Question

In a region, the intensity of an electric field is given by \( \vec{E} = (2\hat{i} + 3\hat{j} + \hat{k}) \) NC\(^{-1} \). The electric flux through a surface of area \( 10 \hat{i} \) m\(^2 \) in the region is

• A. \( 5 \) Nm\(^2 \)C\(^{-1} \)
• B. \( 10 \) Nm\(^2 \)C\(^{-1} \)
• C. \( 15 \) Nm\(^2 \)C\(^{-1} \)
• D. \( 20 \) Nm\(^2 \)C\(^{-1} \)

Hint:

The electric flux is the scalar product (dot product) of the electric field vector and the area vector. The area vector's direction is normal to the surface. Perform the dot product of the given electric field and area vector components to find the electric flux.

Answer 1

The Correct Option is D

The electric flux \( \Phi_E \) through a surface is given by the dot product of the electric field vector \( \vec{E} \) and the area vector \( \vec{A} \): $$ \Phi_E = \vec{E} \cdot \vec{A} $$ The electric field is given as \( \vec{E} = (2\hat{i} + 3\hat{j} + \hat{k}) \) NC\(^{-1} \).
The surface area is given as \( 10 \hat{i} \) m\(^2 \).
The area vector \( \vec{A} \) is perpendicular to the surface and its magnitude is the area of the surface.
Since the area vector is given as \( 10 \hat{i} \), it means the surface is oriented such that its normal is along the x-axis, and the magnitude of the area is 10 m\(^2 \).
So, \( \vec{A} = 10 \hat{i} \) m\(^2 \).
Now, we can calculate the electric flux: $$ \Phi_E = (2\hat{i} + 3\hat{j} + \hat{k}) \cdot (10 \hat{i}) $$ The dot product of the unit vectors are: \( \hat{i} \cdot \hat{i} = 1 \), \( \hat{j} \cdot \hat{i} = 0 \), \( \hat{k} \cdot \hat{i} = 0 \).
$$ \Phi_E = (2 \times 10) + (3 \times 0) + (1 \times 0) $$ $$ \Phi_E = 20 + 0 + 0 $$ $$ \Phi_E = 20 \) Nm\(^2 \)C\(^{-1} \) $$ The electric flux through the surface is 20 Nm\(^2 \)C\(^{-1} \).