The cube shown in Fig. 2.119 has sides of length `L = 10.0 cm` . The electric field is uniform , has a magnitude `E = 4.00 xx 10^(3) NC^(-1) ` and is parallel to the `xy` plane at an angle of `37^(@)` measured from the `+ x - axis` toward the `+ y - axis` .

Electric flux passing through surface `S_(6)` is

Options

A. `- 24 Nm^(2) C^(-1)`
B. `- 24 Nm^(2) C^(-1)`
C. `32 Nm^(2) C^(-1)`
D. `- 32 Nm^(2) C^(-1)`

Answer

D

Explanation

`Phi = vec(E ) . vec(A) = EA cos theta` , where , `vec(A) = A hat(n)`
`hat(n)_(S_(1) = - hat(j) `("l eft"),
`Phi_(S_1))= -( 4 xx 10^(3) NC^(-1)) (0.1 m)^(2) cos (90^(@) - 37^(@)) = - 24 Nm^(2) C^(-1)`
` hat (n)_(S_(2)) = + hat(k)` (top),
`Phi_(S_(2)) = - (4 xx 10^(3) NC^(-1)) (0.1 m)^(2) cos 90^(@) = 0`
`hat (n)_(S_(3)) = + hat(j)` ( right),
`Phi_(S_(3)) = + (4 xx 10^(3) NC^(-1)) (0.1 m)^(2) cos(90^(@) - 37^(@))`
`= + 24 Nm^(2) C^(-1)`
`hat(n) _(S_(4)) = - hat(k)` (bottom),
`Phi_(S_(4)) = ( 4 xx 10^(3) NC^(-1)) (0.1 m)^(2) cos 90^(@) = 0`
`hat(S_(5)) = + hat(i) ` (front),
`Phi_(S_(5)) = + 4 ( 4 xx 10^(3) NC^(-1)) (0.1 m)^(2) cos 37^(@) = 32 Nm^(2) C^(-1)`
`hat(n)_(S_(6)) = - hat(i) ` (back),
`Phi_(S_(6)) = - ( 4 xx 10^(3) NC^(-1)) (0.1 m)^(2) cos 37^(@) = 32 Nm^(2) C^(-1)`
`hat(n)_(S_(6)) = - ( 4 xx 10^(3) NC^(-1)) (0.1 m)^(2) cos 37^(@) = - 32 Nm^(2) C^(-1)`
As the field is uniform , the total flux through the cube must be zero , any flux entering the cube must also leave it.