## Proving properties of electric fields using Gauss’s Theorem

**Author: CHRISTINE ADELLE L. RICO**

** **Use Gauss’s theorem and to prove the following:

(a) Any excess charge placed on a conductor must lie entirely on its surface. (A conductor by definition contains charges capable of moving freely under the action of applied electric fields.)

Solution:Suppose that the field were initially nonzero. Since this is a conductor, any charges in the interior would move in response to the field. After a time, this process stops since the moving charges produce currents which dissipate energy. The final configuration would then have charges arranged so that the interior is zero. Recall that in equilibrium, the electric field inside a conductor is zero. Since everywhere inside the conductor, then from Gauss’s Law, , the charge density everywhere in the interior. Therefore, every point inside a conductor has zero charge, and any excess charge can only reside on the surface of the conductor.

(b) A closed, hollow conductor shields its interior from fields due to charges outside, but does not shield its exterior from the fields due to charges placed inside it.

Solution:

Part 1.Consider the charge exterior to the conductor which produces an electric field, as shown in the figure.The electric field in the conductor is zero, with induced charge densities on the exterior and interior surfaces of the conductor.

i)Imagine moving a charge on the interior surface from point A to point B along path 2 which goes throughout the conductor itself. Since in the conductor, along this path.

ii)Move the same charge from A to B along path 1, in the interior cavity of the conductor. Since the electrostatic field is conservative, along its path.This must be true also for any path in the interior. So generally, in the interior. Therefore the conductor shields its interior from field due to charge placed outside.

Part 2. Consider a positive charge placed inside a hollow conductor as shown in the figure.The charge induces a charge density in the interior surface of the conductor in such a way that the electric field in the interior of the conductor is zero. Assuming that the conductor is charge neutral, this means that there is an induced charge density on the exterior surface of total charge Q. If we apply Gauss’s Law to the Gaussian surface G surrounding the conductor, the total charge enclosed is still . Therefore, there is an electric field outside the conductor.

(c) The electric field at the surface of a conductor is normal to the surface and has a magnitude , where is the charge density per unit area on the surface.

Solution:

Note that in equilibrium, the field at exterior surface must be normal to the surface, so that the tangential component is zero. The magnitude of the field is derived using Gauss’s Law with a Gaussian pillbox which cuts through the surface. The electric field is zero on the conducting side of the pillbox. So,

, with the area on the surface.

Rearranging, we get

.

Define as the charge per unit area ,

.

References: Classical Electrodynamics, John David Jackson, 3rd Edition, Chapter 1. Introduction to Classical Electrodynamics, David Griffiths, Chapter 2. University Physics, Young and Freedman, 11th Edition, Chapter 24. Faraday’s cage, wikipedia.com. Gauss’s Law, wikipedia.com.

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Adelle is currently pursuing her MS Physics degree at the Mindanao State University- Iligan Institute of Technology in Iligan City.

April 13th, 2012 at 9:12 am

philippine information…[…]Proving properties of electric fields using Gauss’s Theorem | Quantum Science Philippines[…]…

April 17th, 2013 at 2:02 am

a combination of mathematics and physics

I need a remedial school

May 18th, 2014 at 10:03 pm

Your figure for part 2 is incorrect. The positive charge Q on the outside surface is uniformly distributed, irrespective of the position of the charge Q inside the hollow conductor.

January 28th, 2015 at 12:40 am

Your “proof” for part I of (b) holds verbatim even in the case that there is charge *interior* to the conductor, and therefore can not be correct.

I am guessing that the mistake is due to singularities in crossing over the boundary of the interior surface of the conductor