Question

The electric field in a region is given by \[\overrightarrow{E}=\dfrac{3}{5}{{E}_{0}}\hat{j}\] with \[{{E}_{0}}=2\times {{10}^{3}}N{{C}^{-1}}\]. Find the flux of this field through a rectangular surface of area \[0.2{{m}^{2}}\] parallel to the Y-Z plane.
A. \[320N{{m}^{2}}{{C}^{-1}}\]
B. \[240N{{m}^{2}}{{C}^{-1}}\]
C. \[400N{{m}^{2}}{{C}^{-1}}\]
D. None of these

Answer 1

Hint: Apply Gauss’s law. In GAUSS’S law the direction of the area-vector is along the normal to the corresponding surface, so find the direction of area vector( x, y or z direction) then use \[\hat{j}\cdot \hat{i}=0\].

Complete step by step answer:
Apply Gauss’s law
\[\phi =\overrightarrow{E}\cdot \overrightarrow{\Delta S}\]
Where, \[\phi \]= flux of electric field through chosen surface
\[\overrightarrow{E}\]= electric field
\[\overrightarrow{\Delta S}\]= area vector
Given data
\[{{E}_{0}}=2\times {{10}^{3}}N{{C}^{-1}}\]
\[\overrightarrow{E}=\dfrac{3}{5}{{E}_{0}}\hat{j}\]
Rectangle surface of area \[0.2{{m}^{2}}\] parallel to the Y-Z plane

Area vector is in x direction because area vector is perpendicular to given plane and x is perpendicular to y-z plane
\[\overrightarrow{\Delta S}=0.2\hat{i}\]
\[\phi =\overrightarrow{E}\cdot \overrightarrow{\Delta S}\]
\[\begin{align}
  & \phi =\overrightarrow{E}\cdot \overrightarrow{\Delta S} \\
 & \overrightarrow{E}=\dfrac{3}{5}{{E}_{0}}\hat{j} \\
 & \phi =\dfrac{3}{5}{{E}_{0}}\hat{j}\cdot 0.2\hat{i} \\
 & (\hat{j}\cdot \hat{i}=0) \\
 & \phi =0 \\
\end{align}\]
Hence, option D is correct.

Note: Remember the direction of the area-vector is always along the normal to the corresponding surface. If the electric field is perpendicular to the surface, it is parallel to the area vector. If the electric field is parallel to the surface, the area vector is perpendicular to the electric field.