---
title: "A student analyzes the electric field of an isolated positive point charge \\(q\\) using Gauss’s law. The student considers two closed Gaussian surfaces centered on the charge: a spherical surface of radius \\(r\\) and a cylindrical surface of radius \\(r\\) and height \\(2r\\). Which of the following best explains why the spherical surface easily yields the electric field magnitude \\(E\\), whereas the cylindrical surface does not?"
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url: "https://nerd-notes.com/ubq/117912/"
date_modified: "2026-08-04T08:02:33+00:00"
---

# A student analyzes the electric field of an isolated positive point charge \(q\) using Gauss’s law. The student considers two closed Gaussian surfaces centered on the charge: a spherical surface of radius \(r\) and a cylindrical surface of radius \(r\) and height \(2r\). Which of the following best explains why the spherical surface easily yields the electric field magnitude \(E\), whereas the cylindrical surface does not?

A student analyzes the electric field of an isolated positive point charge \(q\) using Gauss's law. The student considers two closed Gaussian surfaces centered on the charge: a spherical surface of radius \(r\) and a cylindrical surface of radius \(r\) and height \(2r\). Which of the following best explains why the spherical surface easily yields the electric field magnitude \(E\), whereas the cylindrical surface does not?

![On the left, a single point labeled q sits at the center of a dashed circle representing a spherical Gaussian surface of radius r. Straight arrows labeled E extend radially outward from q, passing perpendicularly through the dashed sphere. On the right, the same point charge q sits at the geometric center of a vertical dashed cylinder of radius r and height 2r. Straight arrows labeled E radiate outward from q, intersecting the top end cap, bottom end cap, and curved side wall of the cylinder at varying distances and oblique angles. No other labels, lines, text, or axes appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1785830553-myeKFh.jpg)

- **A.** The cylindrical surface cannot be used because the total electric flux through it is zero, whereas the total electric flux through the spherical surface is \(\dfrac{q}{\varepsilon_0}\).
- **B.** The spherical surface is chosen because the electric field lines are parallel to the spherical surface at all points, whereas field lines intersect the cylindrical surface at non-zero angles.
- **C.** The cylindrical surface is not useful because the electric field magnitude is uniform across its entire surface, which prevents the enclosed charge from being determined.
- **D.** The spherical surface is chosen because the electric field magnitude is constant over the surface and everywhere perpendicular to it, allowing \(E\) to be factored out of the flux integral, whereas the field magnitude and angle vary across the cylindrical surface.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/117912/*