Using Gauss’s law, derive expression for intensity of electric field at any point near the infinitely long straight uniformly charged wire.
The electric field components in the following figure are `E_(x) = alphax, E_(y) = 0, E_(z) = 0, " in which " alpha = 400 N//C` m. Calculate (i) the electric flux through the cube, and (ii) the charge within the cube assume that a = 0.1m.

Explanation

Accroding the Gauss' law-
`ointN,dvecs=(1)/(eo){q}`
`intvecEvec(ds_(1))+intvecEvec(ds_(2))+intvec(dS_(3))=(1)/(in o)[lambdaL]`
`intEds_(1)Cos0 +intEds_(2)Cos 90^(@)+intEds_(3)Cos90^(@)=(lambdaL)/(in 0)`
`E int ds_(1)=(lambdaL)/(in 0)`
`Exx2pirL=(lambdaL)/(in 0)`
`E=(lambda)/(2 pi in 0 r)`
`vecE=(lambda)/(2 pi in e r) hatr`

`:. E_(x)=propx=400x`
`E_(y)=E_(z)=0`
Hence flux will exist onlt on left and right faces of cubt as `E_(x)ne0`
`:.vecE_(L)*a^(2)(n_(2))+vecE_(R)*a^(2)hat(n_(R))=(1)/(in 0) {qin}= phi`
`-E_(L)*a^(2)hat((n_(2)))+a^(2)E_(R)=pi_("Net")`
`phi_("Net")=-(400a)a^(2)+a^(2)(400xx2a)`
`-=400a^(3)+800a^(3)`
`=400a^(3)`
`=400xx(.1)^(3)`
`phi_("Net")=0.4Nm^(2)c^(-1)`
`:. phi_("Net") =(1)/(in 0){qin}`
`:. qin = in 0 phi_("Net")`
`8.85xx10^(-12)xx0.4`
`=3.540xx10^(-12)c`