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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.

Text Solution

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

Using Gauss’ law, derive an expression for the electric field at a point near an infinitely long straight uniformly charged wire.

Knowledge Check

  • The electric components in the figure are E_(x)=alphax^(1//2) ,E_(y)=0,E_(z)=0 where alpha=800N//m^(2) if a=0.1m is the side of cube then the charge with in the cube is

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    `6xx10^(-12)C`
    C
    `2.5xx10^(-12)C`
    D
    Zero
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