### Video Transcript

The electric field in a region is
given by π equals π divided by π plus ππ₯ in the π’-direction, where π equals
200 newton-meters per coulomb, π equals 2.0 meters, and π equals 2.0. What is the net charge enclosed by
the shaded volume shown?

We can call this net enclosed
charge capital π. And to start solving for it, we can
recall a mathematical relationship between charge and electric field π. Gaussβs law tells us that if you
have a closed surface, a shape with some volume and area, then if you add up that
entire area multiplying each part by the electric field that goes through it, then
that surface integral over a closed surface is equal to the charge enclosed by the
surface over the constant π naught, the permittivity of free space. Weβll treat that constant as having
an exact value of 8.85 times 10 to the negative 12 farads per meter.

As we consider how to apply this
relationship to our scenario, we take a look at our shape, our surface, which is a
box. The box has a height of one and a
half meters, a width of 1.0 meters, and a length of 2.0 meters. Also the box is a closed
surface. So that means itβs a candidate to
use Gaussβs law to solve for the enclosed charge π. Letβs take a closer look at the dot
product of the two vectors involved in this calculation. We have a vector, the electric
field, dotted with an area element, ππ΄, which is also a vector which will point a
normal or perpendicular to the area element it represents.

We know that the electric field in
this example only points in one direction, in the positive π’-direction, which means
we can label our axes π₯, π¦, and π§ for the vertical. What about the area elements of our
box? Which ways do their area vectors
point? Like any box, ours has six faces,
each with its own area vector pointing normal to that face. Now we can recall that the dot
product between two coplanar vectors, weβll call them π΄ and π΅, is equal to the
product of their magnitude times the cosine of the angle between them.

When we consider this dot product
for our electric field, π, pointed in a positive π’-direction and the six faces of
our surface, since the cosine of 90 degrees is zero, that dot product for the four
faces that donβt face in either the positive or negative π₯-direction is zero. So the only faces we need to
account for are the front face, weβll call that π΄ sub one, and then the back face,
which weβll call π΄ sub two. So our surface integral simplifies
to two terms, the electric field π first dotted with π΄ one plus the electric field
π dotted with π΄ two.

Now because the electric field is
position dependent β that is, it has an π₯ variable in its equation β it can be
helpful to write these two terms separately and calculate them separately to add
them together to solve for π over π naught. We can calculate these two terms
one by one. Letβs start with π dot π΄ one. The magnitude of the area π΄ one is
equal to 1.0 meter times 1.5 meters. And π΄ one points in the positive
π₯- or positive π’-direction. The electric field π at this
position is equal to 200 newton-meters per coulomb over 2.0 meters plus 2.0 times
the π₯-coordinate of the plane π΄ sub one, which is 2.0 meters.

π also points in the positive
π’-hat or π₯- direction. So when we take the dot product of
these two terms, we can simply multiply them together then find itβs equal to 50
newton-meters squared per coulomb. Next we move on to calculate π dot
π΄ two. In this case, the magnitude of the
area is the same as before, but now it points in the negative π’-direction. And π now has a value of zero
meters in its π₯- component because weβre in the π§π¦-plane. When we dot these values together,
the result we find is negative 150 newtons-meter squared per coulomb.

When we add these two values on the
left side of our equation together, we find that π over π naught is equal to
negative 100 newtons-meter squared per coulomb. Or π is equal to π naught times
negative 100 newtons- meter squared per coulomb. When we plug-in for π naught and
enter these values on our calculator, we find that π, to two significant figures,
is negative 8.9 times 10 to the negative 10th coulombs. Thatβs the amount of charge
enclosed in this volume.