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MrMuscle
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How to find an expression for bound charge densities in griffiths
- Context: Graduate
- Thread starter MrMuscle
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- Bound Charge Expression Griffiths
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SUMMARY
The discussion focuses on deriving expressions for bound charge densities in the context of electrostatics, specifically using the identity for differentiation with respect to source coordinates. The identity $$\nabla\frac{1}{|\vec{r}-\vec{r'}|}=-\nabla'\frac{1}{|\vec{r}-\vec{r'}|}$$ is highlighted as crucial for understanding the behavior of potential fields. Additionally, the correct application of the divergence identity $$\nabla\cdot (f\vec{A})=f\nabla\cdot\vec{A}+\vec{A}\cdot\nabla f$$ is emphasized for differentiating with respect to source coordinates, particularly when applied to the polarization vector $\vec{P}$ and the potential function.
PREREQUISITES- Understanding of electrostatics and bound charge densities
- Familiarity with vector calculus identities
- Knowledge of differentiation with respect to source coordinates
- Basic concepts of electric polarization
- Study the application of the divergence theorem in electrostatics
- Learn about the role of polarization in bound charge density calculations
- Explore advanced vector calculus techniques in electrostatics
- Investigate the implications of source coordinates in potential theory
Students and professionals in physics, particularly those specializing in electromagnetism, as well as researchers working on theoretical aspects of electrostatics and charge distributions.
Attempt:
- #2
Delta2
Homework Helper
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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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