Electric Potential voltage and conservation of energy

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Homework Help Overview

The discussion revolves around a problem involving electric potential and the conservation of energy, specifically focusing on a proton being accelerated by a uniform electric field and the resulting kinetic energy after a potential decrease of 100 V.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to relate the change in potential energy to kinetic energy but arrives at a negative value for kinetic energy, which raises questions about the validity of their calculations. Some participants suggest reconsidering the interpretation of potential energy and kinetic energy in the context of the problem.

Discussion Status

Participants are actively engaging with the original poster's reasoning, with some providing clarifications on the relationship between potential difference and work done. There is an exploration of different interpretations regarding potential energy and kinetic energy, but no explicit consensus has been reached.

Contextual Notes

The problem involves a proton, which is a charged particle, and the discussion includes considerations of electric fields and potential energy changes. The original poster's calculations and assumptions about energy conservation are under scrutiny, particularly regarding the signs of energy values.

therealkellys
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Hi,

I'm a little stuck on a problem involving voltage and conservation of energy. The answer that I came up with involved a negative value for kinetic energy, which doesn't make sense to me because mass and v^2 are never negative.

Problem: A proton is accelerated from rest by a uniform electric field. The proton experiences a potential decrease of 100 V. Find its final kinetic energy.

What I've done so far:
KE_i = 0 because accelerated from rest
V_b - V_a = -100 V
Change in voltage = Work/charge = [-q(int)E dot ds] / q = Change in potential energy

-100 = -Change in PE --> 100 J = change in potential energy

Therefore, under law of conservation of energy... change in kinetic energy must equal -100 J. Since KE_i = 0 J, KE_f= -100 J. :confused:

Thanks in advance for the help.
 
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Bump... I'm still really unsure about my answer. If anybody could point out anything that I might have overlooked, that would help a lot. Thanks!
 
therealkellys said:
The answer that I came up with involved a negative value for kinetic energy, which doesn't make sense to me because mass and v^2 are never negative.
Your suspicions are correct. Change is usually expressed as final value minus the initial value, hence leading to a positive value for the change in potential energy. Just for reference a perhaps easier method would be to consider the definition of potential difference or voltage; which is work done per unit charge, thus;

V = \frac{W}{q}

From which we can obtain the expression W = Vq, substituting your values in we obtain W = 100 \times 1 = 100\; J.
 
Last edited:
q \Delta V = -\Delta E_{kin}
 
Since your proton is at a lower potential, it has a negative potential energy (-100J), so its kinectic energy must be +100J.
Think of it in terms of mechanics. You have a mass of 1kg at a height of 10m. You have a potential energy of 100J (actually 98) in reference to ground. If you drop the mass, when it hits the ground its potential energy will be zero and the kinectic energy 98J.
 

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