1
$\begingroup$

I'd like to know whether my sense of Gauss's Law is correct here.

Suppose I have a cube of side length $L$ filled uniformly with electric charge. Let's say I use as my integration surface a cube enveloping the cube of charge that has side length $2L$.

In this situation, Gauss's Law holds with my choice of integration surface, and I believe the choice is sensible as well. As long as my surface encloses the charge density I'm trying to find the electric field of and only that, the dimensions of the surface doesn't matter. Does this sound reasonable?

Could one use Gauss's Law in this scenario to solve for the electric field at points outside the charged cube? My intuition is no, as Gauss's Law is only useful when considering the electric fields for regions enveloping a charge density. I'd need to draw a Gaussian surface around the point, and I would reason that since there is no charge within that region then there is no electric field. However, this could be very wrong or right for the wrong reasons, and I'd like to clarify this in my brain.

$\endgroup$
3
$\begingroup$

Gauss's law is used when you want to find the electric field strength at a given distance from a charge. This law applies to ALL closed Gaussian surfaces that contain a charge, but some surfaces are much easier to integrate than others. Examples are:

For a point charge, the electric field is "radiating" from the charge in spherical wave fronts. This means that the proper choice of Gaussian surface is a sphere with the charge located at the center of the sphere. Such a Gaussian surface leads to an integral that is very easy (in fact trivial) to integrate, due to the symmetry of the problem, and the fact that the surface area of a sphere is already a well known equation.

For an infinitely long line of charge, the electric field is "radiating" from the charge in cylindrical wave fronts. This means that the proper choice of Gaussian surface is a cylinder with the line charge located down the axis of the cylinder. Such a Gaussian surface is once again easy to work with, and the area of such a Gaussian surface is just the area of the walls of the cylinder that has a radius equal to the distance from the line charge.

For a cubical shape, you can indeed use a cubical Gaussian surface, but it is highly doubtful that electric fields would "radiate" away from such a shape in cubical wave fronts. This means that the integral that is used to set up this problem probably can't be integrated.

Regarding the solution of the electric field strength outside the Gaussian surface, note that the Gaussian surface is a mathematical construct, and it doesn't actually exist in the real world. This means that once you solve an electric field problem using Gauss's law, you will have an equation that will allow you to calculate electric field strength at ANY distance from the associated charge, regardless of the distance from the Gaussian surface that you originally used.

$\endgroup$
2
  • $\begingroup$ Doesn't knowing which Gaussian surface to use require some a priori knowledge of the electric field you're trying to find though? To my knowledge, we are finding the electric field due to a charged object at any point in space (ultimately) by seeing if $\vec E$ can be taken outside the integral and thus solved for. But doesn't this then require us to know the orientation of the field to find a suitable Gaussian surface? $\endgroup$
    – sangstar
    Jan 29 '19 at 20:32
  • $\begingroup$ Solving a problem with Gauss's law requires you to know that the geometry of your problem will produce electric field wave fronts that have a particular shape. This leads to a choice of Gaussian surface that guarantees that the electric field everywhere on that surface has the same value of electric field strength. If and only if you can meet the "constant electric field strength" constraint can you pull "E" across the integral. $\endgroup$ Jan 29 '19 at 20:36
1
$\begingroup$

Gauss' Law has basically three different, but very similar meanings in electro dynamics. It is in general equivalent to Coulomb's law, and so has as much applicability.

Gauss' law in differential form allows for the integration of the electric flux on a surface to match it against the contained charge. Sufficiently high symmetry like those scenarios David White mentions allows one to quickly determine the electric field.

Thirdly, Already knowing the electric field, some relationships regarding flux take on a simplified form.

The integration of the Maxwell Stress Tensor on a plane is closely related to symmetric Gauss's Law problems. Would recommend playing with those to get a feel for some more subtle applications of the law.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.