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
`vec(E) = 200 hat(i) N//C for x gt 0 and vec(E) = -200 hat(i) N//C` for `x lt 0`. A right circular cylinder of length 20 cm and raidus 5cm has its center 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 the side of the cylinder ? (b) What is the net outward flux through the cyclinder ? (c) what is net charge inside the cylinder ?

Explanation

(a) We can see from the figure that on the left face E and DS are
`phi_(L)=EDeltaS=-200 hati-DeltaS`
`=+200 DeltaS, "since" hatiDeltaS=-DeltaS`
`=+200xxpi(00.05^(2))=+1.57Nm^(2)C^(-1)`
On the right point face E and `DeltaS` are parallel and therefore
`phi_(R)=E.DeltaS=+1.57 Nm^(2)C^(-1)`.
(b) For any point on the side of the cylinder E is perpendicular to `DeltaS` and hence `E.DeltaS = 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=3.14Nm^(2)C^(-1`

(d) The net charge within the cylinder can be found by using Gauss’s law which gives
`q=epsilon_(0)phi`
`=3.14xx8.854xx10^(-12)C`
`2.78xx10^(-11)C` .