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 flux passing through the surface `S_(5)` will be

Options

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

Explanation

To solve the problem of finding the electric flux through the surface \( S_5 \) of the cube, we can follow these steps:

### Step 1: Understand the Electric Field and Surface Area
The electric field is given by:
\[ \vec{E} = -5 \, \text{N/C} \, \hat{i} + 3 \, \text{N/C} \, \hat{k} \]
The cube has a side length \( L = 0.300 \, \text{m} \). The surface \( S_5 \) is one of the faces of the cube, specifically the face that is perpendicular to the \( x \)-axis.

### Step 2: Determine the Area Vector
For surface \( S_5 \), which is facing in the negative \( x \)-direction, the area vector \( \vec{A} \) can be expressed as:
\[ \vec{A} = -A \hat{i} \]
where \( A \) is the area of the face. The area of one face of the cube is:
\[ A = L^2 = (0.300 \, \text{m})^2 = 0.09 \, \text{m}^2 \]
Thus, the area vector becomes:
\[ \vec{A} = -0.09 \hat{i} \, \text{m}^2 \]

### Step 3: Calculate the Electric Flux
The electric flux \( \Phi \) through the surface is given by the dot product of the electric field and the area vector:
\[ \Phi = \vec{E} \cdot \vec{A} \]
Substituting the values:
\[ \Phi = \left(-5 \hat{i} + 3 \hat{k}\right) \cdot \left(-0.09 \hat{i}\right) \]
Calculating the dot product:
\[ \Phi = -5 \cdot (-0.09) + 3 \cdot 0 = 0.45 \, \text{N m}^2/\text{C} \]

### Step 4: Finalize the Result
The final electric flux through the surface \( S_5 \) is:
\[ \Phi = 0.45 \, \text{N m}^2/\text{C} \]

### Summary
The electric flux passing through the surface \( S_5 \) of the cube is \( 0.45 \, \text{N m}^2/\text{C} \).

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