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 ctrcular cyllnder 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 cyllnder?

$(a)$ We can see from the figure that on the left face $E$ and $\Delta S$ are parallel. Therefore, the outward flux is

$\phi_{L}= E \cdot \Delta S =-200 \hat{ i } \cdot \Delta S$

$=+200 \Delta S, \text { since } \hat{ i } \cdot \Delta S =-\Delta S$

$=+200 \times \pi(0.05)^{2}=+1.57 \,N\, m ^{2} \,C ^{-1}$

On the right face, $E$ and $\Delta S$ are parallel and therefore

$\phi_{R}= E \cdot \Delta S =+1.57 \,N\,m ^{2} \,C ^{-1}$

$(b)$ For any point on the side of the cylinder $E$ is perpendicular to $\Delta s$ and hence $E \cdot \Delta S =0 .$ Therefore, the flux out of the side of the cylinder is zero.

$(c)$ Net outward flux through the cylinder

$\phi=1.57+1.57+0=3.14 \,N\, m ^{2} \,C ^{-1}$

$(d)$ The net charge within the cylinder can be found by using Gauss's law which gives

$q=\varepsilon_{0} \phi$

$=3.14 \times 8.854 \times 10^{-12}\, C$

$=2.78 \times 10^{-11}\, C$