Question

An electric field of 1N/C along Y direction. The flux passing through the square of 1m placed in the XY plane inside the electric field is?
A.$1.0\;{\rm{N}}{{\rm{m}}^{\rm{2}}}{\rm{/C}}$
B. $10\;{\rm{N}}{{\rm{m}}^{\rm{2}}}{\rm{/C}}$
C. $2.0\;{\rm{N}}{{\rm{m}}^{\rm{2}}}{\rm{/C}}$
D. Zero

Answer 1

Hint: We can find the electric field orientation using Gauss's law. Electric flux is the number of field lines passing through a given surface. Gauss's law deals with the electric flux through a closed surface.

Complete step by step answer:
We all know that the Electric flux is defined as the total number of electric field lines passing through a given area in a unit time. When the plane is inclined at an angle $\theta $ to the electric field lines, the total flux $\phi $ is given $\phi = EA\cos \theta $ where E is the magnitude of the electric field, A is the area of the plane surface, and $\theta $ is the angle at which the plane is inclined to the electric field line.
Here, the area vector is along the z-axis, perpendicular to the XY plane and hence the value $\theta = 90^\circ $.
Therefore, the flux becomes,
$\begin{array}{l}
\phi = EA\cos 90^\circ \\
 = 0\\
\end{array}$
Therefore, the flux passing through the square is zero, and the correct option is (D).

Additional Information:
The Gaussian surface can be of any shape and size because it's only the amount of flux coming out of the charge distribution. We have to concentrate, which will be the same if allowed to pass through any shape and surface. It's like a cloud coming out from any three-dimensional surface. We use this gauss law to find the electric field's magnitude at any point due to charge distribution.

Note:We can say that the electric flux is nothing, just measuring the electric field's distribution through an area. Electric flux is only used to tell the electric field strength at any distance from the charge responsible for the field.