Question

The electric field in a region is given by $ \vec E = 5.0k{\text{ N/C}} $ . Calculate the electric flux through a square of side $ 10.0cm $ in the following cases
(A) The square is along the XY plane
(B) The square is along XZ plane
(C) The normal to the square makes an angle of $ {45^ \circ } $ with the Z axis.

Answer 1

Hint:
To solve this question, we have to divide the system into three systems, each having a pair of two charges. Then we need to find the potential energy for each of the three systems, and finally we need to add the three of them to get the total potential energy of the system.
Formula used: The formula used in solving this question is
 $ \varphi = EA\cos \theta $ , here $ \varphi $ is the electric flux through a surface of area $ A $ due to an electric field of magnitude $ E $ , and $ \theta $ is the angle between the electric field and the normal to the surface.

Complete step by step answer:
According to the question, $ \vec E = 5.0k{\text{ N/C}} $
The electric field vector indicates that it is parallel to the unit vector along the Z-axis.
Hence, it is along the Z-axis.
We know that the electric flux is given by
 $ \varphi = EA\cos \theta $
According to the question, the side of the square is
  $ a = 10cm $
 $ a = \dfrac{{10}}{{100}}m = 0.1m $
The area $ A $ of the square is
 $ A = {a^2} $
 $ A = {(0.1)^2} = 0.01{m^2} $
(i) Since the square is along the XY plane, so the normal to the square is parallel to the Z-axis.
Since the electric field is parallel to the Z-axis, so the normal to the square is parallel to the electric field, i.e. $ \theta = {0^ \circ } $
Therefore, the flux in this case is
 $ \varphi = EA\cos {0^ \circ } $
Substituting $ E = 5.0N/C $ and $ A = 0.01{m^2} $
 $ \varphi = 5.0 \times 0.01N{m^2}/C $
 $ \varphi = 5.0 \times {10^{ - 2}}N{m^2}/C $
Hence the electric flux in this case is equal to $ 5.0 \times {10^{ - 2}}N{m^2}/C $
(ii) Now the square is along the XZ plane, so the normal to the square is perpendicular to the electric field, that is, $ \theta = {90^ \circ } $
Therefore, the flux in this case is
 $ \varphi = EA\cos {90^ \circ } $
Since $ \cos {90^ \circ } = 0 $
 $ \therefore \varphi = 0 $
Hence, the flux is equal to zero in this case.
(iii)
In this case, the normal to the square makes an angle of $ {45^ \circ } $ with the Z axis, or equivalently with the electric field, so $ \theta = {45^ \circ } $
So, the flux in this case is
 $ \varphi = EA\cos {45^ \circ } $
Substituting $ E = 5.0N/C $ and $ A = 0.01{m^2} $
 $ \varphi = 5.0 \times 0.01 \times 0.707 $
 $ \varphi = 3.5 \times {10^{ - 2}}N{m^2}/C $
Hence the electric flux in this case is equal to $ 3.5 \times {10^{ - 2}}N{m^2}/C $ .

Note:
Do remember to take the angle of the electric field vector with the normal vector to the surface. Do not make the mistake of taking the angle with the surface itself. Otherwise, it will produce incorrect results.