Question

A square surface of side L m is in the plane of the paper. A uniform electric field \(\rightarrow\) E \(\bigg(\frac{V}{m}\bigg)\), also in the plane of the paper, is limited only to the lower half of the square surface, (see figure). The electric flux in SI units associated with the surface is :

• A. \(\frac{\text{EL}^2}{(2\epsilon_0)}\)
• B. \(\frac{\text{EL}^2}{2}\)
• C. Zero
• D. EL²

Answer 1

The Correct Option is C

To solve this problem, we need to determine the electric flux through a square surface under a specific condition.

Concept: The electric flux \(\Phi\) through a surface is given by the formula:

\[\Phi = \int \mathbf{E} \cdot d\mathbf{A}\]

where \(\mathbf{E}\) is the electric field vector and \(d\mathbf{A}\) is the differential area vector perpendicular to the surface. Electric flux essentially measures the flow of the electric field through a given area.

Details of the Problem:

• The surface is a square with side \(L\).
• An electric field \(\mathbf{E}\) is present only in the lower half of the square. The electric field vector is in the plane of the paper.
• Since the electric field is parallel to the surface (as it is in the plane of the paper), it does not pass through the surface perpendicularly.

Analysis:

• If the electric field is parallel to the surface, the angle \(\theta\) between the field direction \(\mathbf{E}\) and the area vector \(\mathbf{A}\) is 90 degrees.
• For electric flux calculation, \( \cos(90^\circ) = 0 \).
• This means the dot product \(\mathbf{E} \cdot d\mathbf{A} = E \cdot A \cdot \cos(90^\circ) = 0\).

Thus, since \(\cos(90^\circ) = 0\), the electric flux through the square surface is indeed Zero.

Conclusion: The correct answer is that the electric flux associated with the surface is Zero.