1 \begin{problem*}{24.21}
2 Figure~P24.21 represents the top view of a cubic gaussian surface in a
3 uniform electric field \vect{E} oriented parallel to the top and
4 bottom faces of the cube. The field makes an angle $\theta$ with side
5 $a$, and the area of each face is $A$. In sumbolic form, find the
6 electric flux through \Part{a} face $a$, \Part{b} face $b$, \Part{c}
7 face $c$, \Part{d} face $d$, and \Part{e} the top and bottom faces of
8 the cube. \Part{f} What is the net electric flux through the
9 cube? \Part{g} How much charge is enclosed within the gaussian
23 vector_field(width=2.4u, height=2.4u, EField(dir=theta));
24 draw(scale(2*u)*shift((-0.5,-0.5))*unitsquare, dashed);
25 label("$a$", (0,u), align=N);
26 label("$b$", (u,0), align=E);
27 label("$c$", (0,-u), align=S);
28 label("$d$", (-u,0), align=W);
30 Angle a = Angle(c+(0,1), c, c+dir(theta), fill=currentpen, L="$\theta$");
37 The electric flux through a surface $S$ is
39 \Phi_{ES} = \int_S \vect{E}\cdot\vect{\dd A} = \int_S E \dd A\cos(\theta) \;.
41 Where $\theta$ is the angle between \vect{E} and the perpendicular
42 \vect{\dd A}. For a uniform field and flat surface, $E$ and $\theta$
43 are constants, so we can pull them of the integral:
45 \Phi_{ES} = E\cos(\theta)\int_S \dd A = EA\cos(\theta) \;.
47 For this problem, that means we only need to find the appropriate
48 expression for $\cos(\theta)$ to solve each part.
51 \Phi_{Ea} &= \ans{EA\cos(\theta)} \\
52 \Phi_{Eb} &= \ans{-EA\sin(\theta)} \\
53 \Phi_{Ec} &= \ans{-EA\cos(\theta)} \\
54 \Phi_{Ed} &= \ans{EA\sin(\theta)} \\
55 \Phi_{E\text{top}} &= \ans{0} \\
56 \Phi_{E\text{bottom}} &= \ans{0} \;
60 Summing the flux through each face (above), we have
62 \Phi_E = EA\cos(\theta) - EA\sin(\theta) - EA\cos(\theta) + EA\sin(\theta)
65 In other words, all the flux that comes into one part of the cube goes
66 out through some other part of the cube.
71 \Phi_E &= \frac{q_\text{in}}{\varepsilon_0} \\
72 q_\text{in} &= \Phi_E \varepsilon_0 = \ans{0} \;.