Why is the same equation used for both particles in the initial state?

Click For Summary

Discussion Overview

The discussion revolves around the application of conservation of energy in a problem involving two positive particles and their potential and kinetic energies. Participants explore the reasoning behind using a single equation for potential energy in the context of both particles and the implications for calculating their final speeds after they are distant from each other.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant uses the conservation of energy to equate the initial potential energy of the system to the final kinetic energies of the two particles, questioning why the same equation is applied to both particles.
  • Another participant clarifies that the equation for potential energy, Q².k/d, represents the total potential energy of the system, not an individual particle's energy.
  • A participant expresses confusion about why the total potential energy is not doubled, suggesting that it should account for both charges.
  • It is noted that the definition of k in the equation is arbitrary but convenient, as it avoids additional prefactors.
  • One participant argues that using Q².k/d for the total energy is incorrect since it pertains to only one charge, prompting further discussion on the nature of potential energy in the system.
  • Another participant suggests that the total work done to bring the two charges together is represented by Q².k/d, emphasizing that this is the potential energy due to the conservative electric field.
  • There is a reiteration that a single charge does not possess potential energy by itself; rather, it is the interaction between the two charges that generates potential energy.
  • A later reply mentions the need to consider conservation of momentum in addition to energy conservation for a complete analysis.

Areas of Agreement / Disagreement

Participants exhibit disagreement regarding the application of potential energy equations and whether the total energy should be considered for both charges. The discussion remains unresolved as different interpretations of the potential energy concept are presented.

Contextual Notes

Participants express uncertainty about the definitions and implications of potential energy in relation to individual charges versus the system as a whole. There are also unresolved questions regarding the mathematical steps involved in the energy calculations.

kent davidge
Messages
931
Reaction score
56
I'm trying to solve an exercise that asks what is the final speed of two positive particles after they are very distants each other. Then I've used the conservation of energy to solve it. The initial energy of the system is the potential energy = Q².k/d The final energy is the 2 knetic energy of them. So Q².k/d = Ke1 + Ke2
My doubt is why use the same equation for the two particles in the init? Why not 2.Q².k/d since both of them have acumulated same amount of energy? (sorry my bad english)
 
Physics news on Phys.org
Q2 k / d is the total potential energy in the system. It is not a property of an individual particle, so there is no reason to double it.
 
yes I know that. But the thing is why only one equation Q²K/d is the total potential energy...?
 
k is defined that way. The definition is arbitrary, but this is the easiest one as you don't need additional prefactors.
 
But since we are adding the energy of each charge to get the total energy of the system, I guess it's wrong to use Q².k/d becous it's the equation for only one charge. Am I right?
 
kent davidge said:
But since we are adding the energy of each charge to get the total energy of the system
That is a questionable approach, but if you do it both charges account for 1/2 of Q2k/d.
 
Think of it this way. Say these two pieces were already at infinitely far away. To get one piece to a certain distance d to the other, work of the magnitude q^2k/d has to be done on that one particle. The same amount of work is needed to move the first piece near the second. Therefore the total work done to set the system is q^2k/d, which is the potential energy due to the electric field being a conservative force, and thus no matter how the two positive pieces got there it has that potential energy.. That is why the total kinetic energy can only be equal to q^2k/d
 
Same thing. Doesnt matter the way you used to calculate the potential energy. The fact is that Q².k/d is the acumulated energy on one charge... so when we added the total energy of the system composed by the two charges, if would be 2.Q².k/d (sorry my bad english)
 
kent davidge said:
But since we are adding the energy of each charge to get the total energy of the system, I guess it's wrong to use Q².k/d becous it's the equation for only one charge. Am I right?
No, you are not. A single charge by itself has no potential energy.
Only the interaction between the two charges "produces" some potential energy.

I suppose you are confusing the potential of the field of one charge with the potential energy of the system of two charges.
If you were to calculate the potential of the field of the two charges you will indeed add the potential produced by each individual charge.
 
  • #10
I tried to translate the exercise for you see the situation. (sorry my bad translation)

http://[ATTACH=full]199884[/ATTACH]
 

Attachments

  • azf9ds.jpg
    azf9ds.jpg
    23.6 KB · Views: 221
Last edited by a moderator:
  • #11
For (a), you'll also have to take conservation of momentum into account. Apart from that, everything else said above applies.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 64 ·
3
Replies
64
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 22 ·
Replies
22
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 24 ·
Replies
24
Views
2K
  • · Replies 7 ·
Replies
7
Views
698
  • · Replies 6 ·
Replies
6
Views
11K
  • · Replies 31 ·
2
Replies
31
Views
4K
  • · Replies 8 ·
Replies
8
Views
964