Question:
A large charged plane having surface charge density \(4.9 \times 10^{-6} \, \text{C/m}^2\) lies in the x-y plane. A circular plane of radius 1 cm is lying completely in the region where x, y, and z coordinates are all positive. When the plane’s normal makes an angle \(60^\circ\) with the z-axis, find the electric flux through the circular plane. \(\left(\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2\right)\)
A large charged plane having surface charge density \(4.9 \times 10^{-6} \, \text{C/m}^2\) lies in the x-y plane. A circular plane of radius 1 cm is lying completely in the region where x, y, and z coordinates are all positive. When the plane’s normal makes an angle \(60^\circ\) with the z-axis, find the electric flux through the circular plane. \(\left(\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2\right)\)
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Electric flux through a tilted surface: \(\Phi = EA \cos \theta\), with \(E\) from plane formula \(E = \sigma/(2 \epsilon_0)\).
Updated On: Jul 18, 2026
- 43.56 N m\(^2\)/C
- 48.36 N m\(^2\)/C
- 36.76 N m\(^2\)/C
- 32.56 N m\(^2\)/C
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The Correct Option is A
Solution and Explanation
Step 1: Recall electric flux formula.
Electric flux through a surface: \(\Phi = \vec{E} \cdot \vec{A} = EA \cos \theta\), where \(\theta\) is angle between field and normal to surface.
Step 2: Electric field due to large plane.
\[ E = \frac{\sigma}{2 \epsilon_0} = \frac{4.9 \times 10^{-6}}{2 \times 8.854 \times 10^{-12}} \approx 2.765 \times 10^5 \, \text{N/C} \]
Step 3: Area of circular plane.
\[ A = \pi r^2 = \pi (0.01)^2 = 3.1416 \times 10^{-4} \, \text{m}^2 \]
Step 4: Angle with electric field.
\(\theta = 60^\circ\), so flux: \(\Phi = E A \cos \theta\)
Step 5: Calculate flux.
\[ \Phi = 2.765 \times 10^5 \times 3.1416 \times 10^{-4} \times \cos 60^\circ \]
\[ \Phi \approx 43.56 \, \text{N m}^2/\text{C} \]
Step 6: Final conclusion.
Hence, the electric flux through the circular plane is:
\[ \boxed{43.56 \, \text{N m}^2/\text{C}} \]
Electric flux through a surface: \(\Phi = \vec{E} \cdot \vec{A} = EA \cos \theta\), where \(\theta\) is angle between field and normal to surface.
Step 2: Electric field due to large plane.
\[ E = \frac{\sigma}{2 \epsilon_0} = \frac{4.9 \times 10^{-6}}{2 \times 8.854 \times 10^{-12}} \approx 2.765 \times 10^5 \, \text{N/C} \]
Step 3: Area of circular plane.
\[ A = \pi r^2 = \pi (0.01)^2 = 3.1416 \times 10^{-4} \, \text{m}^2 \]
Step 4: Angle with electric field.
\(\theta = 60^\circ\), so flux: \(\Phi = E A \cos \theta\)
Step 5: Calculate flux.
\[ \Phi = 2.765 \times 10^5 \times 3.1416 \times 10^{-4} \times \cos 60^\circ \]
\[ \Phi \approx 43.56 \, \text{N m}^2/\text{C} \]
Step 6: Final conclusion.
Hence, the electric flux through the circular plane is:
\[ \boxed{43.56 \, \text{N m}^2/\text{C}} \]
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