# How to find an expression for bound charge densities in griffiths

MrMuscle

Attempt:

Delta2

Homework Helper
Gold Member
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'}|}##.

Last edited:
vanhees71 and MrMuscle