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

[ehrenfest]

[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

$$\bf -p_1 \cdot E_2 - p_2 \cdot E_1$$

?

Homework Equations

The Attempt at a Solution

 

[nrqed]

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

$$\bf -p_1 \cdot E_2 - p_2 \cdot E_1$$

?

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

 

[ehrenfest]

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!

 

[nrqed]

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.