MrMuscle
- 12
- 1
Attempt:
How to find an expression for bound charge densities in griffiths
- Context: Graduate
- Thread starter MrMuscle
- Start date
-
- Tags
- Bound Charge Expression Griffiths
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 replies · 1K views
Attempt:
- 6,002
- 2,629
The minus sign is lost because the differentiation is with respect to the source coordinates ##r'##. It is a well known identity that $$\nabla\frac{1}{|\vec{r}-\vec{r'}|}=-\nabla'\frac{1}{|\vec{r}-\vec{r'}|}$$
where in the above the second ##\nabla## (the one in the right) has a ##'## next to it which means just that the differentiation is with respect to ##r'## coordinates.
For the rest you got the wrong identity involving curl and cross product. You should have the identity involving divergence and dot product which is as follows
$$\nabla\cdot (f\vec{A})=f\nabla\cdot\vec{A}+\vec{A}\cdot\nabla f$$
this identity holds for differentiating with source coordinates as well that is with ##\nabla'## in the place of ##\nabla##. Apply this Identity for ##\vec{A}=\vec{P}## and ##f=\frac{1}{|\vec{r}-\vec{r'}|}##.
where in the above the second ##\nabla## (the one in the right) has a ##'## next to it which means just that the differentiation is with respect to ##r'## coordinates.
For the rest you got the wrong identity involving curl and cross product. You should have the identity involving divergence and dot product which is as follows
$$\nabla\cdot (f\vec{A})=f\nabla\cdot\vec{A}+\vec{A}\cdot\nabla f$$
this identity holds for differentiating with source coordinates as well that is with ##\nabla'## in the place of ##\nabla##. Apply this Identity for ##\vec{A}=\vec{P}## and ##f=\frac{1}{|\vec{r}-\vec{r'}|}##.
Last edited: