---
title: "A flat rectangular sheet of surface area \\(A\\) is placed in a region of uniform electric field of magnitude \\(E\\). The plane of the sheet makes an angle of \\(30^\\circ\\) with the direction of the electric field vector \\(\\vec{E}\\). Which of the following correctly gives the electric flux \\(\\Phi_E\\) through the sheet and the physical justification for how orientation affects the flux?"
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url: "https://nerd-notes.com/ubq/117894/"
date_modified: "2026-08-04T08:02:25+00:00"
---

# A flat rectangular sheet of surface area \(A\) is placed in a region of uniform electric field of magnitude \(E\). The plane of the sheet makes an angle of \(30^\circ\) with the direction of the electric field vector \(\vec{E}\). Which of the following correctly gives the electric flux \(\Phi_E\) through the sheet and the physical justification for how orientation affects the flux?

A flat rectangular sheet of surface area \(A\) is placed in a region of uniform electric field of magnitude \(E\). The plane of the sheet makes an angle of \(30^\circ\) with the direction of the electric field vector \(\vec{E}\). Which of the following correctly gives the electric flux \(\Phi_E\) through the sheet and the physical justification for how orientation affects the flux?

![A 2D schematic diagram showing a uniform electric field represented by five evenly spaced horizontal arrows pointing from left to right, labeled with LaTeX glyph \vec{E}. A flat rectangular sheet is shown edge-on, tilted such that its surface makes a 30-degree angle with the horizontal electric field lines. A dashed vector labeled \vec{A} extends perpendicularly outward from the surface of the sheet, forming a 60-degree angle with the horizontal field lines. The angle of 30 degrees between the surface line and the horizontal field line is clearly arc-labeled as 30^{\circ}. No other labels, lines, text, or axes appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1785830545-8lboTc.jpg)

- **A.** \(\Phi_E = EA \cos(30^\circ)\), because electric flux is proportional to the component of the electric field parallel to the plane of the sheet.
- **B.** \(\Phi_E = EA \sin(30^\circ)\), because electric flux is proportional to the component of the electric field perpendicular to the plane of the sheet.
- **C.** \(\Phi_E = EA \cos(30^\circ)\), because electric flux is proportional to the component of the electric field perpendicular to the plane of the sheet.
- **D.** \(\Phi_E = EA\), because the electric flux through a fixed area in a uniform field is independent of the orientation of the surface.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/117894/*