7. A closed surface with dimensions $a = b = 0.40$ m and $c = 0.60$ m is located as shown in the diagram. The electric field throughout the region is nonuniform and given by $\vec{E} = (3.0 + 2.0x)\hat{i}$ N/C, where x is in meters.
(a) Calculate the net electric flux leaving the closed surface.
(b) What net (induced) charge is enclosed by the surface?
a)
$\Phi_E = \Phi_{right} - \Phi_{left}$
$x = a + c$
$E_{right} = 3 + 2 \cdot 0.60 = 3.8$ N/C
$A = a \cdot c = 0.40 \times 0.60 = 0.24$ m$^2$
$\Phi_{right} = E_{right} \cdot A = 3.8 \times 0.24 = 0.912$ N m$^2$/C
$x = a$
$E_{left} = 3$ N/C
$\Phi_{left} = E_{left} \cdot A = 3 \times 0.24 = 0.72$ N m$^2$/C
$\Phi_E = \Phi_{right} - \Phi_{left} = 0.912 - 0.72 = 0.192$ N m$^2$/C
b)
$q_{encl}$
$\Phi_E = \frac{q_{encl}}{\epsilon_0} \implies q_{encl} = \Phi_E \cdot \epsilon_0 = (0.192)(8.85 \times 10^{-12}) = 1.6992 \times 10^{-12}$ C