A hollow cylindrical box of length 0.5m and area of cross-section `20cm^(2)` is placed in a three dimensional coordinate system as shown in the figure. The electric field in the region is given by `vec E = 20 x hat i`, where E is `NC^(-1)` and x is in meters, Find

(i) Net flux through the cylinder.
(ii) Charge enclosed in the cylinder.

Explanation

(i) Given, `vec E = 20 x hat i`
and `Delta s = 20cm^(2)=20 xx 10^(-4)m^(2)`
As the electric field is only along the x-axis, so flux will pass only through the cross-section of cylinder.
Magnitude of electric field at A,
`E_(A)=20xx0.5 = 10N//C`
Magnitude of electric field at cross-selection B.
`E_(B) = 20 xx 1 = 20NC^(-1)`

The corresponding electric fluxes are
`phi_(A)=vec E. Delta vecs`
`=10xx 20xx 10^(-4) xx cos 180^(@) = - 0.02 Nm^(2)C^(-1)`
`phi_(B)=vec E. Delta vec s `
`=20xx 20xx10^(-4) xx cos 0^(@) = 0.04Nm^(2)C^(-1)`
So, the net flux through the cylinder,
`phi = phi_(A)+phi_(B)`
`= -0.02 +0.04=0.02 Nm^(2)C^(-1)`
(ii) Using Gauss's Law:
`oint barE.bar(ds)=(q)/(E_(0))`
`0.02 =(q)/(8.85xx10^(-12))`
`q=8.85xx0.02xx10^(-12)`
`q=0.177xx10^(-12)C.`