A cube of side `a` is placed such that the nearest face , which is parallel to the `yz` plane , is at a distance `a` from the origin . The electric field components are
`E_(x) = alpha x ^(1//2) , E_(y) = E_(z) = 0`.
The charge within the cube is

Options

A. `sqrt(2) alpha epsilon_(0) a^(5//2)`
B. `alpha epsilon_(0) a^(5//2)`
C. `( sqrt(2) - 1) alpha epsilon_(0) a^(5//2)`
D. zero

Answer

C

Explanation

`phi_(1) = E_(1) a^(2) = - alpha a^(1//2) a^(2) = - alpha a^(5//2)`
and `phi_(2) = E_(2) a^(2) = alpha( 2a)^(1//2) a^(2) = sqrt(2) alpha a^(5//2)`
`phi _(n et) = ( sqrt(2) - 1) alpha a^(5//2)`
Using the Gauss theorem , we get
`phi = (q_(in))/( epsilon_(0))`
or `q_(in) = phi epsilon_(0) = ( sqrt(2) - 1) epsilon_(0) alpha a^(5//2)`