# Electric Potential Difference and the Hydrogen Atom

## Метаданные

- **Канал:** Flipping Physics
- **YouTube:** https://www.youtube.com/watch?v=xBhgLnblgIQ
- **Дата:** 13.04.2026
- **Длительность:** 4:21
- **Просмотры:** 1,022

## Описание

In this lesson, we explore the electric potential difference between two locations in the electric field of a point charge — one of the most essential concepts in AP Physics and introductory electrostatics. Using a positive point charge as our model, we examine whether the electric potential difference is positive or negative when moving between two points, and we prove it three different ways: using electric field line logic, algebraic substitution with Coulomb's law, and work-energy reasoning. We also tackle a critical conceptual question, “Does the path between two points affect the electric potential difference?”, and connect the answer to conservative forces and equipotential surfaces. Whether you're studying for the AP Physics exam or just trying to build deep intuition for electricity, this video will give you the clarity you need.

Want Lecture Notes? → http://www.flippingphysics.com/electric-potential-difference-hydrogen.html

Thank you to Gerardo Garcia and the rest of my wonderful Patreon supporters. Please consider supporting me monthly at http://patreon.com/FlippingPhysics

Thank you to Christopher Becke and Julie Langenbrunner for being my Quality Control Team for this video. http://flippingphysics.com/quality-control.html

Chapters:
0:00 The Problem
1:20 Solving the Problem
3:06 Equipotential Surfaces and Conservative Forces

#electricpotentialdifference #bohrmodel #equipotentialsurfaces

## Содержание

### [0:00](https://www.youtube.com/watch?v=xBhgLnblgIQ) The Problem

Good morning. Bobby, please read the problem. Billy, please translate the problem. And Bo, please solve the problem. — Absolutely. — Okay. Sure. — Flipics. [singing] — The bore model of an atom has one proton in the nucleus and one electron orbiting the proton. The electron can only exist at certain orbital radi. The smallest orbital radius of the electron is 0. 529 * 10 -10 m. The next possible radius of the electron is 2. 116 * 10 -10 m. Part A, what is the absolute value of the electric potential difference between the two orbits? Part B, which orbit is at the higher electric potential? All right, our known values are the charge of a proton is 1. 6 6 * 109 kum because that is the elementary charge and a proton has a charge value equal to the positive value of the elementary charge which is 1. 60 * 109 kum. We also know the initial radius is 0. The final radius is 2. And we are solving for the electric potential difference. So delta capital V equals question mark. Okay, we know the electric potential which surrounds a

### [1:20](https://www.youtube.com/watch?v=xBhgLnblgIQ&t=80s) Solving the Problem

point charge equals the kulum constant times the charge of the charge divided by the distance from the charge to the location of the electric potential. And the electric potential difference equals electric potential final minus electric potential initial. We can substitute in the equation for the electric potential which surrounds a point charge with distance final and distance initial. We can factor out the coolum constant charge and substitute in numbers we get 8. 99 * 10 9th * 1. 60 60 * 10 -19 * the quantity the inverse of 2. 116 * 10 -10 minus the inverse of 0. 529 * 10 -10. That works out to be -220. 393 or -20. 4 volts with three sigfigs. But part A asks for the absolute value of that which is 20. 4 volts. What about part B? Am I supposed to do that part? Mr. P asked you to solve the problem and part B is a part of the problem. Sure. We know that electric potential increases as locations get closer to positive charges. Therefore, the smaller orbit is at a higher potential. Also, the electric potential we determined is negative when going from the smaller to the larger radius. And that means the electric potential decreases when moving from a smaller to a larger orbital radius. That confirms that the smaller orbital radius has a higher potential.

### [3:06](https://www.youtube.com/watch?v=xBhgLnblgIQ&t=186s) Equipotential Surfaces and Conservative Forces

— Yep. Well done everybody. Now realize each of the orbital radiates an orbital shell which has a uniform electric potential which is called an electro equipotential surface. [gasps] And the electrostatic force is a conservative force which means the work done by the electrostatic force is independent of path taken by the object. In other words, it does not matter where the initial point is on the smaller orbital shell. Nor does it matter where the final point is on the larger orbital shell. Nor does it matter the path taken from the smaller to larger orbital shell. The electric potential difference is always the same. When moving from the smallest orbital shell to the next orbital shell of the hydrogen atom, the electric potential difference is always -20. 4 vol. Actually, it does not even matter that it is an electron which is moving from the smaller to larger orbital shell because electric potential is independent of the object that may or may not be moving between the two locations. That is why we used zero information about the electron when solving this problem. Thank you very much for learning with me today. I enjoy learning with you.

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*Источник: https://ekstraktznaniy.ru/video/49326*