Page 339 - Intro to Tensor Calculus

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since only the ends of the cylinder contribute to the above surface integral. On the sides of the cylinder we
will have ˆ n ·± b e n = 0 and so the surface integral over the sides of the cylinder is zero. By equating the
µ ∗
results from equations (2.6.24) and (2.6.25) we obtain the result that β = and consequently we can write
2 0
µ ∗
~
E = b e n where b e n represents one of the normals to the surface.
2 0
∗
Note an electric field will always undergo a jump discontinuity when crossing a surface charge µ . As in
µ ∗ µ ∗
~
~
the above example we have E up = b e n and E down = − b e n so that the difference is
2 0 2
µ ∗ i (1) i (2) µ ∗
~
~
E up − E down = b e n or E n i + E n i + =0. (2.6.26)
0 0
It is this difference which causes the jump discontinuity.
EXAMPLE 2.6-3.
Calculate the electric field associated with a uniformly charged sphere of radius a.
Solution: We proceed as in the previous example. Let µ denote the uniform charge distribution over the
∗
surface of the sphere and let b e r denote the unit normal to the sphere. The total charge then is written as
ZZ
∗
2 ∗
q = µ dσ =4πa µ . If we construct a sphere of radius r> a around the charged sphere, then we have
S a
by the Gauss theorem
ZZZ q
~
E · b e r dσ = Q e = . (2.6.27)
0 0
S r
~
Again, we can assume symmetry for E and assume that it points radially outward in the direction of the
~
~
surface normal b e r and has the form E = β b e r for some constant β. Substituting this value for E into the
equation (2.6.27) we find that
ZZ ZZ q
~
2
E · b e r dσ = β dσ =4πβr = . (2.6.28)
0
S r S r
1 q
~
This gives E = b e r where b e r is the outward normal to the sphere. This shows that the electric field
4π 0 r 2
outside the sphere is the same as if all the charge were situated at the origin.
3
2
i
1
i
For S a piecewise closed surface enclosing a volume V and F = F (x ,x ,x ) i =1, 2, 3, a continuous
vector field with continuous derivatives the Gauss divergence theorem enables us to replace a flux integral
i
i
of F over S by a volume integral of the divergence of F over the volume V such that
ZZ ZZZ ZZ ZZZ
i
i
~
~
F n i dσ = F dτ or F · ˆ n dσ = divFdτ. (2.6.29)
,i
S V S V
i
If V contains a simple closed surface Σ where F is discontinuous we must modify the above Gauss divergence
theorem.
EXAMPLE 2.6-4.
We examine the modification of the Gauss divergence theorem for spheres in order to illustrate the
concepts. Let V have surface area S which encloses a surface Σ. Consider the figure 2.6-4 where the volume
V enclosed by S and containing Σ has been cut in half.