PH 222
Homework 2, Thursday
Due: 14 Apr, 2016
Problem Statement
Consider a right-triangular prism, as shown, sitting in a uniform 400 N/C electric field
directed perpendicularly into the sloped face of the prism. Find the electric fluxes
through each of prism’s five faces.
Solution
H
~ · dA.
~
Let’s start with the sloped-face, which I’ll call F1 . The flux is ΦE,F1 = F1 E
Because the electric field is everywhere antiparallel (i.e., along the same axis, but in the
~ ~
opposite direction) to the surface normal, we
H have E · dA = −E dA (recall that E is the
~ so the flux is ΦE,F1 = −
magnitude of E),
E dA.
F1
In principle the electric field could vary as we move along the surface, in which case we
might have some difficulty evaluating the integral. But, in this case, the field is uniform
on the surface. We can therefore pull it out of the integral as a constant to find
I
I
ΦE,F1 = −
E dA = −E
dA = −EA1 .
F1
F1
The last step is to find A1 , the area of F1 . Recalling that sin 30◦ = sin π/6 = 1/2, the
length of the “unknown” side is 2.0 m/ sin 30◦ = 2.0 m/1/2 = 4.0 m. Hence, A1 = 16 m2 ,
giving a net flux
ΦE,F1 = −6400 Nm2/C.
Observe that this is negative, which we expect because it is flux into the surface.
18 April, 2016
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PH 222
Homework 2, Thursday
Due: 14 Apr, 2016
Turning to the triangular faces, we see that the electric field is everywhere perpendicular
~ · dA
~ = 0 and there is no flux through either of these
to the surface normal. Hence, E
sides.
Let’s look now at the upright rectangular surface, which I’ll call F2 . Here, the normal
is horizontal to the left. We find the dot-product by considering the angle between this
normal and the electric field as it emerges from F2 . Because the surface F1 makes a 30◦
~ (which is perpendicular to it) makes
angle with the left-ward horizontal, we know that E
◦
~ · dA
~ = E dA cos 60◦ = 1/2E dA.
a −60 angle with the left-ward horizontal. Hence, E
~ and dA
~ is constant
Note that this relies on our observation that the angle between E
over the whole surface; if it varied, we’d have to account for that in our integral.
But it doesn’t vary, and again we see that the field strength is uniform along our surface,
so we get
I
I
1
1
1
E dA = E
dA = EA2 .
ΦE,F2 =
2
2
F1
F1 2
Of course, A2 = 8.0 m2 , so, with E = 400 N/C we conclude
ΦE,F2 = 1600 Nm2/C.
Finally, we consider the bottom face, F3 . We can get the flux here easily and directly
from Gauss’s law: there is no charge contained inside our surface, so the net flux must
be zero. That is,
ΦE,F1 + ΦE,F2 + ΦE,F3 = 0,
which implies
ΦE,F3 = −ΦE,F1 − ΦE,F2
= 6400 Nm2/C − 1600 Nm2/C
= 4800 Nm2/C.
√
~ · dA
~ = E dA cos 30◦ = 3 E dA.
To check, let’s compute it like the others. In this case, E
2
As before, the electric field is uniform, so we need only find the area of the base. Recalling
the the sloped edge has length 4.0
√ m, we see that the “unknown” edge
√ of the base must
◦
have length 4.0 cos 30 m = 2.0 3 m, giving a total area A3 = 8.0 3 m. The flux is
then
√
√
I
3
3
ΦE,F1 =
E
dA =
EA3
2
2
F3
= E 12 m2 = 4800 Nm2/C
as expected.
18 April, 2016
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