Define electric flux. Write its SI units.
(b) The electric field components due to a charge inside the cube of side `0.1 m ` area as shown in `E_x =alpha x , ` where `alpha =500 N//C m, E_Y =0 and E_Z =0.`
Calculate (i) the flux through the cube , and
(ii) the charges inside the cube.

Explanation

(i) As `E_y and E_Z ` are zero and E =`E_x = alpha x ,` hence electric flux is linked only with two faces of the cube lying in
y-z plane (i.e., perpendicular to `oversetto (E_x) `)
At the position of left face of cube `x= 0.1 m , "hence" E_x = alpha x =500 xx 0.1 = 50 N//C ` and surface area of face `s= (0.1) ^(2) =0.001 m^(2) `
` therefore " " ` Flux on this face ` phi_1 =E_x s= 50 xx0.01 = 0.5 N m^(2) C ^(-1) `(inward )
Again on the opposite face of cube (i.e., right face ) x= 0.2 m and
hence ` " " E_x. =alpha x= 500 xx 0.2 = 100 N//C `
Flux on this face `phi_2 =E_x.s =100 xx 0.01 =1 N m^(2) C^(-1) ` (outward )
` therefore ` Net electric flux through the cube ` phi= 0.5 N m^(2) C^(-1) ("inward") + 1.0N m^(2) C^(-1) ("outward") = +0.5 N m^(2) C^(-1) ` (outward)
(ii) From Gauss.s law ` phi= (q)/( in_0) `
`rArr " " ` Charge inside the cube ` q= phi in_0 =+ 0.5 xx 8.85 xx 10 ^(-12) =+ 4.42 xx 10 ^(-12) C `