Question

A square surface of side $L$ meter in the plane of the paper is placed in a uniform electric field $E(volt/m)$ acting along the same plane at an angle $q$ with the horizontal side of the square as shown in figure. The electric flux linked to the surface, in units of volt $m$ is

• A. $ EL^2 $
• B. $ EL^2 \cos \theta$
• C. $ EL^2 \sin \theta$
• D. zero

Answer 1

The Correct Option is D

To solve this problem, we need to determine the electric flux through the square surface. The electric flux, denoted by $\Phi_E$, through a surface is given by the formula:

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

For a flat surface placed in a uniform electric field, this simplifies to:

\[\Phi_E = E \cdot A \cdot \cos(\theta)\]

where:

• $E$ is the magnitude of the electric field,
• $A$ is the area of the surface, and
• $\theta$ is the angle between the electric field and the normal (perpendicular) to the surface.

Now, let's analyze the given situation:

• The square surface has sides of length $L$. Hence, its area is $A = L^2$.
• The electric field $E$ is given to be in the plane of the paper.
• Since the electric field is along the same plane as the square, the angle $\theta$ between the electric field and the normal to the surface is 90 degrees, i.e. $\theta = 90^\circ$.
• The cosine of 90 degrees is zero: $\cos(90^\circ) = 0$.

Thus, substituting these values into the flux formula, we get:

\[\Phi_E = E \cdot L^2 \cdot \cos(90^\circ) = E \cdot L^2 \cdot 0 = 0\]

This calculation shows that the electric flux linked to the surface is zero.

Therefore, the correct answer is zero.