A cube has sides of length `L = 0.300 m`. It is placed with one corner at the origin ( refer to Fig.2.119).The electric field is not uniform but is given by `vec( E) = ( -5.00 NC^(-1)) x hat(i) + (3.00 NC^(-1)) z hat(k)`.
The total flux passing through the cube is

Options

A. `- 0.135 Nm^(2) C^(-1)`
B. `- 0.054 Nm^(2) C^(-1)`
C. `0.081 Nm^(2) C^(-1)`
D. zero

Answer

B

Explanation

For `S_(1) , vec(A) = L^(2) ( - hat(j)) , vec(E ) = - 5 x hat(i) + 3 z hat(k)`.
Here `vec(E ) _|_ vec(A)` , so `phi_(s_(1)) = 0`.
For `S_(2) , vec(A) = L^(2) hat (k) , vec (E ) = - 5 x hat(i) + 3L hat(k)`.
Here `- 5 x hat(i)` is perpendicular to `vec(A)` , so no flux due to this component.
`phi_(s_(2)) = ( 3L hat(k)) . L^(2) hat (k) = 3 L^(3) = 81 xx 10^(-3) Nm^(2) C^(-1)`
For `S_(3) , vec(A) = L^(2) hat (j) , vec( E) = - 5 x hat(i) + 3 z hat(k)`.
Here `vec(E ) _|_ vec(A)` , so `phi_(s_(3)) = 0`.
For `S_(4) , vec(A) = - L^(2) hat(k) , vec(E ) = - 5 x hat (i) + 0 hat(k)` (since `z = 0`).
Here `vec _|_ vec(A)` , so `phi_(S_(4)) = 0`.
For `S_(5) , vec(A) = L^(2) hat (i) , vec (E ) = - 5 L hat(i) + 3z hat(k)`.
No flux due to `3 z hat (k) ` component.
`phi_(S_(5)) = ( - 5 L hat (i)) . L^(2) hat(i) = - 5 L^(3) = - 135 xx 10^(-3) Nm^(2) C^(-1)`
For` S_(6), vec(A) = - L^(2) hat(i) , vec( E) = 0 hat (i) + 3 z hat(k)` ( since `x = 0`).
Here `vec(E ) _|_ vec(A)` , so `phi_(s_(6)) = 0`.
Total flux through the cube is `phi = phi (S_(1) + S_(2) + S_(3) + S_(5) + S_(6))`
`= 81 xx 10^(-3) - 135 xx 10^(-3) = - 54 xx 10^(-3) Nm^(2) C^(-1)`
Now , `phi = ( q_(in))/( epsilon_(0)) or q_(in) = phi epsilon_(0) = - 0.054 epsilon_(0) C`