Question

An electric field is given by \( \vec{E} = (6\hat{i} + 5\hat{j} + 3\hat{k}) \, \text{N/C} \). The electric flux through a surface area \( 30\hat{i} \, \text{m}^2 \) lying in the YZ-plane (in SI units) is:

• A. \( 180 \)
• B. \( 90 \)
• C. \( 150 \)
• D. \( 60 \)

Hint:

When calculating electric flux, only the components of the electric field vector that align with the area vector contribute to the dot product.

Answer 1

The Correct Option is A

An electric field is given by \( \vec{E} = (6\hat{i} + 5\hat{j} + 3\hat{k}) \, \text{N/C} \). The electric flux through a surface area \( 30\hat{i} \, \text{m}^2 \) lying in the YZ-plane (in SI units) is:

Solution:

The electric flux (\( \Phi \)) through a surface is calculated using the dot product of the electric field vector (\( \vec{E} \)) and the area vector (\( \vec{A} \)):

\( \Phi = \vec{E} \cdot \vec{A} \)

Given:

• Electric field: \( \vec{E} = (6\hat{i} + 5\hat{j} + 3\hat{k}) \, \text{N/C} \)
• Area vector: \( \vec{A} = 30\hat{i} \, \text{m}^2 \)

The area vector \( 30\hat{i} \, \text{m}^2 \) indicates that the surface lies in the YZ-plane, and the vector is directed along the X-axis (perpendicular to the YZ-plane).

Now, calculate the dot product:

\( \Phi = (6\hat{i} + 5\hat{j} + 3\hat{k}) \cdot (30\hat{i}) \)

Recall that:

• \( \hat{i} \cdot \hat{i} = 1 \)
• \( \hat{j} \cdot \hat{i} = 0 \)
• \( \hat{k} \cdot \hat{i} = 0 \)

Therefore:

\( \Phi = (6 \times 30)(\hat{i} \cdot \hat{i}) + (5 \times 30)(\hat{j} \cdot \hat{i}) + (3 \times 30)(\hat{k} \cdot \hat{i}) \)

\( \Phi = (6 \times 30)(1) + (5 \times 30)(0) + (3 \times 30)(0) \)

\( \Phi = 180 \, \text{N m}^2/\text{C} \)

Thus, the electric flux through the surface is 180 N m²/C.

Final Answer:

\( \Phi = 180 \, \text{N m}^2/\text{C} \)