### Video Transcript

The cross sections of the closed
surfaces (a), (b), (c), and (d) are shown in the accompanying diagrams. Find the electric flux through the
closed surface (a). Find the electric Flux through the
closed surface (b). Find the electric flux through the
closed surface (c). Find the electric flux through the
closed surface (d).

If we represent electric flux in
general with a Greek symbol capital π· with an π¬ subscript, then we want to solve
for this value in the situations of (a), (b), (c), and (d). Weβll call the values of electric
flux for each of these cases π· sub π¬ (a) π· sub π¬ (b) π· sub π¬ (c) and π· sub π¬
(d) respectively.

As we get started solving for π·
sub π¬ (a), we can recall Gaussβs law. This law tells us that given a
closed surface like we have in each of these four examples, if we integrate over the
total area of that surface, multiplying each area element by the electric field that
moves through that area element, then that surface integral will be equal to the
total charge π enclosed by the surface divided by π naught, the permittivity of
free space.

π naught is a constant value. And weβll treat it as exactly 8.85
times 10 to the negative 12 farads per meter. Now if we look back at the
left-hand side of Gaussβs law, this is an expression where weβre multiplying an area
element times the electric field. This can remind us of the
relationship for electric flux. π· sub π¬, the electric flux, is
equal to electric field multiplied by the area through which that field line
passes.

Because weβre considering, in each
of our four cases, a closed surface and weβre integrating over the entire thing, we
can combine Gaussβs law with a relationship defining electric flux to write that the
electric flux in any of these four scenarios is equal to the charge enclosed by the
volume over π naught. Knowing that, letβs move ahead to
solve for π· sub π¬ (a) considering scenario (a).

π· sub π¬ (a) equals the charge
enclosed by our shape, weβll call it π sub π, over π naught. And as we look at situation (a), we
see the amount of charge enclosed. Itβs 3.0 times 10 to the negative
eighth coulombs. We see there is a charge outside
our shape. But since itβs on the outside, we
donβt include it in our calculation for π· sub π¬ (a). So we plug in our value for π
naught and calculate this fraction. And we find that, to two
significant figures, π· sub π¬ (a) is 3.4 times 10 to the third newtons-meter
squared per coulomb. Thatβs the electric flux through
the closed surface in (a).

Next, we move to scenario (b) to
calculate π· sub π¬ (b). And as we look at the shape in
scenario (b), we see that it encloses no charge. That means that π sub π, the name
we can give to the charge enclosed by the shape in scenario (b), is equal to
zero. And therefore, π· sub π¬ (b) is
equal to zero as well. Now on to scenario (c) where this
time our shape does enclose the charge. In this case π sub π, the name we
can give to the charge enclosed by the shape in scenario (c), is equal to negative
2.0 times 10 to the negative sixth coulombs.

When we insert our value for π
naught and calculate this fraction, we find that, to two significant figures, π· sub
π¬ (c) is negative 2.3 times 10 to the fifth newtons-meter squared per coulomb. Thatβs the electric Flux through
the surface in scenario (c). And finally, we turn to scenario
(d) where our volume, in this case, encloses part of the conductor which extends
infinitely far up and down. If π sub π is the charge enclosed
in our volume in this case, then as we look at scenario (d), that charge π will
equal the product of the surface charge density π and the area of the end cap of
our volume.

In other words, π· sub π¬ (d) is
equal to π, the surface charge density in the conductor, multiplied by the area, we
can call it π΄, of the end cap. Weβre told the value for π as well
as the value for the area of the end cap. And when we plug in those values
along with our values for π naught and calculate this fraction, we find that, to
two significant figures, it equals 90 newtons-meter squared per coulomb. Thatβs the electric flux through
the volume in scenario (d).