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title: "An electron of mass \\(m\\) and charge magnitude \\(e\\) is released from rest in a uniform electric field of magnitude \\(E\\) directed to the right. The electron accelerates under the influence of the electric field through a distance \\(d\\). What is the change in electric potential energy \\(\\Delta U\\) of the electron-field system after the electron travels distance \\(d\\), and what is the speed \\(v\\) of the electron at that point?"
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url: "https://nerd-notes.com/ubq/118073/"
date_modified: "2026-08-04T08:04:58+00:00"
---

# An electron of mass \(m\) and charge magnitude \(e\) is released from rest in a uniform electric field of magnitude \(E\) directed to the right. The electron accelerates under the influence of the electric field through a distance \(d\). What is the change in electric potential energy \(\Delta U\) of the electron-field system after the electron travels distance \(d\), and what is the speed \(v\) of the electron at that point?

An electron of mass \(m\) and charge magnitude \(e\) is released from rest in a uniform electric field of magnitude \(E\) directed to the right. The electron accelerates under the influence of the electric field through a distance \(d\). What is the change in electric potential energy \(\Delta U\) of the electron-field system after the electron travels distance \(d\), and what is the speed \(v\) of the electron at that point?

![A horizontal region showing four parallel, evenly spaced horizontal arrows pointing to the right, labeled \vec{E}. A circle with a minus sign, representing an electron, is located near the right end of the field region. A dashed horizontal arrow points to the left from the electron along its path of motion, labeled d. No other labels, lines, text, or axes appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1785830698-53LHeA.jpg)

- **A.** \(\Delta U = +eEd\) and \(v = \sqrt{\dfrac{eEd}{m}}\)
- **B.** \(\Delta U = +eEd\) and \(v = \sqrt{\dfrac{2eEd}{m}}\)
- **C.** \(\Delta U = -eEd\) and \(v = \sqrt{\dfrac{eEd}{m}}\)
- **D.** \(\Delta U = -eEd\) and \(v = \sqrt{\dfrac{2eEd}{m}}\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/118073/*