---
title: "A uniform electric field of magnitude \\(E\\) is directed parallel to the central axis of symmetry of an open hemispherical surface of radius \\(R\\), entering through the flat circular opening and exiting through the curved surface. What is the magnitude of the electric flux through the curved hemispherical surface?"
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url: "https://nerd-notes.com/ubq/117892/"
date_modified: "2026-08-04T08:02:24+00:00"
---

# A uniform electric field of magnitude \(E\) is directed parallel to the central axis of symmetry of an open hemispherical surface of radius \(R\), entering through the flat circular opening and exiting through the curved surface. What is the magnitude of the electric flux through the curved hemispherical surface?

A uniform electric field of magnitude \(E\) is directed parallel to the central axis of symmetry of an open hemispherical surface of radius \(R\), entering through the flat circular opening and exiting through the curved surface. What is the magnitude of the electric flux through the curved hemispherical surface?

![A 3D line drawing showing an open hemispherical bowl with radius R. The circular flat base of the hemisphere lies horizontally, and the curved dome bulges upward. Exactly four vertical parallel arrows, representing a uniform electric field vector \vec{E}, point straight upward, passing vertically through the flat circular open base and exiting through the top of the curved dome. A vertical dashed line marks the central axis of symmetry. A straight horizontal line segment from the center of the circular base to its outer rim is labeled R. The electric field arrows are labeled \vec{E}. No other labels, lines, text, or axes appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1785830544-d723Vm.jpg)

- **A.** \(0\)
- **B.** \(E \pi R^2\)
- **C.** \(2\pi E R^2\)
- **D.** \(4\pi E R^2\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/117892/*