To understand the problem, recall the formula for electric flux ([tex]\( \Phi_E \)[/tex]). The electric flux through a surface is given by:
[tex]\[
\Phi_E = E \cdot A \cdot \cos(\theta)
\][/tex]
where:
- [tex]\( E \)[/tex] is the electric field strength,
- [tex]\( A \)[/tex] is the area of the surface,
- [tex]\( \theta \)[/tex] is the angle between the electric field and the normal (perpendicular) to the surface.
Let's solve the problem in two parts:
### (i) When the electric field is perpendicular to the surface:
When the electric field is perpendicular to the surface, the angle [tex]\( \theta = 0^\circ \)[/tex]. The cosine of 0 degrees is 1 ([tex]\( \cos(0^\circ) = 1 \)[/tex]).
Hence, the electric flux [tex]\( \Phi_E \)[/tex] in this situation can be calculated as:
[tex]\[
\Phi_E = E \cdot A \cdot \cos(0^\circ) = E \cdot A \cdot 1 = E \cdot A
\][/tex]
Given:
- [tex]\( E = 6.2 \times 10^5 \, \text{N/C} \)[/tex]
- [tex]\( A = 3.2 \, \text{m}^2 \)[/tex]
[tex]\[
\Phi_E = 6.2 \times 10^5 \, \text{N/C} \times 3.2 \, \text{m}^2 = 1984000 \, \text{N} \cdot \text{m}^2 / \text{C}
\][/tex]
So, the electric flux when the electric field is perpendicular to the surface is:
[tex]\[
1.98 \times 10^6 \, \text{N} \cdot \text{m}^2 / \text{C}
\][/tex]
### (ii) When the electric field is parallel to the surface:
When the electric field is parallel to the surface, the angle [tex]\( \theta = 90^\circ \)[/tex]. The cosine of 90 degrees is 0 ([tex]\( \cos(90^\circ) = 0 \)[/tex]).
Hence, the electric flux [tex]\( \Phi_E \)[/tex] in this case can be calculated as:
[tex]\[
\Phi_E = E \cdot A \cdot \cos(90^\circ) = E \cdot A \cdot 0 = 0
\][/tex]
So, the electric flux when the electric field is parallel to the surface is:
[tex]\[
0 \, \text{N} \cdot \text{m}^2 / \text{C}
\][/tex]
Combining both parts, the answer is (D):
[tex]\[
\boxed{1.98 \times 10^6 \, \text{N} \cdot \text{m}^2 / \text{C} ; 0}
\][/tex]
