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Physics
Classical Physics
Electromagnetism
How to find an expression for bound charge densities in griffiths
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[QUOTE="Delta2, post: 6242460, member: 189563"] 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'}|}##. [/QUOTE]
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Forums
Physics
Classical Physics
Electromagnetism
How to find an expression for bound charge densities in griffiths
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