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 hati N//C` and `E = - 200 hati N/C `for `x lt 0`. A right circular cylinder of length 20 cm and radius 5 cm has its centre at the origin and its axis along the *-axis so that one face is at jc = + 10 cm and the other is at x = - 10 cm as shown in figure, (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 ?

Explanation

`A_(1) = A_(2) = pir^(2)`
`=(3.14) (0.05)^(2)`
`=7.85 xx 10^(-3) m^(2)`
(a) Electric flux linked with face-1 is,
`phi_(1) =A_(1)E_(1) cos theta_(1)`
`=(7.85 xx 10^(-3)) (200) cos0^(@)`
`therefore phi_(1) = 1.57 (Nm^(2))/C`
Electric flux linked with face-2 is,
`ph_(2) = A_(2)E_(2) cos theta_(2)`
`=(7.85 xx 10^(-3)) (200) cos 0^(@)`
`therefore phi_(2) = 1.57 (Nm^(2))/C`
(b) Here curved surface of given cylinder is parallel to electric field and so no electric flux passes through this curved surface. Hence, `phi_(3)=0`
(c) Net electric flux passing through given cylinder is,
`phi = phi_(1) + phi_(2) + phi_(3)`
`=1.57 + 1.57 + 0`
`therefore phi = 3.14 (Nm^(2))/C`
(d) If q is the net charge enclosed by above cylinder then according to Gauss theorem,
`phi = q/(epsilon_(0))`
`therefore q = phi epsilon_(0)`
`=(3.14)(8.85 xx 10^(-12))`
`therefore q = 2.779 xx 10^(-11) C`