Week 2: Continuous Charge and Gauss’s Law 77
a
θ
b
a a’
S’
∆S
∆
θ
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a’
E
Figure 15: Geometry of the flux integral over a small surface areathat the continuous field is approximately uniform across it (we will eventually make it differentially
small, of course, so this is no problem).
Since the field is uniform and at right angles to the field, the flux through just this little chunk
is easy to evaluate. It is just:
∆φe=|E~|∆S=|E~|ab (90)That was easy enough! Let’s make things a little more complicated.
Suppose that we consider a rectangular surface ∆S′=a′bthat istippedwith respect to the first
surface at an angleθ, that shares the lengthbof the first surface, and that has a lengtha′that is long
enough that it precisely subtends the same “stream” of the vector fieldE~as shown. Basically, all
the field lines that pass through the first surface pass through the second surface, and again we are
assuming that the field is continuous and we can make the picture as small as we like (differentially
small in the limit) so that a conservedE~doesn’t change itsmagnitude or directionin between the
two surfaces.
Note thata=a′cos(θ), so that:∆S′=a′b= ab
cos(θ)(91)
If we just multiply|E~|by ∆S′, we see that we’ll get ∆φ′e= ∆φe/cos(θ), right? And we’d like to
get the same thing, as we’d like the flux integral tomeasurethe continuity and conservation of the
electric field across the tiny region between the two surfaces. So we multiply by cos(θ) on top to