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The electric field in a region is radial...

The electric field in a region is radially outward with magnitude `E = ar`. If `a = 100 Vm^(-2)` and `R = 0.30m`, then the value of charge contained in a sphere of radius R centred at the origin is `W xx 10^(-10)C`. Find W.

Text Solution

Verified by Experts


Given `E=alphar`, when `r=R`
`ER=alphaR`
So `phi=E_(R)("area")=alphaR4piR^(2)`
by gauss's theorem the net electric flux is
`(1)/(epsilon_(0))` charged enclosed
`alphaR4piR^(2)=(1)/(epsilon_(0))Q_("enclosed")`
`thereforeQ_("enclosed")=(4piepsilon_(0))alphaR^(3)`
Given `R=0.30m,alpha=100Vm^(-2)`
`Q_("enclosed")=(1)/(9xx10^(9))xx100xx(0.30)^(3)=3xx10^(-10)C`
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Knowledge Check

  • The electric field in a region is radially outward with magnitude E = (A)/(gamma) . The charge contained in a sphere of radius gamma_(0) centered at the origin is

    A
    `(1)/(4piepsi_(0)) Agamma_(0)^(2)`
    B
    `4piepsi_(0)Agamma_(0)`
    C
    `(4piepsi_(0)A)/(gamma_(0))`
    D
    `(1)/(4piepsi_(0)) (A)/(gamma_(0))`

  • The electric field in a region is radially outward with magnitude E=Agamma_(0) . The charge contained in a sphere of radius gamma_(0) centered at the origin is

    A
    `1/(4piepsilon_(0)) Agamma_(0)^(3)`
    B
    `4 piepsilon_(0)Agamma_(0)^(3)`
    C
    `(4piepsilon_(0)A)/gamma_(0)`
    D
    `1/(4piepsilon_(0))A/(gamma_(0)^(3)`

  • The electric field in a region is radially outward with magnitude E=Ar_(0) . The charge contained in a sphere of radius r_(0) centred at the origin is

    A
    `(Ar_(0)^(3))/(4piepsilon_(0))`
    B
    `(4piepsilon_(0)A)/(r_(0))`
    C
    `4piepsilon_(0)Ar_(0)^(3)`
    D
    `(A)/(4pir^(3)epsilon_(0))`

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