Physics Asked on February 27, 2021
Especially in electrodynamics, the integral theorems of Gauss and Stokes (in connection with the Maxwell equations) are often used to compute electric or magnetic fields. To compute the resulting integrals, one must then choose suitable surfaces.
This part causes me problems. In textbooks this is often not explained exactly; often there is only the hint that the surface was chosen "according to the symmetry of the problem".
For example:
An infinitely long, infinitely thin hollow cylinder of radius $R$ is homogeneously charged with the
electrical surface charge density $sigma_0$. The cylinder rotates at the angular velocity $vec{omega}=omegavec{e_z}$
around its axis of symmetry, so that the surface current density is $vec{K}=sigma_0omega Rvec{e_varphi}$.
As indicated above, I get from the theorem of Stokes and the Maxwell equations
$$int_{partial F} vec{H}(vec{r})cdot dvec{l}=int_{F} vec{j}(vec{r})cdot dvec{S}$$
with a magnetic field that has only a z-component, i.e. $vec{H}(vec{r})=H(r)vec{e_z}$.
How do you choose the area $F$ here to calculate the two integrals?
(I would also be happy, if anybody can explain this with another example. Perhaps someone knows a book or a website with an overview of suitable choices for several different geometric situations?)
One doesn't need to use specific surfaces to compute these fields. Gauss's Law works for any surface. The only requirement for Gauss's Law to hold is that the interaction caused by the field must be inverse square with respect to position, such as Coulomb's Law:
$$overrightarrow{F}=frac{q_i,q_j}{4piepsilon_0,r^2}hat{r}$$
(where the unit vector points from the field source and towards the probe charge), and superposition must hold in order to sum or integrate the source contributions.
Note that another field that satisfies these requirements is the gravitational field of Newtonian Mechanics:
$$overrightarrow{F}=-frac{GmM}{r^2}hat{r}$$
Gauss's Law for this field is:
$$oint_{partial V}overrightarrow{E} cdot overrightarrow{dA}, =, int_V-4pi Grho,dV $$
As to why certain surfaces with geometric symmetry are chosen for problems- even though any surface geometry is fine- it comes down to vast simplification in achieving the result. For example, for a uniform sphere of charge, note how the field is always radial to its source. While you could enclose the sphere in any type of Gaussian surface, if you enclose it in a sphere, which is geometrically symmetric to to the physical sphere, the dot product on the left-hand side of the equation goes away. This is because the radial symmetry always allows the field vectors at the surface and the normal vectors of the surface area to be parallel; thus resulting in the dot product always simplifying to:
$$oint_{partial V}E, dA, =, int_Vfrac{rho}{epsilon_0},dV $$
If the field vectors and surface norms were not parallel because of symmetry, the flux integral could still be correctly evaluated, but is usually MUCH harder to calculate. There are instances were it may not be solvable analytically. One other benefit that can occur from properly-chosen symmetry is creating a Gaussian surface that has a constant-magnitude field everywhere on its surface; which is clearly the case for the spherical example. Thus, E is now a constant, which leads to:
$$Eoint_{partial V}, dA, =, int_Vfrac{rho}{epsilon_0},dV $$
$$E,times (Surface Area), =, int_Vfrac{rho}{epsilon_0},dV $$
$$E,=, frac{1}{epsilon_0},int_Vfrac{rho}{(Surface Area)},dV $$
As for which geometry to choose for a given problem, it will depend on the problem. Try to look for a surface that is radially symmetric and, if possible, yields a constant field strength over the entire surface.
Hope this helped!
Answered by Brian Opatosky on February 27, 2021
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