Answer:
The angle between the electric field and sheet's normal vector is 80.96 degrees.
Explanation:
Given that,
Area of the flat sheet, [tex]A=3.8\ m^2[/tex]
Electric field, E = 10 N/C
Electric flux of the sheet, [tex]\phi=6\ Nm^2/C[/tex]
The electric flux is through the sheet is given by the dot product of electric field and the area vector. It is given by :
[tex]\phi=E{\cdot} A[/tex]
or
[tex]\phi=EA\ cos\theta[/tex]
[tex]\theta[/tex] is the angle between electric field and sheet's normal vector
So,
[tex]cos\theta=\dfrac{\phi}{EA}[/tex]
[tex]cos\theta=\dfrac{6}{10\times 3.8}[/tex]
[tex]\theta=cos^{-1}(0.157)[/tex]
[tex]\theta=80.96^{\circ}[/tex]
So, the angle between the electric field and sheet's normal vector is 80.96 degrees. Hence, this is the required solution.
