Griffiths Problem 4.8 [SOLVED]: Why Don't You Need to Add Energy?

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Homework Help Overview

This discussion revolves around a problem from Griffiths' Electromagnetism textbook concerning the energy associated with two dipoles and their interactions. Participants are exploring why the total energy does not simply sum the energies of each dipole in the system.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the reasoning behind not needing to add the energies of both dipoles when calculating the total energy of the system. There is an exploration of the analogy with point charges and the energy required to assemble the dipoles from infinity.

Discussion Status

The discussion is ongoing, with participants providing insights into the energy considerations of the dipole system. Some guidance has been offered regarding the assembly process and the work done, but there is still uncertainty about the implications of holding the first dipole at rest.

Contextual Notes

Participants are grappling with the definitions of work in physics and how they apply to the scenario presented in the problem. There is a focus on the assumptions made regarding energy contributions during the assembly of the dipole system.

ehrenfest
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[SOLVED] Griffiths Problem 4.8

Homework Statement


This question refers to Griffiths E and M book.

Why is the answer not 2 times that?
Why don't you need to add the energy of dipole 1 to the energy of dipole 2 like

[tex]\bf -p_1 \cdot E_2 - p_2 \cdot E_1[/tex]

?

Homework Equations


The Attempt at a Solution

 
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ehrenfest said:

Homework Statement


This question refers to Griffiths E and M book.

Why is the answer not 2 times that?
Why don't you need to add the energy of dipole 1 to the energy of dipole 2 like

[tex]\bf -p_1 \cdot E_2 - p_2 \cdot E_1[/tex]

?


Homework Equations





The Attempt at a Solution


For the same reason that the potential energy of a pair of two point charges is k q_1 q_2/r.

Imagine assembling the system, starting with the two dipoles at infinite distances. Bringing the first dipole from infinity requires no energy. Bringing the second dipole from infinity (while holding the first one at rest) requires an energy equal to - p2 dot E_1. That's the total energy stored in the system since it's the total energy that was required to assemble the system
 
nrqed said:
For the same reason that the potential energy of a pair of two point charges is k q_1 q_2/r.

Imagine assembling the system, starting with the two dipoles at infinite distances. Bringing the first dipole from infinity requires no energy. Bringing the second dipole from infinity (while holding the first one at rest) requires an energy equal to - p2 dot E_1. That's the total energy stored in the system since it's the total energy that was required to assemble the system

But it also requires energy to hold the first dipole at rest, doesn't it? Well I guess there is no work done doing that. But it seems like you don't get that for free!
 
Last edited:
ehrenfest said:
But it also requires energy to hold the first dipole at rest, doesn't it? Well I guess there is no work done doing that. But is seems like you don't get that for free!

as you say, there is no work involved doing that. I know what you mean but there is a very clear definition of work in physics and we have to stick to it.
 

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