15c. Charges and flux

[math] \newcommand{\mathds}{\mathbb}[/math]

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Time for electricity. In analogy with the usual study of gravity, let us start with:

Definition

Given charges [math]q_1,\ldots,q_k\in\mathbb R[/math] located at positions [math]x_1,\ldots,x_k\in\mathbb R^3[/math], we define their electric field to be the vector function

[[math]] E(x)=K\sum_i\frac{q_i(x-x_i)}{||x-x_i||^3} [[/math]]

so that their force applied to a charge [math]Q\in\mathbb R[/math] positioned at [math]x\in\mathbb R^3[/math] is given by [math]F=QE[/math].

More generally, we will be interested in electric fields of various non-discrete configurations of charges, such as charged curves, surfaces and solid bodies. Indeed, things like wires or metal sheets or solid bodies coming in all sorts of shapes, tailored for their purpose, play a key role, so this extension is essential. So, let us go ahead with:

Definition

The electric field of a charge configuration [math]L\subset\mathbb R^3[/math], with charge density function [math]\rho:L\to\mathbb R[/math], is the vector function

[[math]] E(x)=K\int_L\frac{\rho(z)(x-z)}{||x-z||^3}\,dz [[/math]]

so that the force of [math]L[/math] applied to a charge [math]Q[/math] positioned at [math]x[/math] is given by [math]F=QE[/math].

With the above definitions in hand, it is most convenient now to forget about the charges, and focus on the study of the corresponding electric fields [math]E[/math].

These fields are by definition vector functions [math]E:\mathbb R^3\to\mathbb R^3[/math], with the convention that they take [math]\pm\infty[/math] values at the places where the charges are located, and intuitively, are best represented by their field lines, which are constructed as follows:

Definition

The field lines of an electric field [math]E:\mathbb R^3\to\mathbb R^3[/math] are the oriented curves [math]\gamma\subset\mathbb R^3[/math] pointing at every point [math]x\in\mathbb R^3[/math] at the direction of the field, [math]E(x)\in\mathbb R^3[/math].

As a basic example here, for one charge the field lines are the half-lines emanating from its position, oriented according to the sign of the charge:

[[math]] \begin{matrix} \nwarrow&\uparrow&\nearrow\\ \leftarrow&\oplus&\rightarrow\\ \swarrow&\downarrow&\searrow \end{matrix}\qquad\qquad\qquad\qquad \begin{matrix} \searrow&\downarrow&\swarrow\\ \rightarrow&\ominus&\leftarrow\\ \nearrow&\uparrow&\nwarrow \end{matrix} [[/math]]

For two charges now, if these are of opposite signs, [math]+[/math] and [math]-[/math], you get a picture that you are very familiar with, namely that of the field lines of a bar magnet:

[[math]] \begin{matrix} \nearrow&\ \ \nearrow&\rightarrow&\rightarrow&\rightarrow&\rightarrow&\searrow\ \ \ &\!\!\!\searrow\\ \nwarrow&\!\!\!\uparrow&\nearrow&\rightarrow&\rightarrow&\searrow&\downarrow&\!\!\!\swarrow\\ \leftarrow&\!\!\!\oplus&\rightarrow&\rightarrow&\rightarrow&\rightarrow&\ominus&\!\!\!\leftarrow\\ \swarrow&\!\!\!\downarrow&\searrow&\rightarrow&\rightarrow&\nearrow&\uparrow&\!\!\!\nwarrow\\ \searrow&\ \ \searrow&\rightarrow&\rightarrow&\rightarrow&\rightarrow&\nearrow\ \ \ &\!\!\!\nearrow \end{matrix} [[/math]]

If the charges are [math]+,+[/math] or [math]-,-[/math], you get something of similar type, but repulsive this time, with the field lines emanating from the charges being no longer shared:

[[math]] \begin{matrix} \leftarrow\ &\!\!\!\!\!\nwarrow&\nwarrow&&&\nearrow&\ \ \ \nearrow&\rightarrow\\ &\uparrow&\nearrow&&&\nwarrow&\uparrow&\\ \leftarrow&\oplus&\ &\ &\ \ \ \ \ \ \ &\ &\oplus&\rightarrow\\ &\downarrow&\searrow&&&\swarrow&\downarrow&\\ \leftarrow\ &\!\!\!\!\!\swarrow&\swarrow&&&\searrow&\ \ \ \searrow&\rightarrow \end{matrix} [[/math]]

These pictures, and notably the last one, with [math]+,+[/math] charges, are quite interesting, because the repulsion situation does not appear in the context of gravity. Thus, we can only expect our geometry here to be far more complicated than that of gravity.

The field lines, as constructed in Definition 15.25, obviously do not encapsulate the whole information about the field, with the direction of each vector [math]E(x)\in\mathbb R^3[/math] being there, but with the magnitude [math]||E(x)||\geq0[/math] of this vector missing. However, say when drawing, when picking up uniformly radially spaced field lines around each charge, and with the number of these lines proportional to the magnitude of the charge, and then completing the picture, the density of the field lines around each point [math]x\in\mathbb R[/math] will give you then the magnitude [math]||E(x)||\geq0[/math] of the field there, up to a scalar.

Let us summarize these observations as follows:

Proposition

Given an electric field [math]E:\mathbb R^3\to\mathbb R^3[/math], the knowledge of its field lines is the same as the knowledge of the composition

[[math]] nE:\mathbb R^3\to\mathbb R^3\to S [[/math]]

where [math]S\subset\mathbb R^3[/math] is the unit sphere, and [math]n:\mathbb R^3\to S[/math] is the rescaling map, namely:

[[math]] n(x)=\frac{x}{||x||} [[/math]]

However, in practice, when the field lines are accurately drawn, the density of the field lines gives you the magnitude of the field, up to a scalar.

Show Proof

We have two assertions here, the idea being as follows:

(1) The first assertion is clear from definitions, with of course our usual convention that the electric field and its problematics take place outside the locations of the charges, which makes everything in the statement to be indeed well-defined.

(2) Regarding now the last assertion, which is of course a bit informal, this follows from the above discussion. It is possible to be a bit more mathematical here, with a definition, formula and everything, but we will not need this, in what follows.

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Let us introduce now a key definition, as follows:

Definition

The flux of an electric field [math]E:\mathbb R^3\to\mathbb R^3[/math] through a surface [math]S\subset\mathbb R^3[/math], assumed to be oriented, is the quantity

[[math]] \Phi_E(S)=\int_S \lt E(x),n(x) \gt dx [[/math]]

with [math]n(x)[/math] being unit vectors orthogonal to [math]S[/math], following the orientation of [math]S[/math]. Intuitively, the flux measures the signed number of field lines crossing [math]S[/math].

Here by orientation of [math]S[/math] we mean precisely the choice of unit vectors [math]n(x)[/math] as above, orthogonal to [math]S[/math], which must vary continuously with [math]x[/math]. For instance a sphere has two possible orientations, one with all these vectors [math]n(x)[/math] pointing inside, and one with all these vectors [math]n(x)[/math] pointing outside. More generally, any surface has locally two possible orientations, so if it is connected, it has two possible orientations. In what follows the convention is that the closed surfaces are oriented with each [math]n(x)[/math] pointing outside.

As a first illustration, let us do a basic computation, as follows:

Proposition

For a point charge [math]q\in\mathbb R[/math] at the center of a sphere [math]S[/math],

[[math]] \Phi_E(S)=\frac{q}{\varepsilon_0} [[/math]]

where the constant is [math]\varepsilon_0=1/(4\pi K)[/math], independently of the radius of [math]S[/math].

Show Proof

Assuming that [math]S[/math] has radius [math]r[/math], we have the following computation:

[[math]] \begin{eqnarray*} \Phi_E(S) &=&\int_S \lt E(x),n(x) \gt dx\\ &=&\int_S\left \lt \frac{Kqx}{r^3},\frac{x}{r}\right \gt dx\\ &=&\int_S\frac{Kq}{r^2}\,dx\\ &=&\frac{Kq}{r^2}\times 4\pi r^2\\ &=&4\pi Kq \end{eqnarray*} [[/math]]

Thus with [math]\varepsilon_0=1/(4\pi K)[/math] as above, we obtain the result.

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More generally now, we have the following result:

Theorem

The flux of a field [math]E[/math] through a sphere [math]S[/math] is given by

[[math]] \Phi_E(S)=\frac{Q_{enc}}{\varepsilon_0} [[/math]]

where [math]Q_{enc}[/math] is the total charge enclosed by [math]S[/math], and [math]\varepsilon_0=1/(4\pi K)[/math].

Show Proof

This can be done in several steps, as follows:

(1) Before jumping into computations, let us do some manipulations. First, by discretizing the problem, we can assume that we are dealing with a system of point charges. Moreover, by additivity, we can assume that we are dealing with a single charge. And if we denote by [math]q\in\mathbb R[/math] this charge, located at [math]v\in\mathbb R^3[/math], we want to prove that we have the following formula, where [math]B\subset\mathbb R^3[/math] denotes the ball enclosed by [math]S[/math]:

[[math]] \Phi_E(S)=\frac{q}{\varepsilon_0}\,\delta_{v\in B} [[/math]]

(2) By linearity we can assume that we are dealing with the unit sphere [math]S[/math]. Moreover, by rotating we can assume that our charge [math]q[/math] lies on the [math]Ox[/math] axis, that is, that we have [math]v=(r,0,0)[/math] with [math]r\geq0[/math], [math]r\neq1[/math]. The formula that we want to prove becomes:

[[math]] \Phi_E(S)=\frac{q}{\varepsilon_0}\,\delta_{r \lt 1} [[/math]]

(3) Let us start now the computation. With [math]u=(x,y,z)[/math], we have:

[[math]] \begin{eqnarray*} \Phi_E(S) &=&\int_S \lt E(u),u \gt du\\ &=&\int_S\left \lt \frac{Kq(u-v)}{||u-v||^3},u\right \gt du\\ &=&Kq\int_S\frac{ \lt u-v,u \gt }{||u-v||^3}\,du\\ &=&Kq\int_S\frac{1- \lt v,u \gt }{||u-v||^3}\,du\\ &=&Kq\int_S\frac{1-rx}{(1-2xr+r^2)^{3/2}}\,du \end{eqnarray*} [[/math]]

(4) In order to compute the above integral, we will use spherical coordinates for the unit sphere [math]S[/math], which are as follows, with [math]s\in[0,\pi][/math] and [math]t\in[0,2\pi][/math]:

[[math]] \begin{cases} x\!\!\!&=\ \cos s\\ y\!\!\!&=\ \sin s\cos t\\ z\!\!\!&=\ \sin s\sin t \end{cases} [[/math]]

We recall that the corresponding Jacobian, computed before, is given by:

[[math]] J=\sin s [[/math]]

(5) With the above change of coordinates, our integral from (3) becomes:

[[math]] \begin{eqnarray*} \Phi_E(S) &=&Kq\int_S\frac{1-rx}{(1-2xr+r^2)^{3/2}}\,du\\ &=&Kq\int_0^{2\pi}\int_0^\pi\frac{1-r\cos s}{(1-2r\cos s+r^2)^{3/2}}\cdot\sin s\,ds\,dt\\ &=&2\pi Kq\int_0^\pi\frac{(1-r\cos s)\sin s}{(1-2r\cos s+r^2)^{3/2}}\,ds\\ &=&\frac{q}{2\varepsilon_0}\int_0^\pi\frac{(1-r\cos s)\sin s}{(1-2r\cos s+r^2)^{3/2}}\,ds \end{eqnarray*} [[/math]]

(6) The point now is that the integral on the right can be computed with the change of variables [math]x=\cos s[/math]. Indeed, we have [math]dx=-\sin s\,ds[/math], and we obtain:

[[math]] \begin{eqnarray*} \int_0^\pi\frac{(1-r\cos s)\sin s}{(1-2r\cos s+r^2)^{3/2}}\,ds &=&\int_{-1}^1\frac{1-rx}{(1-2rx+r^2)^{3/2}}\,dx\\ &=&\left[\frac{x-r}{\sqrt{1-2rx+r^2}}\right]_{-1}^1\\ &=&\frac{1-r}{\sqrt{1-2r+r^2}}-\frac{-1-r}{\sqrt{1+2r+r^2}}\\ &=&\frac{1-r}{|1-r|}+1\\ &=&2\delta_{r \lt 1} \end{eqnarray*} [[/math]]

Thus, we are led to the formula in the statement.

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More generally now, we have the following key result, due to Gauss:

Theorem (Gauss law)

The flux of a field [math]E[/math] through a surface [math]S[/math] is given by

[[math]] \Phi_E(S)=\frac{Q_{enc}}{\varepsilon_0} [[/math]]

where [math]Q_{enc}[/math] is the total charge enclosed by [math]S[/math], and [math]\varepsilon_0=1/(4\pi K)[/math].

Show Proof

This basically follows from Theorem 15.29, or even from Proposition 15.28, by adding to the results there a number of new ingredients, as follows:

(1) Our first claim is that given a closed surface [math]S[/math], with no charges inside, the flux through it of any choice of external charges vanishes:

[[math]] \Phi_E(S)=0 [[/math]]

This claim is indeed supported by the intuitive interpretation of the flux, as corresponding to the signed number of field lines crossing [math]S[/math]. Indeed, any field line entering as [math]+[/math] must exit somewhere as [math]-[/math], and vice versa, so when summing we get [math]0[/math].

(2) In practice now, in order to prove this rigorously, there are several ways. A standard argument, which is quite elementary, is the one used by Feynman in , based on the fact that, due to [math]F\sim 1/d^2[/math], local deformations of [math]S[/math] will leave invariant the flux, and so in the end we are left with a rotationally invariant surface, where the result is clear.

(3) The point now is that, with this and Proposition 15.28 in hand, we can finish by using a standard math trick. Let us assume indeed, by discretizing, that our system of charges is discrete, consisting of enclosed charges [math]q_1,\ldots,q_k\in\mathbb R[/math], and an exterior total charge [math]Q_{ext}[/math]. We can surround each of [math]q_1,\ldots,q_k[/math] by small disjoint spheres [math]U_1,\ldots,U_k[/math], chosen such that their interiors do not touch [math]S[/math], and we have:

[[math]] \begin{eqnarray*} \Phi_E(S) &=&\Phi_E(S-\cup U_i)+\Phi_E(\cup U_i)\\ &=&0+\Phi_E(\cup U_i)\\ &=&\sum_i\Phi_E(U_i)\\ &=&\sum_i\frac{q_i}{\varepsilon_0}\\ &=&\frac{Q_{enc}}{\varepsilon_0} \end{eqnarray*} [[/math]]

(4) To be more precise, in the above the union [math]\cup U_i[/math] is a usual disjoint union, and the flux is of course additive over components. As for the difference [math]S-\cup U_i[/math], this is by definition the disjoint union of [math]S[/math] with the disjoint union [math]\cup(-U_i)[/math], with each [math]-U_i[/math] standing for [math]U_i[/math] with orientation reversed, and since this difference has no enclosed charges, the flux through it vanishes by (2). Finally, the end makes use of Proposition 15.28.

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So long for the Gauss law. We will be back to it in a moment, with a new, better proof, by using some advanced 3D calculus, that we will have to learn first.

General references

Banica, Teo (2024). "Calculus and applications". arXiv:2401.00911 [math.CO].