phy1020.DVI

HerenOis a unit vector perpendicular to surfaceSandAis the total area ofS.IfSis a curved surface instead
of a plane, then the electric flux is more generally

ˆED

Z

S

EnOdA (16.5)

wheredAis an infinitesimally small piece of area ofS, and the integral is over the entire area ofS. In other
words, we imagine dividing surfaceSinto many tiny squares, each of which has areadA. For each square,
we draw a normal unit vectornOat that square, and we computeEnO, which is the component of the electric
fieldEthat is perpendicular to that square. We then multiply that result by the area of the squaredAto get
the electric flux through areadA. We add each of those fluxes together over the entire areaAof surfaceS.

16.5 Gauss’s Law

One important application of electric flux is its appearance inGauss’s law, which is one of the four funda-
mental equations of classical electromagnetism known asMaxwell’s equations. Gauss’s law states that if
we draw an imaginaryclosedsurface in space, then the total electric flux through that closed surfaceSis
proportional to the total amount of charge enclosed inside that surface:

ˆED

I

S

EnOdAD

qencl

" 0

: (16.6)

Here the circle on the integration sign indicates that the integral is over theclosedsurfaceS.
While Gauss’s law is generally true, one of its important practical uses is that it allows the quick determi-
nation of the electric field of a symmetrical distribution of charges. For example, it allows the electric field
of a spherical or cylindrical charge to be determined very easily.
As a simple example, suppose we wish to find the electric fieldEat a distancerfrom a point chargeq.
We would imagine drawing an imaginary spherical surface of radiusrcentered onq, so that the sphere passes
through the point at which we wish to calculateE. Then on the left-hand side of Eq. (16.6), the electric flux
is the electric fieldEtimes the area of the sphere:ˆED

H

EdADE

H

dADE.4r^2 /. Since the total
charge inside the sphere isq, the right-hand side becomesq=" 0. Gauss’s law then gives4r^2 EDq=" 0 ,or
EDq=.4r^2 " 0 /, in agreement with Coulomb’s law.
Both Coulomb’s law and Gauss’s law allow us to determine the electric field due to an arbitrary distri-
bution of charge. The difference between them is that, forsymmetricalcharge distributions, Gauss’s law
provides a shortcut that allows us to compute the electric field much more easily than using Coulomb’s law.
We can always use Coulomb’s law—Gauss’s law is just much less work when we have a symmetrical charge
distribution. For irregular charge distributions, though, we may have no choice but to “do it the hard way”
and resort to Coulomb’s law.

16.6 Electric Fields of Conductors

If an electrical conductor holds a net charge, then it has a number of important properties. If the conductor is
in electrostatic equilibrium (i.e. all charges have stopped moving), then:

• The electric field inside the conductor is zero.The conductor has free electrons throughout its interior.
  If there were an electric field inside the conductor, then there would be a force on those free elec-
  trons, causing them to accelerate, in violation of the assumption that the conductor is in equilibrium.
  ThereforeED 0 inside a conductor.