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Gauss's Law; What is it?

  1. Oct 2, 2012 #1
    My professor explained it and class and I've tried reading the book but I just cannot wrap my mind around what Gauss's law is.

    I understand that it has to do with electric fields and surfaces; the amount of something passing through a surface.

    I think my main problem is that I can't think of a good practical application of Gauss's law. Can anyone help me?
  2. jcsd
  3. Oct 2, 2012 #2
    The integral form of Gauss's law is the 3d version of something you already should have a grasp of: the fundamental theorem of calculus.

    What is the fundamental theorem of calculus? Here's a refresher:

    [tex]\int_a^b F'(x) \; dx = F(b) - F(a)[/tex]

    For any differentiable function [itex]F(x)[/itex] whose derivative is [itex]F'(x)[/itex]. From this, we get the idea that integration and differentiation naturally cancel one another in a particular way.

    Let's think about this theorem another way, though. Key to this theorem is the interval [itex][a,b][/itex]. This could be thought of a a segment of a number line--a nice, continuously connected stretch. It has a "directionality" or "orientation" that tells us to go from a to b and not the other way around. It also has a boundary: the point [itex]b[/itex] is the upper boundary and [itex]a[/itex] is the lower boundary. We sum or "integrate" the function [itex]F(x)[/itex] on this boundary--it's just an integral over two points is pretty trivial.

    The fundamental theorem is extended to 2d and 3d and above in the following way: on some region [itex]V[/itex], the integral of a function over the boundary of [itex]V[/itex] is equal to the integral of the derivative over the entirety of [itex]V[/itex].

    Gauss's law is the 3d case for a vector field. The surface integral is the integral of the function over the boundary. The volume integral is the integral of the derivative over the whole of the volume.

    To help people think about this in an "intuitive" way, we have the concept of flux--where the electric field is perpendicular to the surface, there is more flux passing through the surface, similar to fluid passing through a net or light through a window. And we say that flux related to the charge enclosed--the charge being the volume integral of charge density, which is itself the derivative of the electric field.

    The integral form of Gauss's law allows us, through symmetries of the problem, to find the electric field from a charge distribution without going through a brute-force calculation of summing over a bunch of differential elements and treating them all like point charges.
  4. Oct 3, 2012 #3


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    What Murphrid describes is usually known as "Gauß's Theorem" and refers to a mathematical theorem in vector calculus in threedimensional euclidean space. It says
    [tex]\int_{V} \mathrm{d}^3 \vec{x} \vec{\nabla} \cdot \vec{A}(\vec{x})=\int_{\partial V} \mathrm{d}^2 \vec{F} \cdot \vec{A}(\vec{x}).[/tex]
    Here [itex]\vec{A}[/itex] is a sufficiently well-behaved vector field, [itex]V[/itex] a volume with sufficiently "nice" shape and boundary in threedimensional Euclidean space, and [itex]\partial V[/itex] its boundary. The orientation of the surface-normal vectors [itex]\mathrm{d}^2 \vec{F}[/itex] is such that they point out of the volume [itex]V[/itex].

    Gauß's Law is one fundamental equation in electromagnetism and thus part of Maxwell's equations. It says (in Heaviside-Lorentz Units)
    [tex]\vec{\nabla} \cdot \vec{E}=\rho.[/tex]
    Here [itex]\vec{E}[/itex] are the electric components of the electromagnetic field (mostly just called "the electric field") and [itex]\rho[/itex] the density of electric charges.

    Using Gauß's theorem, you get the integral form of the same law,
    [tex]\int_{\partial_V} \mathrm{d}^2 \vec{F} \cdot \vec{E} = \int_V \mathrm{d}^3 \vec{x} \rho(\vec{x})=Q_{V}.[/tex]
    Again the orientation of the surface vectors is as explained above. It tells you that the electric flux, i.e., the surface integral over the electric field on the left-hand side of the equation always equals the total charge contained in the volume, bounded by the surface.

    Sometimes, if a problem is sufficiently symmetric, you can use the integral form to determine the electric field. The most simple case is the field of a point charge [itex]Q[/itex], which we put for simplicity in the origin of the coordinate system. Due to spherical symmetry, we expect that the electric field is radial and its magnitude only depends on the distance from the charge, i.e., we make the ansatz
    [tex]\vec{E}=E(r) \vec{e}_r,[/tex]
    where [itex]\vec{e}_r=\vec{x}/|\vec{x}|[/itex] is the radial unit vector.

    Now take a sphere [itex]S_R[/itex] of radius [itex]R[/itex] around the origin and use spherical coordinates for the surface integral. The surface normal vectors in standard spherical coordinates [itex](r,\vartheta,\varphi)[/itex] are
    [tex]\mathrm{d}^2 \vec{F}=\mathrm{d} \vartheta \; \mathrm{d}\varphi \; \sin \vartheta \vec{e}_r.[/tex]
    From this you get for Gauß's Law
    [tex]4 \pi R^2 E_r(R)=Q[/tex]
    and thus immediately the Coulomb-field solution
    [tex]E_r(R)=\frac{Q}{4 \pi R^2}.[/tex]
    It is pretty easy to show that indeed
    [tex]\vec{\nabla} \cdot \vec{E}=0 \quad \text{for} \quad \vec{x} \neq 0.[/tex]
    That the factor is correct, we have just proven with the integral form of Gauß's Law. At this step you necessarily need the integral form, because a point charge is a singularity, because the corresponding charge density is given by a Dirac [itex]\delta[/itex] distribution,
    [tex]\rho(\vec{x})=Q \delta^{(3)}(\vec{x}).[/tex]
  5. Oct 3, 2012 #4
    Basically the flux of electric field coming out of a surface has to match the amount of electric charge confined within this surface. This applies to any surface.
  6. Oct 3, 2012 #5
    A problem in gauss law is that you should have a certain symmetry situation so that the direction of electric field can be guessed otherwise you will have to solve the corresponding poisson eqn for the given charge distribution and then take derivatives of potential so obtained to get the electric field.
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