An electric field is given by `(6 hat i + 5 hat j + 3 hat k)`N/C. The electric flux through a surface area `30 hat i m^2` lying in YZ-plane (in SI unit) is :

Options

A. 90
B. 150
C. 180
D. 60

Answer

To find the electric flux through a surface area in the YZ-plane, we can use the formula for electric flux, which is given by:

\[ \Phi_E = \vec{E} \cdot \vec{A} \]

where:
- \(\Phi_E\) is the electric flux,
- \(\vec{E}\) is the electric field vector,
- \(\vec{A}\) is the area vector.

### Step 1: Identify the Electric Field Vector
The electric field is given as:

\[ \vec{E} = 6 \hat{i} + 5 \hat{j} + 3 \hat{k} \, \text{N/C} \]

### Step 2: Identify the Area Vector
Since the surface area is lying in the YZ-plane, the area vector will be perpendicular to the surface and will point along the X-axis. The magnitude of the area is given as \(30 \, \text{m}^2\). Therefore, the area vector can be expressed as:

\[ \vec{A} = 30 \hat{i} \, \text{m}^2 \]

### Step 3: Calculate the Dot Product
Now we can calculate the electric flux using the dot product of the electric field vector and the area vector:

\[ \Phi_E = \vec{E} \cdot \vec{A} = (6 \hat{i} + 5 \hat{j} + 3 \hat{k}) \cdot (30 \hat{i}) \]

Calculating the dot product:

\[ \Phi_E = 6 \cdot 30 + 5 \cdot 0 + 3 \cdot 0 = 180 \, \text{N m}^2/\text{C} \]

### Step 4: Conclusion
The electric flux through the surface area is:

\[ \Phi_E = 180 \, \text{N m}^2/\text{C} \]