Quantitative Use of Gauss's Law



" It has been said that democracy is the worst form of government except all the others that have been tried."
Winston Churchill

 
   elec gauss 3    
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SPHERICAL SYMMETRY
elec gauss figure 5Consider a single point charge +Q and a spherical surface, S,  of radius r and center at the location of +Q.  From the symmetry of this situation we can conclude that, everywhere on the surface S, E has the same value and is directed radially outwards (normal to the surface).  This is the same as the direction of dA.  Therefore,
elec gauss eqn5
so that,
elec gauss eqn6
which is exactly Coulomb's Law !!

As has already been stated - Gauss's Law and Coulomb's Law are different statements of the same physical principle.


  • Spherical Charge Distribution with Uniform Charge Density
Charge is distributed uniformly throughout the volume of the sphere (this means that the sphere must be a non-conductor since as we have seen the charge on a conductor must reside on the surface) such that the total charge Q is given by,

elec gauss eqn7

where ρ is the (volume) charge density, in units of Coulombs/m3.

What is the electric field at any point either outside or inside the sphere ?
Due to the symmetry of this configuration we can conclude that E is directed radially outwards everywhere and can (at most) depend only on the (radial) distance from the center of the sphere.  There are two distinct regions to consider:

elec gauss figure 5Outside the sphere,  r > R

Applying Gauss's Law over a Gaussian surface (sphere) of radius r, then,
elec gauss eqn 5

so that,
elec gauss eqn 6

In other words, for points outside the sphere, the sphere behaves as a point charge located the sphere's center.
hot  We saw exactly the same type of behavior when considering the gravitational effect of a spherical mass.

elec gauss figure 7Inside the sphere, r < R

Applying Gauss's Law over a Gaussian surface (sphere) of radius r, then,

elec gauss eqn8
Or in terms of Q and R,

elec gauss eqn9

Note that for r < R only the charge inside a sphere of radius r contributes to E.  The charge between r and R has no effect.

exclamation It is important to realize that without using Gauss's Law, these results could be obtained via Coulomb's Law, but would involve considerably more work - setting up  a non-trivial multiple integral to consider every point charge in the sphere....

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CYLINDRICAL SYMMETRY

  • Infinite Line Charge with Linear Charge Density λ
elec gauss figure8Determine the E field a distance r from the line charge.  (Note that the units of λ are Coulombs/meter)

Symmetry tells us that E can only have a component perpendicular to the line charge, that is perpendicular to the cylindrical surface shown.

Applying Gauss's Law over the cylindrical Gaussian surface, radius r and length l, as shown, there will in principle be three contributions - one from the curved surface and one from each of the two ends.  However, on the ends E and dA are perpendicular, so that E·dA = 0, therefore there is no contribution to the flux through S.  On the curved surface E and dA are parallel, thus,

elec gauss eqn10
so that,
elec gauss eqn11

We can extend this analysis to the case of a uniformly charged infinite cylinder in a similar manner to the extension of the point charge to the spherical charge distribution above.

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RECTANGULAR SYMMETRY
  • Infinite plane of charge
elec gauss figure 9Determine the E field at any distance above or below an infinite plane with charge density σ (Coulombs/m2).

Symmetry dictates the E must be perpendicular to the surface everywhere.

Applying Gauss's Law over the cylindrical surface shown, then the curved surface of the cylinder  contributes nothing to the flux since E and dA are perpendicular.  But on the ends E and dA are parallel.  Therefore,

elec gauss eqn12

so that,
elec gauss eqn13

That is the electric field is constant - it does not depend on how far the field point is from the plane !! 

exclamation Note that this is only true for an infinite plane of charge.  If the distance of the field point from the plane is small compared to the "size" of the plane, the above expression is a good approximation.

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  • In all the above situations the key to using Gauss's Law is SYMMETRY.  There must be enough symmetry in the problem to know the direction of E everywhere in the vicinity of the charge distribution.  Knowing the direction of E the trick is then to choose a Gaussian surface over which to apply Gauss's Law such that E can be "taken out" of the flux integral.  So when using Gauss's Law to determine E there are three key steps:

    1. State what you are assuming about E based on the symmetry of the problem.

    2. State clearly the Gaussian surface(s) you will use - often most easily done by sketching the surface(s) on a diagram.

    3. Evaluate the surface (flux) integral to determine E.  The symmetry of E and choice of Gaussian surface should allow "E" to be "taken out" of the integral and thus be determined.



An engineer friend of mine told me of a group of scientists that were nominated for a Nobel prize. Using dental tools, they were able to sort out the smallest particles that mankind has yet discovered. The group became known as " the Graders of the Flossed Quark."


 

Dr. C. L. Davis
Physics Department
University of Louisville
email: c.l.davis@louisville.edu