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A square surface of side $L$ metres is in the plane of the paper. A uniform electric field $\vec{E}$ (volt/m), 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 :-
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A
$\dfrac{EL^2}{2\varepsilon_0}$
B
$\dfrac{EL^2}{2}$
C
Zero
D
$EL^2$
Detailed Solution
Electric flux through a surface: $\phi = \vec{E}\cdot\vec{S} = ES\cos\theta$, where $\theta$ is the angle between the field and the area vector.
The area vector of the square is perpendicular to its plane, i.e. perpendicular to the plane of the paper.
The electric field lies in the plane of the paper, so it is perpendicular to the area vector: $\theta = 90^\circ$.
$\phi = ES\cos 90^\circ = 0$
The field lines run along the surface and none of them passes through it, so the flux is zero whether the field covers the whole square or only half of it.
The area vector of the square is perpendicular to its plane, i.e. perpendicular to the plane of the paper.
The electric field lies in the plane of the paper, so it is perpendicular to the area vector: $\theta = 90^\circ$.
$\phi = ES\cos 90^\circ = 0$
The field lines run along the surface and none of them passes through it, so the flux is zero whether the field covers the whole square or only half of it.