An electric field is uniform and `E=200 hatiNC^(-1)` along `+x`s and `E=-200hatiNC^(-1)` along `-x` direction. A right circular cylinder of length 0.20 m and radius 0.05 m has its centre at the origin and its axis along the x-axis so that one face is `x=+0.10m` and the other `=x-0.10m`.
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 throught the cylinder?
d. What is the net charge inside the cylinder?

Explanation

a. Electric flux through the left the face
`phi_(L)=E_(L) DeltaScos theta, theta=180^(@), cos 180^(@)=-1`
Hence `phi_(L)=-200xx3.142xx(0.05)^(2)xx-1`
i.e. `phi_(L)=+1.57Vm` (or `Nm^(2)C^(-1)`)
Electric flux through the right flat face
`phi_(R)=E_(R)DeltaScos theta,theta=0^(@),cos 0^(@)=+1`
i.e. `phi_(R)=200xx3.142xx1(0.05)^(2)xx1`
i.e. `phi_(R)=+1.57VM` (or `Nm^(2)C^(-1)`)
b. Electric flux through the canvas or curved surface area,
`phi_(CSA)=E DeltaS cos 90^(@)=0`
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 the relatiion `q= epsilon_(0)phi`.
i.e. `q=8.854xx10^(-12)xx3.14C`
or `q=2.78xx10^(-11)C`