Looking for classes? Ksquare Career Institute, Bengaluru →
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 $\theta$ with the horizontal side of the square as shown in figure. The electric flux linked to the surface, in units of volt-m, is
[add image]
[add image]
A
Zero
B
$EL^2$
C
$EL^2\cos\theta$
D
$EL^2\sin\theta$
Detailed Solution
Electric flux: $\phi = \vec{E}\cdot\vec{S} = ES\cos\alpha$, where $\alpha$ 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 field lies in the plane of the paper, so it is perpendicular to the area vector: $\alpha = 90^\circ$, whatever the angle $\theta$ is.
$\phi = EL^2\cos90^\circ = 0$
The field lines only skim along the surface and none pass through it, so the flux is zero.
The area vector of the square is perpendicular to its plane, i.e., perpendicular to the plane of the paper.
The field lies in the plane of the paper, so it is perpendicular to the area vector: $\alpha = 90^\circ$, whatever the angle $\theta$ is.
$\phi = EL^2\cos90^\circ = 0$
The field lines only skim along the surface and none pass through it, so the flux is zero.