## Physics Friday 38

Electrostatics and Charged Surfaces

In the integral form, Gauss’ Law reads:

or in words, that the flux of an electric field through a closed surface equals the total enclosed electric charge divided by ε0. Now, let us consider some surface with a surface charge density σ(r). Let us pick a point on the surface. Let us call the two sides of the surface side 1 and side 2, and let n be a surface normal at the point, directed from side 1 to side 2. Lastly, let us call the electric field immediately on either side of the surface E1 and E2.

Now, let us make a gaussian surface around our point. Our gaussian “pillbox” will be a small, flat cylinder with ends of area dA and axis parallel to n. Letting the height of the cylinder be small, the flux through the “sides” will vanish, and the net flux will be that through the ends. For the end on side 2, we see the (outward) flux will be, for a small “pillbox”, E2·(dAn)=dAE2·n. Similarly, the flux for the end on side 1 is E1·(-dAn)=-dAE1·n. Thus, the total flux is (E2E1n dA. Now, the total enclosed charge is just σ dA, and so:
.
Thus, a surface charge creates a discontinuity in the field perpendicular to the charged surface.

Now, let us consider a perfect conductor in an electric field (the basis of many electrostatics problems). The requirement for a static situation is that the electric potential be constant on the surface of the conductor, and that the electric field be zero within the conductor (otherwise charges in the conductor will be experiencing net electric forces, and thus no static state). As a result, this generally leads to a distribution of charges on the surface of the sphere (“induced charges;” see this previous problem for an example), as required by the resulting discontinuity in the electric field. Note that since the definition of the electric potential Φ is that E=-Φ, the condition that Φ is constant on the surface of the conductor is equivalent to the requirement that the electric field be normal to the surface.

Now, combining these conditions with our finding about surface charges, we see that at the surface of the conductor, with our conventions as before, and side 1 the interior of the conductor, E1=0, and so σ= ε0E2·n. Writing this in terms of the potential outside the conductor, we have σ=-ε0Φ·n=-ε0nΦ; where nΦ=Φ·n is the directional derivative in the direction of the outward surface normal n. This allows us to find the induced charge on the surface of a conductor from the electric potential outside the conductor.

### One Response to “Physics Friday 38”

1. Physics Friday 121 « Twisted One 151's Weblog Says:

[…] rather than discrete points, the more useful form of this is Gauss’ Law (see here and here for previous uses). In integral form, Gauss’ Law reads: . Using the divergence theorem, this […]