Question

Given a uniform electric field $\vec E = 5 \times {10^3}\hat i\,N{C^{ - 1}}$, find the flux of this field through a square of $10\,cm$ on a side whose plane is parallel to the $YZ$ plane. What would be the flux through the same square if the plane makes a ${30^ \circ }$ angle with the $X$ axis?

Answer 1

Hint
The flux of the electric field is determined by the electric flux on the electric field formula. Initially the electric field parallel to the $YZ$ plane, so the angle of the electric flux is taken as zero, then the electric flux can be determined. Then the same electric field is ${30^ \circ }$ with the $X$ axis, then the opposite angle of ${60^ \circ }$ is taken, then the flux can be determined.
The electric flux in the electric field is given by,
$\Rightarrow F = ES\cos \theta $
Where, $F$ is the electric flux, $E$ is the electric field, $S$ is the area of the square and $\theta $ is the angle of the square.

Complete step by step answer
Given that, The electric field is, $\vec E = 5 \times {10^3}\hat i\,N{C^{ - 1}}$,
The length of the square is, $l = 10\,cm$,
The angle of the square with the $X$ axis, $\theta = {30^ \circ }$
Now, The electric flux in the electric field is given by,
$\Rightarrow F = ES\cos \theta \,.................\left( 1 \right)$
Initially the electric field parallel to the $YZ$ plane, so the angle of the electric flux is taken as zero. $\theta = {0^ \circ }$
By substituting the electric field and the area of the square and the angle in the above equation, then
$\Rightarrow F = 5 \times {10^3} \times {\left( {10 \times {{10}^{ - 2}}} \right)^2}\cos {0^ \circ }$
From the trigonometry, the value of the $\cos {0^ \circ } = 1$, then
$\Rightarrow F = 5 \times {10^3} \times {\left( {10 \times {{10}^{ - 2}}} \right)^2}$
By squaring the terms in the above equation, then
$\Rightarrow F = 5 \times {10^3} \times 100 \times {10^{ - 4}}$
By multiplying the terms in the above equation, then
$\Rightarrow F = 50\,N{C^{ - 1}}{m^2}$
Now the square makes an ${30^ \circ }$ angle with the $X$ axis, then the angle is $\theta = {60^ \circ }$
Substitute the electric field and the area of the square and the angle in the above equation (1), then
$\Rightarrow F = 5 \times {10^3} \times {\left( {10 \times {{10}^{ - 2}}} \right)^2}\cos {60^ \circ }$
From the trigonometry, the value of the $\cos {60^ \circ } = \dfrac{1}{2}$, then
$\Rightarrow F = 5 \times {10^3} \times {\left( {10 \times {{10}^{ - 2}}} \right)^2} \times \dfrac{1}{2}$
By squaring the terms in the above equation, then
$\Rightarrow F = 5 \times {10^3} \times 100 \times {10^{ - 4}} \times \dfrac{1}{2}$
By multiplying the terms in the above equation, then
$\Rightarrow F = 25\,N{C^{ - 1}}{m^2}$
Thus, the electric flux when the square is parallel to the $YZ$ plane is, $F = 50\,N{C^{ - 1}}{m^2}$.
Thus, the electric flux when the square is ${30^ \circ }$ angle with the $X$ axis is, $F = 50\,N{C^{ - 1}}{m^2}$.

Note
The electric flux is directly proportional to the electric field in the square and the area of the square and the angle of the square. If the electric field in the square and the area of the square increases, the electric flux in the square increases.