# Energy in dielectric systems

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1. Sep 29, 2014

### almarpa

Hello all.

I have a doubt about the derivation of energy in dielectrics formula (Griffiths pages 191 - 192).

In a certain step of the formula derivation, we encounter the following operation:

(view formula below).

I do not undertand that operation.

Can someone help me?

Last edited by a moderator: Sep 30, 2014
2. Sep 29, 2014

### USeptim

What means the $\bigtriangleup$? It looks like it only works on the first term of D·E.

3. Sep 29, 2014

### zoki85

Laplace operator

4. Sep 29, 2014

### USeptim

Thanks zoki85, I used to see for the Laplacian $\bigtriangledown^2$.

Almarpa. The link you have set it's a bit out of context. It only shows that you can conmute the lapace operator and the dot product since D and E differ only by a constant $\epsilon$.

5. Sep 29, 2014

### athosanian

2(E.E)=▽2E.E+E.▽2E

you can think the ▽2 as a scalar, but it also is a differential operator like d/dx

6. Sep 30, 2014

### almarpa

It is not the laplacian operator. It represents an incremental variation of the quantity this symbol goes with.

Last edited: Sep 30, 2014
7. Sep 30, 2014

### almarpa

It is not the laplace operator. It is just an increment.

8. Sep 30, 2014

### Meir Achuz

$\nabla^2({\bf E\cdot E})$ is not that simple.

9. Sep 30, 2014

### Meir Achuz

$\Delta$ is just an infinitesimal variation. That step is only valid if $\epsilon$
does not vary with position anywhere in space. Then the step just says
$\Delta({\bf E\cdot D)=E\cdot(\Delta D)+(\Delta E)\cdot D}$.

10. Sep 30, 2014

### zoki85

If so, he should wrote it d , not Δ

11. Sep 30, 2014

### Meir Achuz

$\Delta$ is commonly used, with the limit $d=lim\Delta\rightarrow 0$.

12. Oct 1, 2014

### almarpa

Sorry, but I still do not get it.

What happens with the 1/2 term? It vanishes, but I can not see why.

13. Oct 1, 2014

### Meir Achuz

$\Delta({\bf E\cdot D)=E\cdot(\Delta D)+(\Delta E)\cdot D}=2\epsilon{\bf E\cdot E}$
if $\epsilon$ is constant.