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title: "An electron of mass \\(m\\) and charge \\(-e\\) is projected with speed \\(v_0\\) at an angle \\(\\theta\\) relative to the normal toward a large, uniformly charged nonconducting plate with negative surface charge density \\(-\\sigma\\), where \\(\\sigma > 0\\). The perpendicular stopping distance \\(d\\) is defined as the distance the electron travels toward the plate before the component of its velocity perpendicular to the plate momentarily becomes zero. Which of the following expressions correctly gives the stopping distance \\(d\\), and which statement correctly describes its limiting behavior as the launch angle \\(\\theta\\) approaches \\(\\dfrac{\\pi}{2}\\)?"
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url: "https://nerd-notes.com/ubq/124746/"
date_modified: "2026-09-28T14:09:51+00:00"
---

# An electron of mass \(m\) and charge \(-e\) is projected with speed \(v_0\) at an angle \(\theta\) relative to the normal toward a large, uniformly charged nonconducting plate with negative surface charge density \(-\sigma\), where \(\sigma > 0\). The perpendicular stopping distance \(d\) is defined as the distance the electron travels toward the plate before the component of its velocity perpendicular to the plate momentarily becomes zero. Which of the following expressions correctly gives the stopping distance \(d\), and which statement correctly describes its limiting behavior as the launch angle \(\theta\) approaches \(\dfrac{\pi}{2}\)?

An electron of mass \(m\) and charge \(-e\) is projected with speed \(v_0\) at an angle \(\theta\) relative to the normal toward a large, uniformly charged nonconducting plate with negative surface charge density \(-\sigma\), where \(\sigma > 0\). The perpendicular stopping distance \(d\) is defined as the distance the electron travels toward the plate before the component of its velocity perpendicular to the plate momentarily becomes zero. Which of the following expressions correctly gives the stopping distance \(d\), and which statement correctly describes its limiting behavior as the launch angle \(\theta\) approaches \(\dfrac{\pi}{2}\)?

![A vertical line represents a large flat plate positioned on the right side of the diagram, labeled -\sigma. To the left of the plate, a horizontal dashed line represents the normal axis perpendicular to the plate. A small solid circle representing an electron is located on the normal axis at a distance to the left of the plate, labeled -e. An arrow labeled v_0 originates from the electron and points toward the upper-right toward the plate. An arc indicates the angle \theta between the horizontal normal axis and the arrow v_0. A horizontal double-headed arrow below the dashed line extends from the electron's position to a vertical dotted line closer to the plate, labeled d. No other labels, lines, text, or axes appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1790604591-OlCy6h.jpg)

- **A.** \(d = \dfrac{\varepsilon_0 m v_0^2}{e\sigma}\). As \(\theta \to \dfrac{\pi}{2}\), \(d\) approaches a non-zero constant because electric potential is a scalar quantity depending only on distance from the plate, so all initial kinetic energy must be converted into electric potential energy at the turning point.
- **B.** \(d = \dfrac{\varepsilon_0 m v_0^2 \sin^2\theta}{e\sigma}\). As \(\theta \to \dfrac{\pi}{2}\), \(d\) approaches a maximum value because the launch direction approaches the plane of the plate, maximizing the path length along which the electric field decelerates the electron.
- **C.** \(d = \dfrac{\varepsilon_0 m v_0^2 \cos^2\theta}{e\sigma}\). As \(\theta \to \dfrac{\pi}{2}\), \(d \to 0\) because the initial kinetic energy associated with velocity perpendicular to the plate approaches zero, so zero work by the electric field is required to reverse the perpendicular velocity.
- **D.** \(d = \dfrac{\varepsilon_0 m v_0^2}{e\sigma \cos^2\theta}\). As \(\theta \to \dfrac{\pi}{2}\), \(d \to \infty\) because the trajectory becomes nearly parallel to the plate, requiring an infinite perpendicular distance for the electric force to deflect the electron.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/124746/*