---
title: "An open, thin hemispherical bowl of radius \\(R\\) is situated in a uniform electric field \\(\\vec{E}\\). The field is directed parallel to the central symmetry axis of the bowl, entering through the flat circular rim and exiting through the curved surface. How does the magnitude of the electric flux through the curved surface compare to the magnitude of the electric flux through an imaginary flat circular surface that spans the rim, and what is the correct justification?"
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url: "https://nerd-notes.com/ubq/124491/"
date_modified: "2026-09-28T14:08:17+00:00"
---

# An open, thin hemispherical bowl of radius \(R\) is situated in a uniform electric field \(\vec{E}\). The field is directed parallel to the central symmetry axis of the bowl, entering through the flat circular rim and exiting through the curved surface. How does the magnitude of the electric flux through the curved surface compare to the magnitude of the electric flux through an imaginary flat circular surface that spans the rim, and what is the correct justification?

An open, thin hemispherical bowl of radius \(R\) is situated in a uniform electric field \(\vec{E}\). The field is directed parallel to the central symmetry axis of the bowl, entering through the flat circular rim and exiting through the curved surface. How does the magnitude of the electric flux through the curved surface compare to the magnitude of the electric flux through an imaginary flat circular surface that spans the rim, and what is the correct justification?

![A horizontal dashed ellipse representing the flat circular rim of a bowl, with a dashed horizontal radius line from its center to its right edge labeled \(R\). Below the ellipse, a solid semicircular arc connects the left and right edges, forming a downward-oriented hemispherical surface. A vertical dashed centerline passes through the center of the ellipse and the bottom apex of the bowl. Exactly four vertical, parallel arrows point upward from below the bowl through its curved surface and out the top opening. A text label \(\vec{E}\) is placed to the right of the rightmost vertical arrow. No other labels, lines, text, or axes appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1790604497-8Csz7e.jpg)

- **A.** The magnitude of the flux through the curved surface is greater than that through the flat surface because the curved hemisphere has a surface area of \(2\pi R^2\), which is twice the area of the circular rim.
- **B.** The magnitude of the flux through the curved surface is less than that through the flat surface because the electric field vector is not parallel to the surface normal across most of the hemisphere.
- **C.** The magnitude of the flux through the curved surface is equal to that through the flat surface because the two surfaces together form a closed surface that encloses zero net charge.
- **D.** The magnitude of the flux through the curved surface is equal to that through the flat surface because the electric field lines are everywhere perpendicular to the curved surface.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/124491/*