An electric field is uniform,and in the positive $x$ direction for positive $x,$ and uniform with the same magnitude but in the negative $x$ direction for negative $x$. It is given that $E = 200 \hat{i} \; N/C$ for $x > 0$ and $E = -200 \hat{i} \; N/C$ for $x < 0$. $A$ right circular cylinder of length $20 \; cm$ and radius $5 \; cm$ has its centre at the origin and its axis along the $x$-axis so that one face is at $x = +10 \; cm$ and the other is at $x = -10 \; cm$.
$(a)$ What is the net outward flux through each flat face?
$(b)$ What is the flux through the side of the cylinder?
$(c)$ What is the net outward flux through the cylinder?
$(d)$ What is the net charge inside the cylinder?

(N/A) On the left face $(x = -10 \; cm)$,the electric field is $E = -200 \hat{i} \; N/C$ and the area vector is $\Delta S = -\Delta S \hat{i}$. The outward flux is:
$\phi_{L} = E \cdot \Delta S = (-200 \hat{i}) \cdot (-\Delta S \hat{i}) = 200 \Delta S$
$\phi_{L} = 200 \times \pi (0.05)^2 = 200 \times 3.1416 \times 0.0025 = 1.57 \; N \cdot m^2/C$
On the right face $(x = +10 \; cm)$,the electric field is $E = 200 \hat{i} \; N/C$ and the area vector is $\Delta S = \Delta S \hat{i}$. The outward flux is:
$\phi_{R} = E \cdot \Delta S = (200 \hat{i}) \cdot (\Delta S \hat{i}) = 200 \Delta S = 1.57 \; N \cdot m^2/C$
$(b)$ For any point on the curved side of the cylinder,the electric field $E$ is parallel to the surface and perpendicular to the area vector $\Delta S$. Thus,$E \cdot \Delta S = 0$. The flux through the side is zero.
$(c)$ The net outward flux $\phi$ is the sum of the fluxes through all surfaces:
$\phi = \phi_{L} + \phi_{R} + \phi_{side} = 1.57 + 1.57 + 0 = 3.14 \; N \cdot m^2/C$
$(d)$ Using Gauss's law,the net charge $q$ inside the cylinder is:
$q = \varepsilon_{0} \phi = (8.854 \times 10^{-12} \; C^2/N \cdot m^2) \times (3.14 \; N \cdot m^2/C) \approx 2.78 \times 10^{-11} \; C$