Step 1: Understanding the Concept:
Electric flux ($\Phi$) measures the total number of electric field lines passing perpendicularly through a given surface area. A crucial geometrical detail in electrostatics is that the area vector ($\vec{A}$) is always oriented perpendicular (normal) to the surface plane itself. Therefore, the angle used in the flux calculation must be measured between this normal vector and the electric field lines, not between the flat plane sheet and the field lines.
Step 2: Key Formula or Approach:
The formula for electric flux through a flat surface in a uniform field is:
\[ \Phi = \vec{E} \cdot \vec{A} = E A \cos\theta \]
Where:
- $E$ is the magnitude of the uniform electric field.
- $A$ is the area of the circular sheet ($A = \pi r^2$).
- $\theta$ is the angle between the electric field vector $\vec{E}$ and the normal area vector $\vec{A}$.
Step 3: Detailed Explanation:
First, let's establish the proper angle $\theta$. The problem states that the circular plane sheet itself makes an angle of 60° with the electric field lines.
Since the area vector $\vec{A}$ sits at exactly 90° to the sheet's surface, the angle $\theta$ between $\vec{E}$ and $\vec{A}$ is:
\[ \theta = 90^\circ - 60^\circ = 30^\circ \]
Next, convert the radius of the circular sheet from centimeters to standard SI units (meters):
\[ r = 10 \text{ cm} = 0.1 \text{ m} \]
Calculate the area ($A$) of the circular sheet:
\[ A = \pi r^2 = \pi \times (0.1)^2 = 0.01\pi \text{ m}^2 \approx 3.1416 \times 10^{-2} \text{ m}^2 \]
Now, substitute the values into the electric flux formula:
\[ \Phi = E A \cos 30^\circ \]
\[ \Phi = (5 \times 10^5) \times (3.1416 \times 10^{-2}) \times \frac{\sqrt{3}}{2} \]
Substitute $\frac{\sqrt{3}}{2} \approx 0.866$:
\[ \Phi = (5 \times 10^5) \times (3.1416 \times 10^{-2}) \times 0.866 \]
\[ \Phi = 1.5708 \times 10^4 \times 0.866 \approx 1.3603 \times 10^4 \text{ N m}^2\text{C}^{-1} \]
Step 4: Final Answer:
The electric flux through the sheet is $1.36 \times 10^4$ N m²C$^{-1}$.