Page 338 - Intro to Tensor Calculus

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where Q e represents the total charge enclosed by the surface S with ˆ n the unit outward normal to the surface.
The proof of Gauss’s theorem follows. Consider a single charge q within the closed surface S. The electric
1 q
~
field at a point on the surface S due to the charge q within S is represented E = b e r and so the flux
4π 0 r 2
integral is
ZZ ZZ q b e r · ˆ n q ZZ dΩ q
~
φ E = E · ˆ n dσ = 2 dσ = 2 = (2.6.22)
S S 4π 0 r 4π 0 S r 0
ZZ
b e r · ˆ n cos θdσ dΩ
since = = = dω and dω =4π. By superposition of the charges, we obtain a similar
r 2 r 2 r 2 S
n
X
result for each of the charges within the surface. Adding these results gives Q e = q i . For a continuous
i=1
ZZZ
∗
∗
distribution of charge inside the volume we can write Q e = ρ dτ,where ρ is the charge distribution
V
per unit volume. Note that charges outside of the closed surface do not contribute to the total flux across
the surface. This is because the field lines go in one side of the surface and go out the other side. In this
ZZ
~
case E · ˆ n dσ = 0 for charges outside the surface. Also the position of the charge or charges within the
S
volume does not effect the Gauss law.
The equation (2.6.21) is the Gauss law in integral form. We can put this law in differential form as
follows. Using the Gauss divergence theorem we can write for an arbitrary volume that
ZZ ZZZ ZZZ ρ ∗ 1 ZZZ
~
~
E · ˆ n dσ = ∇· Edτ = dτ = Q e = ρ dτ
∗
S V V 0 0 0 V
which for an arbitrary volume implies
ρ ∗
~
∇· E = . (2.6.23)
0
The equations (2.6.23) and (2.6.7) can be combined so that the Gauss law can also be written in the form
ρ ∗
2
∇ V = − which is called Poisson’s equation.
0
EXAMPLE 2.6-2
Find the electric field associated with an infinite plane sheet of positive charge.
∗
Solution: Assume there exists a uniform surface charge µ and draw a circle at some point on the plane
surface. Now move the circle perpendicular to the surface to form a small cylinder which extends equal
distances above and below the plane surface. We calculate the electric flux over this small cylinder in the
limit as the height of the cylinder goes to zero. The charge inside the cylinder is µ A where A is the area of
∗
the circle. We find that the Gauss law requires that
ZZ µ A
∗
~
E · ˆ n dσ = Q e = (2.6.24)
S 0 0
where ˆ n is the outward normal to the cylinder as we move over the surface S. By the symmetry of the
situation the electric force vector is uniform and must point away from both sides to the plane surface in the
direction of the normals to both sides of the surface. Denote the plane surface normals by b e n and − b e n and
~
~
assume that E = β b e n on one side of the surface and E = −β b e n on the other side of the surface for some
constant β. Substituting this result into the equation (2.6.24) produces
ZZ
~
E · ˆ n dσ =2βA (2.6.25)
S