Does Electric Potential Energy Depend on the Mass of an Ion?

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Homework Help Overview

The discussion revolves around the relationship between electric potential energy (EPE) and the mass of an ion as it accelerates due to a potential difference. The original poster questions whether the EPE at point B is influenced by the ion's mass.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of the equation for electric potential energy, questioning whether mass plays a role in determining EPE. Some examine the units involved, while others challenge the assumption that mass is a determining factor.

Discussion Status

The discussion is active, with participants presenting differing viewpoints on the dependence of EPE on mass. Some provide reasoning based on unit analysis, while others argue against the relevance of mass in this context. No consensus has been reached yet.

Contextual Notes

Participants are navigating the implications of the equations provided and questioning the assumptions underlying the relationship between electric potential, charge, and mass. The original poster's inquiry is framed within the constraints of a homework assignment.

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Homework Statement


An ion, starting from rest, accelerates from point A to point B due to a potential difference between the two points. Does the electric potential energy of the ion at point B depend on it's mass?

Homework Equations


$$\Delta EPE = q \Delta V$$
$$[J]=[C\cdot \frac{J}{C}]$$
$$[J]=[N\cdot m]=[\frac{kg\cdot m^2}{s^2}]$$

The Attempt at a Solution


As can be seen in the third equation, the unit of EPE, [J] is given by it's mass [kg]. So the answer should be yes? Why is it not?
 
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Look at equation 1

" $$\Delta EPE = q \Delta V$$ "

Does the EPE here depend on mass?
 
Stephen Hodgson said:
Look at equation 1

" $$\Delta EPE = q \Delta V$$ "

Does the EPE here depend on mass?
Yes, since EPE depends on V and V depends on mass.

$$[V]=[\frac{J}{C}]=[\frac{N \cdot m}{C}] = [\frac{kg \cdot m^2}{C\cdot s^2}]$$
 
Who says it depends on mass? just because it includes kg in the units doesn't mean it depends on mass. If V=5 JC-1 everywhere, it doesn't matter how much you weigh... it's still 5.
 

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