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. 43.56 N m\(^2\)/C
• B. 48.36 N m\(^2\)/C
• C. 36.76 N m\(^2\)/C
• D. 32.56 N m\(^2\)/C

Hint:

Electric flux through a tilted surface: \(\Phi = EA \cos \theta\), with \(E\) from plane formula \(E = \sigma/(2 \epsilon_0)\).

Answer 1

The Correct Option is A

Step 1: Get the sheet's field straight from the given constant.
Since $\dfrac{1}{4\pi\epsilon_0} = 9\times10^9$, we get $\dfrac{1}{2\epsilon_0} = 2\pi\times9\times10^9 = 5.655\times10^{10}$. The field near a charged plane is then
\[ E = \frac{\sigma}{2\epsilon_0} = (4.9\times10^{-6})(5.655\times10^{10}) \approx 2.771\times10^{5}\ \text{N/C} \]
Step 2: Work out the area and the tilt.
The circular plane has area
\[ A = \pi r^2 = \pi (0.01)^2 = 3.1416\times10^{-4}\ \text{m}^2 \]
and its normal is tilted $60^\circ$ from the field, so only $\cos 60^\circ$ of that area actually intercepts the flux lines.
Step 3: Multiply everything out.
\[ \Phi = EA\cos 60^\circ = (2.771\times10^{5})(3.1416\times10^{-4})(0.5) \approx 43.5\ \text{N m}^2/\text{C} \]
\[ \boxed{43.56\ \text{N m}^2/\text{C}} \]