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Question: Answered & Verified by Expert
A large charged plane having surface charge density $4.9 \times 10^{-6} \mathrm{C} \mathrm{m}^{-2}$ lies in the $\mathrm{x}-\mathrm{y}$ plane. A circular plane of radius of $1 \mathrm{~cm}$ is lying completely in the region where $\mathrm{x}, \mathrm{y}$ and $\mathrm{z}$ coordinates are all positive. When the plane's normal makes an angle $60^{\circ}$ with the z-axis, the electric flux through the circular plane is $\left(\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right)$
PhysicsElectrostaticsAP EAMCETAP EAMCET 2022 (08 Jul Shift 1)
Options:
  • A $43.56 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-1}$
  • B $48.36 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-1}$
  • C $36.76 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-1}$
  • D $32.56 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-1}$
Solution:
2610 Upvotes Verified Answer
The correct answer is: $43.56 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-1}$


$\phi=\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}}$
$\begin{aligned} & =\mathrm{EA} \cos 60^{\circ} \\ & =\frac{\sigma}{2 \varepsilon_0} \mathrm{~A} \cdot \frac{1}{2}=\frac{\sigma \mathrm{A}}{4 \varepsilon_0}=\frac{4.9 \times 10^{-6} \times \pi \times(0.01)^2}{4 \times 8.85 \times 10^{-12}} \\ & =43.56 \mathrm{Nm}^2 \mathrm{C}^{-1}\end{aligned}$

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