} \qquad &\mathrm{0D},\ \frac{1}{4\pi\epsilon_0} \int \frac{\lambda}{\mathfrak{r}} dl' \qquad &\mathrm{1D},\ \frac{1}{4\pi\epsilon_0} \iint \frac{\sigma}{\mathfrak{r}} da' \qquad &\mathrm{2D},\ \frac{1}{4\pi\epsilon_0} \iiint \frac{\rho}{\mathfrak{r}} d\tau' \qquad &\mathrm{3D}.\end{cases}

??? note "Proof:"

Follows from divergence theorem.

Electric work

The work W required to move a charge q from \mathscr{A} to \mathscr{B} in an electric field \mathbf{E} is given by

\begin{align*} W &= -\int_\mathscr{A}^\mathscr{B} \mathbf{F} \cdot d\mathbf{l},\ &= -q \int_\mathscr{A}^\mathscr{B} \mathbf{E} \cdot d\mathbf{l},\ &= q \Big[V\Big]_\mathscr{A}^\mathscr{B}, \end{align*}

with V the electric potential.

As we set the reference point at infinity, then qV could be interpreted as the mechanic potential energy.

Theorem 4: The work W required to construct an assembly of N \in \mathbb{N} charges should adhere to

W = \frac{1}{2} \sum_{i=1}^N q_i V(\mathbf{r}_i),

with V the electric potential for a charge q_i with position \mathbf{r}_i.

??? note "Proof:"

We have 

$$
    \begin{align*}
        W_i &= q_i \sum_j V_j(\mathbf{r}_i),\\ 
            &= q_i \sum_j \frac{1}{4\pi\epsilon_0} \frac{q_j}{\mathfrak{r}_{ij}},
    \end{align*}
$$

such that the total work $W$ is given by

$$
    \begin{align*}
        W &= \sum_{i=1}^N W_i,\\
          &= \frac{1}{2} \sum_{i=1}^N \sum_{j \neq i} \frac{1}{4\pi\epsilon_0} \frac{q_j}{\mathfrak{r}_{ij}},\\
          &=\frac{1}{2} \sum_{i=1}^N q_i V(\mathbf{r}_i).
    \end{align*}
$$

Is the energy stored in the system.

It follows now that for a volume charge density \rho the work W needed to construct the system is given by

W = \frac{1}{2} \int_\mathscr{V} \rho V d\tau,

and the integrals for line and surface charge densities (\lambda,\sigma) are of course similar.

Theorem 5: The work W required to construct the system can be expressed in terms of the electric field \mathbf{E} as

W = \frac{\epsilon_0}{2} \int_{\mathbb{R}^3} |\mathbf{E}|^2 d\tau.

??? note "Proof:"

Rewrite the work in terms of the electric field $\mathbf{E}$:

$$
    \begin{align*}
        W &= \frac{\epsilon_0}{2} \int_\mathscr{V} (\nabla \cdot \mathbf{E}) V d\tau,\\
          &= \frac{\epsilon_0}{2} \Big(-\int_\mathscr{V} \mathbf{E} \cdot (\nabla V) d\tau + \oint_{\partial \mathscr{V}} V \mathbf{E} \cdot d\mathbf{a}\Big),\\
          &= \frac{\epsilon_0}{2} \Big(\int_\mathscr{V} \|\mathbf{E}\|^2 d\tau + \oint_\mathscr{\partial \mathscr{V}} V \mathbf{E} \cdot d\mathbf{a}\Big).
    \end{align*}
$$

If we now set $\mathscr{V} = \mathbb{R}^3$ then the integral over $\partial \mathscr{V}$ goes to zero and we are left with:

$$
    W = \frac{\epsilon_0}{2} \int_{\mathbb{R}^3} \|\mathbf{E}\|^2 d\tau.
$$

That is performing the integral over all of space. Which is mathematically rather nice, but imposes that the construction of point charges requires an infinite amount of energy. A result of the introduction of infinities.

As may be observed, this result does not obey the superposition principle.

Conductors and capacitors

In the electrostatic regime the following properties of conductors are valid:

1. \mathbf{0} = \mathbf{J} = \sigma \mathbf{E} \implies \mathbf{E} = \mathbf{0} inside a conductor.
2. \mathbf{0} = \nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0} \implies \rho = 0 inside a conductor.

Which implies that any net charge resides on the surface, that a conductor is an equipotential and that the electric field is normal to the surface of the conductor.

Since the electric field \mathbf{E} is proportional to the volume charge density \rho or charge Q, so also is V. Which enables us to define a constant of proportionality C = \frac{Q}{V}, the capacitance.

The work required to charge a capacitor can then be expressed as

W = \int_0^Q \frac{q}{C} dq = \frac{1}{2} \frac{Q^2}{C} = \frac{1}{2} C V^2.

Electric potential

The primary task of electrostatics is to find the electric field \mathbf{E} of a given stationary charge distribution \rho. In principle, this purpose is accomplished by Coulombs's law as by the result of Theorem 1. But these integrals are generally difficult to solve for any but the simplest charge distributions. In some case symmetries in the charge distribution allow Gauß' law to be used. But more often than not we require to determine the potential V with the result in Theorem 3 or with Poisson's equation:

\nabla^2 V = -\frac{\rho}{\epsilon_0},

where in the region where \rho = 0 this reduces to Laplace's equation:

\nabla^2 V = 0,

which imposes the following result:

Theorem 6: For an electric potential V: \mathbf{r} \mapsto V(\mathbf{r}) that satisfies \nabla^2 V=0, we have that for a spherical surface S of radius R centered at \mathbf{r}:

V(\mathbf{r}) = \frac{1}{4\pi R^2} \oint_S V da,

the average of surrounding V is V(\mathbf{r}) for S(\mathbf{r},R).

As a consequence we have that:

Corollary 1: an electric potential V: \mathbf{r} \mapsto V(\mathbf{r}) that satisfies \nabla^2 V=0 has no local extrema.

This result brings us to the consideration of the boundary condition for which the following is true:

Theorem 7: An electric potential V: \mathbf{r} \mapsto V(\mathbf{r}) that satisfies \nabla^2 V=0 in some volume \mathscr{V} is unique if V is specified on the boundary surface \partial \mathscr{V}.

and consequentially:

Corollary 2: An electric potential V: \mathbf{r} \mapsto V(\mathbf{r}) that satisfies \nabla^2 V=-\frac{\rho}{\epsilon_0} in some volume \mathscr{V} is unique if \rho: \mathbf{r} \mapsto \rho(\mathbf{r}) is known \forall \mathbf{r} \in \mathscr{V} and V is specified on the boundary surface \partial \mathscr{V}.

The combination of Theorem 7 and Corollary 2 is known as the first uniqueness theorem.

The second uniqueness theorem is relevant when the boundary is defined by conductors:

Theorem 8: An electric potential V: \mathbf{r} \mapsto V(\mathbf{r}) that satisfies \nabla^2 V = -\frac{\rho}{\epsilon_0} in some volume \mathscr{V} that is surrounded by conductors is unique if \rho: \mathbf{r} \mapsto \rho(\mathbf{r}) is known \forall \mathbf{r} \in \mathscr{V} and the total charge on the conductors is known.

The method of images to solve for the electric potential V makes use of these uniqueness theorems. In essence this method uses a solveable image system that obeys the same boundary conditions, then by unicity the same solution should be obtained. Though, the electric work required to construct the system is not always the same.

Another method to solve for Poisson's or Laplace's equation is a seperation of variables often in combination with a Fourier series formulation.

Multipole expansion

Approximation is the de facto option if no exact solution of the electric potential exists. The multipole expansion enables the approximation of the electric potential at large distances. Its fundament is the following result:

Theorem 9: The electric potential V: \mathbf{r} \mapsto V(\mathbf{r}) of a charge distribution \rho: \mathbf{r} \mapsto \rho(\mathbf{r}) may be described as

V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \sum_{n=0}^\infty \frac{1}{r^{n+1}} \iiint (r')^n P_n(\cos \alpha) \rho(\mathbf{r}') d\tau',

with \cos \alpha = \frac{\langle \mathbf{r}, \mathbf{r}'\rangle}{r r'}.

??? note "Proof:"

We start with

$$
    V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \iiint \frac{\rho(\mathbf{r}')}{\mathfrak{r}} d\tau'.
$$

We may write $\mathfrak{r}$ as

$$
    \mathfrak{r}^2 = r^2 + (r')^2 - 2 r r' \cos \alpha,
$$

such that

$$
    \mathfrak{r} = r \sqrt{1+\epsilon},
$$

with 

$$
    \epsilon = \frac{r'}{r} \bigg(\frac{r'}{r} - 2 \cos \alpha\bigg).
$$

Let $\chi = \{\epsilon \ll 1\}$ then

$$
    \frac{1}{\mathfrak{r}} \overset{\chi}{=} \frac{1}{r} \sum_{n=0}^\infty \bigg(\frac{r'}{r}\bigg)^n P_n (\cos \alpha),
$$

which obtains

$$
    V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \sum_{n=0}^\infty \frac{1}{r^{n+1}} \iiint (r')^n P_n(\cos \alpha) \rho(\mathbf{r}') d\tau'.
$$

The monopole (n=0) term of V is then given by

V_0(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \frac{1}{r} \iiint \rho(\mathbf{r}') d\tau',

is zero if the net charge is zero.

The dipole (n=1) term of V is given by

\begin{align*} V_1(\mathbf{r}) &= \frac{1}{4\pi\epsilon_0} \cdot \frac{1}{r^2} \Bigg\langle \frac{1}{r} \mathbf{r}, \iiint \mathbf{r}' \rho(\mathbf{r}') d\tau' \Bigg\rangle,\ &= \frac{1}{4\pi\epsilon_0} \cdot \frac{1}{r^2} \langle \mathbf{e}_r, \mathbf{p} \rangle, \end{align*}

with \mathbf{p} = \iiint \mathbf{r}' \rho(\mathbf{r}') d\tau' the electric dipole moment.

For a collection of N \in \mathbb{N} point charges we have that the electric dipole moment \mathbf{p} is given by

\mathbf{p} = \sum_{i=1}^N q_i \mathbf{r}_i'.

Note that if the net charge of the charge distribution is zero then the electric dipole moment is independent of the choice of origin.

Electrostatics in matter

The electric dipole moment \mathbf{p} of an induced dipole due to an external field \mathbf{E} is given by

\mathbf{p} = \bm{\alpha} \lrcorner \mathbf{E},

with \bm{\alpha} \in \mathscr{T}^2_0 the polarisability.

Definition 3: The induced electric dipole moment \mathbf{p} in a medium may be expressed by the polarisation \mathbf{P} of the medium defined in terms of

\mathbf{p} = \int_\mathscr{V} \mathbf{P} d\tau,

with \mathscr{V} the volume of the medium.

The polarisation \mathbf{P} is in essence a sort of electric dipole moment per unit volume of the medium.

Theorem 10: The polarisation of the medium \mathbf{P} adheres to

\begin{align*} \nabla \cdot \mathbf{P} &= - \rho_b,\ \mathbf{P} \cdot \mathbf{e}_n &= \sigma_b, \end{align*}

with \rho_b: \mathbf{r} \mapsto \rho_b(\mathbf{r}) the bound volume charge density and \sigma_b: \mathbf{r} \mapsto \sigma_b(\mathbf{r}) the bound surface charge density.

??? note "Proof:"

The dipole term of the potential in terms of the polarisation $\mathbf{P}$ is given by

$$
    V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \int_\mathscr{V} \frac{\mathbf{P}(\mathbf{r}') \cdot \mathbf{e}_\mathfrak{r}}{\mathfrak{r}^2} d\tau'.
$$

Now observe that $\nabla' \frac{1}{\mathfrak{r}} = \frac{1}{\mathfrak{r}^2} \mathbf{e}_\mathfrak{r}$, obtains

$$
    V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \int_\mathscr{V} \mathbf{P}(\mathbf{r}') \cdot \nabla' \frac{1}{\mathfrak{r}} d\tau',
$$

and by integration by parts

$$
    V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \Bigg(\int_\mathscr{V} \nabla' \cdot \frac{1}{\mathfrak{r}} \mathbf{P}(\mathbf{r}') d\tau' - \int_\mathscr{V} \frac{1}{\mathfrak{r}} \nabla' \cdot \mathbf{P}(\mathbf{r}') d\tau'\Bigg),
$$

such that by the divergence theorem we obtain

$$
    V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \Bigg(\oint_{\partial \mathscr{V}} \frac{1}{\mathfrak{r}} \mathbf{P}(\mathbf{r}') \cdot d\mathbf{a}' - \int_\mathscr{V} \frac{1}{\mathfrak{r}} \nabla' \cdot \mathbf{P}(\mathbf{r}') d\tau' \Bigg).
$$

Setting $\rho_b = - \nabla \cdot \mathbf{P}$ and $\sigma_b = \mathbf{P} \cdot \mathbf{e}_n$ we obtain

$$
    V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \Bigg(\oint_{\partial \mathscr{V}} \frac{\sigma_b(\mathbf{r}')}{\mathfrak{r}} da' + \int_\mathscr{V} \frac{\rho_b(\mathbf{r}')}{\mathfrak{r}} d\tau'\Bigg),
$$

implying that the potential of a polarised object is the same as that produced by a volume charge density $\rho_b$ plus a a surface charge density $\sigma_b$. 

We may as well define the free volume charge density \rho_f in terms of the volume charge density \rho and the bound volume charge density \rho_b by notion of \rho_f = \rho - \rho_b.

Definition 4: Let the electric displacement \mathbf{D} of the medium be defined as

\mathbf{D} = \epsilon_0 \mathbf{E} + \mathbf{P},

with \mathbf{E} the electric field and \mathbf{P} the polarisation of the medium.

The usefullness of Definition 4 may become apparent with the following result:

Theorem 11: The electric displacment of the medium \mathbf{D} adheres to

\begin{align*} \nabla \cdot \mathbf{D} &= \rho_f,\ \nabla \times \mathbf{D} &= \nabla \times \mathbf{P}, \end{align*}

with \rho_f the free volume charge density of the medium and \mathbf{P} the polarisation of the medium.

??? note "Proof:"

From *Definition 4* we may write the electric field in terms of the electric displacement $\mathbf{D}$ and the polarisation $\mathbf{P}$ of the medium

$$
    \mathbf{E} = \frac{1}{\epsilon_0} \bigg(\mathbf{D} - \mathbf{P}\bigg),
$$

such that with *Axiom 1* we obtain

$$
    \epsilon_0 \nabla \cdot \mathbf{E} = \nabla \cdot \mathbf{D} + \rho_b \implies \nabla \cdot \mathbf{D} = \rho_f,
$$

and

$$
    \epsilon_0 \nabla \times \mathbf{E} = \nabla \times \mathbf{D} - \nabla \times \mathbf{P} \implies \nabla \times \mathbf{D} = \nabla \times \mathbf{P}.
$$

Recall that from the divergence theorem we have

\oint_{\partial \mathscr{V}} \mathbf{D} \cdot d\mathbf{a} = \int_\mathscr{V} \rho_f d\tau = Q_f,

with Q_f the enclosed free charge, if \nabla \times \mathbf{D} = \mathbf{0}.

Linear media

In linear media we have \mathbf{P} = \epsilon_0 \chi_e \mathbf{E} with \chi_e the electric susceptibility of the medium.

Furthermore \mathbf{D} = \epsilon_0 (1 + \chi_e) \mathbf{E} = \epsilon \mathbf{E} with \epsilon the permittivity and \epsilon_r = 1 + \chi_e = \frac{\epsilon}{\epsilon_0} the dielectric constant.