A cube has sides of length `L`. It is placed with one corner at the origin ( refer to Fig.2.119). The electric field is uniform and given by `vec€ = - B hat (i) + C hat (j) - D hat(k)` , where `B , C , and D` are positive constants.
The total flux passing through the cube is

Options

A. `( B + C + D)L^(2)`
B. `2 ( B + C + D) L^(2)`
C. `6 ( B + C + D) L^(2)`
D. zero

Explanation

To find the total electric flux passing through the cube, we will use Gauss's Law, which states that the total electric flux \( \Phi \) through a closed surface is equal to the charge enclosed \( Q \) divided by the permittivity of free space \( \epsilon_0 \). However, in this case, since there is no charge enclosed within the cube, we will calculate the flux through each face of the cube and sum them up.

### Step-by-Step Solution:

1. **Identify the Electric Field and Cube Orientation:**
The electric field is given by:
\[ \vec{E} = -B \hat{i} + C \hat{j} - D \hat{k} \]
The cube has sides of length \( L \) and is placed with one corner at the origin.

2. **Calculate the Area Vector for Each Face:**
The area vector \( \vec{A} \) for each face points outward from the cube. The cube has 6 faces, and we will denote them as follows:
- Face 1 (S1): \( \hat{i} \) (x = 0)
- Face 2 (S2): \( -\hat{i} \) (x = L)
- Face 3 (S3): \( \hat{j} \) (y = 0)
- Face 4 (S4): \( -\hat{j} \) (y = L)
- Face 5 (S5): \( \hat{k} \) (z = 0)
- Face 6 (S6): \( -\hat{k} \) (z = L)

3. **Calculate the Flux Through Each Face:**
The electric flux \( \Phi \) through a surface is given by:
\[ \Phi = \vec{E} \cdot \vec{A} \]
- For Face 1 (S1):
\[ \Phi_1 = \vec{E} \cdot (L^2 \hat{i}) = (-B \hat{i} + C \hat{j} - D \hat{k}) \cdot (L^2 \hat{i}) = -B L^2 \]
- For Face 2 (S2):
\[ \Phi_2 = \vec{E} \cdot (-L^2 \hat{i}) = (-B \hat{i} + C \hat{j} - D \hat{k}) \cdot (-L^2 \hat{i}) = B L^2 \]
- For Face 3 (S3):
\[ \Phi_3 = \vec{E} \cdot (L^2 \hat{j}) = (-B \hat{i} + C \hat{j} - D \hat{k}) \cdot (L^2 \hat{j}) = C L^2 \]
- For Face 4 (S4):
\[ \Phi_4 = \vec{E} \cdot (-L^2 \hat{j}) = (-B \hat{i} + C \hat{j} - D \hat{k}) \cdot (-L^2 \hat{j}) = -C L^2 \]
- For Face 5 (S5):
\[ \Phi_5 = \vec{E} \cdot (L^2 \hat{k}) = (-B \hat{i} + C \hat{j} - D \hat{k}) \cdot (L^2 \hat{k}) = -D L^2 \]
- For Face 6 (S6):
\[ \Phi_6 = \vec{E} \cdot (-L^2 \hat{k}) = (-B \hat{i} + C \hat{j} - D \hat{k}) \cdot (-L^2 \hat{k}) = D L^2 \]

4. **Sum the Fluxes:**
Now, we sum all the fluxes through the six faces:
\[ \Phi_{\text{total}} = \Phi_1 + \Phi_2 + \Phi_3 + \Phi_4 + \Phi_5 + \Phi_6 \]
Substituting the values:
\[ \Phi_{\text{total}} = (-B L^2) + (B L^2) + (C L^2) + (-C L^2) + (-D L^2) + (D L^2) \]
Simplifying:
\[ \Phi_{\text{total}} = 0 \]

### Final Answer:
The total electric flux passing through the cube is:
\[ \Phi_{\text{total}} = 0 \]