Physics equations/Integrals for Maxwell's equations

Wright State University Lake Campus/2019-9

Examples of integral as Riemann sum that can be understood without using calculus. On the left $\int _{0}^{L}y(x)dx\approx \sum y_{j}\Delta x={\tfrac {L^{2}}{2}}$is the area of a triangle. On the right$\int _{0}^{L}y(x)dx\approx \sum y_{j}\Delta x=LH$ is the area of a rectangle.

A great deal of insight into electrodynamics can be obtained using surface, volume and line integrals so simple they are often not even mentioned in a calculus course. These integrals assume an a priori knowledge of the field, except for an unknown constant. For our purposes, we shall denote these constants with the zero subscript $(E_{0},B_{0}),\,$ for electric and magnetic fields, respectively. If the same integral is useful for both electric and magnetic fields, we denote this constant as $F_{0}$.

In all cases, the field is constrained by symmetries that render the integrals so trivial that one can visualize the answer. It will be assumed that the reader is familiar with the circumference and area of a circle $(2\pi R,\pi R^{2}),\,$ respectively, as well as the surface area and volume of a sphere $(4\pi R^{2}),{\tfrac {4}{3}}\pi R^{3})\,.$ The student needs to fully understand how these pairs for the circle and sphere are related via #Onion_proof and #Surface_area

Line integrals

Circular line integral with axymuthal symmetry

Vector field in azimuthal direction depends only on r. The differential path element ${\color {blue}d{\vec {\ell }}}$ is shown in blue.

The line integral can be a circle situated on the inside of this toroid

This line integral is first encountered in Ampere's Law as applied to a long straight wire. The field lines are circles centered on an infinite line and depend only on the distance to the z axis. The path is any arc of a circle that is centered on the z axis, provided the circle is aligned so that its surface is perpendicular to the z axis:

$\int _{\theta _{1}{\text{ on arc}}}^{\theta _{2}{\text{ on arc}}}{\vec {F}}(r,z,\theta )\cdot d{\vec {\ell }}$

Define ${\hat {\theta }}$ to be a unit vector that points in the direction of increasing $\theta$. Both the field ${\vec {F}}$ and the differential $d{\vec {\ell }}$ are parallel to this unit vector:

• ${\vec {F}}(r,z,\theta )=F_{0}{\hat {\theta }}$ where $F_{0}=|{\vec {F}}(r,z,\theta )|$ is constant at $r=R$, where R is the radius of the circle defined by the arc's path.
• $d{\vec {\ell }}={\hat {\theta }}d\ell ={\hat {\theta }}Rd\theta$, since the $\theta$ (in radians) is defined so that the arclength and angle are defined so that $d\ell =Rd\theta$.

Since ${\hat {\theta }}\cdot {\hat {\theta }}=1$, we have:

${\vec {F}}(r,z,\theta )\cdot d{\vec {\ell }}=F_{0}\,R\;d\theta$

and,

$\int _{\theta _{1}{\text{ on arc}}}^{\theta _{2}{\text{ on arc}}}{\vec {F}}(r,z,\theta )\cdot d{\vec {\ell }}=\left(\theta _{2}-\theta _{1}\right)R\,F_{0}$

Integrating all the way around the circle leads to:

$\oint {\vec {F}}(r,z,\theta )\cdot d{\vec {\ell }}=2\pi R\,F_{0}$

provided the closed loop is a circle of radius R, centered on the z-axis.

It is important to appreciate that this result depends only on the value of ${\vec {F}}(x,y,z)$ on the circle of radius R. This is depicted in the top figure to the right with a circle at the center. Conditions inside this circle are important, but only because electromagnetism dictates that this line integral is related to currents and/or fields that exist inside the circle. The evaluation of the line integral as 2πRF_0 involves only the field on the circle of radius R.

Discontinuity between two uniform fields

In the first figure, the field is uniform and in the ${\hat {z}}({\text{ or }}{\hat {k}})$ direction. We use $(F_{1},F_{2})$ to denote the constant magnitudes of the field in the two regions. The minus sign in the contribution from $F_{2}$ arises from the fact that ${\vec {F}}_{2}$ is antiparallel to $d{\vec {\ell }}$.

$\oint {\vec {F}}\cdot d{\vec {\ell }}=\left(F_{1}-F_{2}\right)L$

Note that the line integral vanishes along the paths bc and da because the ${\vec {F}}\perp d{\vec {\ell }}$. The field is uniform and parallel to the path along ab and cd.

If the solenoid is not infinitely long, the line integral deviates from this simple formula because the field is not uniform inside the solenoid. And, the field is not exactly parallel to the axis everywhere inside the solenoid due to end effects.

Simple integrals of a vector field over a closed surface

Closed surfaces (left) and surfaces with boundary curve shown in red (right)

For more information about open and closed surfaces, visit #Closed_surfaces.

Sphere with spherical symmetry

Simple illustration of uniform electric flux out of a sphere or cylinder

The surface area of a sphere of radius $R$ is $4\pi R^{2}$. If the charge distribution is spherically symmetric, then

$\oint {\vec {E}}\cdot d{\vec {A}}=4\pi R^{2}E$

Cylindar with field lines pointing away from the axis

The same figure (shown to the right) can be used to calculate the flux through a cylinder if the electric field points away from the z axis and varies in magnitude only by the distance from the z axis (no dependence on z or θ is permitted.) In that case, the figure to the right displays a top-down view of a Gaussian surface that is a cylinder of length L. And, the integral of the electric flux is:

$\oint {\vec {E}}\cdot d{\vec {A}}=2\pi RLE$.

Pillbox for one-dimensional (slab) geometry

$\oint {\vec {E}}\cdot d{\vec {A}}=(E_{2}-E_{1})A$

The pillbox does not need to be a cylindar

The pillbox is a useful Gaussian surface for infinite or very large slabs of charge. An important example is the parallel plate capacitor where each linear dimension of the plates is much larger than the plate separation.

Typically, charge is located within the pillobx. If the pillbox is taken from an isolated plain of charge, the fields point in opposite directions, with $E_{1}=-E_{2}$, so that,

$\oint {\vec {E}}\cdot d{\vec {A}}=2E_{2}A$

But for a capacitor, it is typically assumed that $E_{1}=0$ so that,

$\oint {\vec {E}}\cdot d{\vec {A}}=EA$

The pillbox method is only an approximation if the plates are not very large.

Footnotes

• See also file:Gauss.wiki.20150921.pdf