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MrMuscle

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- Thread starter MrMuscle
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- #1

MrMuscle

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- 1

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- #2

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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'}|}##.

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