78 Week 2: Continuous Charge and Gauss’s Law
compensate and get:
∆φ′e = |E~|cos(θ)a′b= |E~|cos(θ)ab
cos(θ)
= |E~|ab
= ∆φe (92)We can interpret this as meaning (in words) “IfE~is a continuous, constant vector field in the
region between ∆Sand ∆S′, then ∆φ′e= ∆φeand the flux through the two surfaces is conserved.”
Note that|E~|=E~·nˆand|E~|cos(θ) =E~·ˆn′, so that we can write:lim
∆S→ 0∆φe = E~·ˆn∆Sdφe = E~·ˆndS (93)whichdoes not varyfor any possible tipping of the surfacedS. The dot product precisely compensates
for the increase in the area ofdSas it tips relative to the direction ofE~.
n
n
q
S
E
∆S’
dΩ
θ
r
∆S
Figure 16: Point charge inside a closed surfaceS. Note that the flux through the tipped differential
piece of the surface ∆S′=r^2 dΩ/cosθis equal to that through theuntippedspherical piece of the
surface ∆S=r^2 dΩ that is subtended by the same solid angledΩ and osculates the tipped surface.
Now suppose that we have a point charge surrounded by aclosed surfaceS. This basically means
thatSis a topological deformation of a soap bubble – itcontainsa volumeVwith no openings. We
can then imagine that the electric field of this charge is “radiated” away in all directions according
to the point charge rule:
E~=keqrˆ
r^2(94)
This situation is pictured in figure 16.
From the above, we know that if we evaluate the flux across the small patch ∆Sof thespherical
surface indicated (an osculating distancerfrom the charge) the fieldE~will beexactlyconstant and
exactlyperpendicular to that patch. In fact, the flux through that surface patch is:
∆φe=E~·nˆ∆S=|E~|r^2 ∆Ω (95)where ∆Ω is thesolid anglesubtended by the cone formed by the charge and the boundary of
∆S=r^2 ∆Ω on the surface.
We’ve just shown that if we consider thetippedpatch ∆S′that osculates (kisses) ∆Sone end, is
tipped up through an angleθso it is actually a part of the blob shaped “arbitrary” closed surface