I Electric field due to polarized object

AI Thread Summary
The discussion centers on the conditions necessary for the electric fields of two objects, one polarized and the other with surface and volume charge densities, to be equal. It emphasizes the importance of satisfying specific polarization conditions, which are supported by a standard proof involving the potential of a dipole. The mathematical derivation illustrates how the dipole potential can be expressed in terms of the dipole moment per unit volume and the integration over the source charge distribution. Key concepts include the definitions of surface charge density and volume charge density related to polarization. The explanation ultimately clarifies the relationship between polarization and electric fields in the context of electrostatics.
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So I am given the conditions necessary for the electric field to be equal for given two objects of same size and shape, one of which is polarized and the other has some surface and volume charge densities as shown in the diagram
16377788745095814392662940793349.jpg
16377789157864281810467250891417.jpg
 
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VVS2000 said:
Summary:: So I am given the conditions necessary for the electric field to be equal for given two objects of same size and shape, one of which is polarized and the other has some surface and volume charge densities as shown in the diagram

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So you are. Is there a question you wish to ask?
 
kuruman said:
So you are. Is there a question you wish to ask?
Yeah, they have mentioned the given 2 conditions regarding polarization must be satisfied for electric fields of the given 2 objects to be equal
I still don't know how
 
The justification is a standard proof. You might have come across the potential of a dipole ##\phi(\mathbf{r}) = \dfrac{\mathbf{p} \cdot \mathbf{r}}{4\pi \epsilon_0 r^3}##. Therefore if a material has a dipole moment per unit volume ##\mathbf{P} = \dfrac{d\mathbf{p}}{dV}##, the dipole potential is\begin{align*}
\phi(\mathbf{r}) &= \dfrac{1}{4\pi \epsilon_0} \int_{\mathscr{V}} \dfrac{\mathbf{P}(\mathbf{R}) \cdot (\mathbf{r} - \mathbf{R})}{|\mathbf{r} - \mathbf{R}|^3} d^3 \mathbf{R}
\end{align*}where ##\mathbf{R}## is the integration variable over the source charge distribution (contained with in ##\mathscr{V}##). Letting ##\nabla_{\mathbf{R}} = \left( \frac{\partial}{\partial R_x}, \frac{\partial}{\partial R_y}, \frac{\partial}{\partial R_z} \right)## denote differentiation with respect to the coordinates of ##\mathbf{R}##, notice now that\begin{align*}
\nabla_{\mathbf{R}} \cdot \dfrac{\mathbf{P}(\mathbf{R})}{|\mathbf{r} - \mathbf{R}|} &= \sum_i \dfrac{\partial}{\partial R_i} \dfrac{P_i(\mathbf{R})}{|\mathbf{r} - \mathbf{R}|} \\
&= \sum_i \left[ \dfrac{1}{|\mathbf{r} - \mathbf{R}|} \dfrac{\partial P_i}{\partial R_i} + P_i \dfrac{\partial}{\partial R_i} \dfrac{1}{|\mathbf{r} - \mathbf{R}|} \right] \\
&= \sum_i \left[ \dfrac{1}{|\mathbf{r} - \mathbf{R}|} \dfrac{\partial P_i}{\partial R_i} + P_i \dfrac{(\mathbf{r} - \mathbf{R})_i}{|\mathbf{r} - \mathbf{R}|^3} \right] \\
&= \dfrac{\nabla_{\mathbf{R}} \cdot \mathbf{P}}{|\mathbf{r} - \mathbf{R}|} + \dfrac{\mathbf{P} \cdot (\mathbf{r} - \mathbf{R})}{|\mathbf{r} - \mathbf{R}|^3}
\end{align*}and therefore\begin{align*}
\phi(\mathbf{r}) &= \dfrac{1}{4\pi \epsilon_0} \int_{\mathscr{V}}\left[ \nabla_{\mathbf{R}} \cdot \dfrac{\mathbf{P}}{|\mathbf{r} - \mathbf{R}|} - \dfrac{\nabla_{\mathbf{R}} \cdot \mathbf{P}}{|\mathbf{r} - \mathbf{R}|} \right] d^3 \mathbf{R} \\
&= \dfrac{1}{4\pi \epsilon_0} \int_{\partial \mathscr{V}} \dfrac{\mathbf{P}}{|\mathbf{r} - \mathbf{R}|} \cdot \hat{\mathbf{n}} dS- \dfrac{1}{4\pi \epsilon_0} \int_{\mathscr{V}} \dfrac{\nabla_{\mathbf{R}} \cdot \mathbf{P}}{|\mathbf{r} - \mathbf{R}|} d^3 \mathbf{R} \\

&= \dfrac{1}{4\pi \epsilon_0} \int_{\partial \mathscr{V}} \dfrac{\sigma_{\mathrm{b}}}{|\mathbf{r} - \mathbf{R}|} dS + \dfrac{1}{4\pi \epsilon_0} \int_{\mathscr{V}} \dfrac{\rho_{\mathrm{b}}}{|\mathbf{r} - \mathbf{R}|} d^3 \mathbf{R}
\end{align*}upon defining ##\sigma_{\mathrm{b}} \equiv \mathbf{P} \cdot \hat{\mathbf{n}}## and ##\rho_{\mathrm{b}} \equiv - \nabla_{\mathbf{R}} \cdot \mathbf{P}##.
 
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ergospherical said:
The justification is a standard proof. You might have come across the potential of a dipole ##\phi(\mathbf{r}) = \dfrac{\mathbf{p} \cdot \mathbf{r}}{4\pi \epsilon_0 r^3}##. Therefore if a material has a dipole moment per unit volume ##\mathbf{P} = \dfrac{d\mathbf{p}}{dV}##, the dipole potential is\begin{align*}
\phi(\mathbf{r}) &= \dfrac{1}{4\pi \epsilon_0} \int_{\mathscr{V}} \dfrac{\mathbf{P}(\mathbf{R}) \cdot (\mathbf{r} - \mathbf{R})}{|\mathbf{r} - \mathbf{R}|^3} d^3 \mathbf{R}
\end{align*}where ##\mathbf{R}## is the integration variable over the source charge distribution (contained with in ##\mathscr{V}##). Letting ##\nabla_{\mathbf{R}} = \left( \frac{\partial}{\partial R_x}, \frac{\partial}{\partial R_y}, \frac{\partial}{\partial R_z} \right)## denote differentiation with respect to the coordinates of ##\mathbf{R}##, notice now that\begin{align*}
\nabla_{\mathbf{R}} \cdot \dfrac{\mathbf{P}(\mathbf{R})}{|\mathbf{r} - \mathbf{R}|} &= \sum_i \dfrac{\partial}{\partial R_i} \dfrac{P_i(\mathbf{R})}{|\mathbf{r} - \mathbf{R}|} \\
&= \sum_i \left[ \dfrac{1}{|\mathbf{r} - \mathbf{R}|} \dfrac{\partial P_i}{\partial R_i} + P_i \dfrac{\partial}{\partial R_i} \dfrac{1}{|\mathbf{r} - \mathbf{R}|} \right] \\
&= \sum_i \left[ \dfrac{1}{|\mathbf{r} - \mathbf{R}|} \dfrac{\partial P_i}{\partial R_i} + P_i \dfrac{(\mathbf{r} - \mathbf{R})_i}{|\mathbf{r} - \mathbf{R}|^3} \right] \\
&= \dfrac{\nabla_{\mathbf{R}} \cdot \mathbf{P}}{|\mathbf{r} - \mathbf{R}|} + \dfrac{\mathbf{P} \cdot (\mathbf{r} - \mathbf{R})}{|\mathbf{r} - \mathbf{R}|^3}
\end{align*}and therefore\begin{align*}
\phi(\mathbf{r}) &= \dfrac{1}{4\pi \epsilon_0} \int_{\mathscr{V}}\left[ \nabla_{\mathbf{R}} \cdot \dfrac{\mathbf{P}}{|\mathbf{r} - \mathbf{R}|} - \dfrac{\nabla_{\mathbf{R}} \cdot \mathbf{P}}{|\mathbf{r} - \mathbf{R}|} \right] d^3 \mathbf{R} \\
&= \dfrac{1}{4\pi \epsilon_0} \int_{\partial \mathscr{V}} \dfrac{\mathbf{P}}{|\mathbf{r} - \mathbf{R}|} \cdot \hat{\mathbf{n}} dS- \dfrac{1}{4\pi \epsilon_0} \int_{\mathscr{V}} \dfrac{\nabla_{\mathbf{R}} \cdot \mathbf{P}}{|\mathbf{r} - \mathbf{R}|} d^3 \mathbf{R} \\

&= \dfrac{1}{4\pi \epsilon_0} \int_{\partial \mathscr{V}} \dfrac{\sigma_{\mathrm{b}}}{|\mathbf{r} - \mathbf{R}|} dS + \dfrac{1}{4\pi \epsilon_0} \int_{\mathscr{V}} \dfrac{\rho_{\mathrm{b}}}{|\mathbf{r} - \mathbf{R}|} d^3 \mathbf{R}
\end{align*}upon defining ##\sigma_{\mathrm{b}} \equiv \mathbf{P} \cdot \hat{\mathbf{n}}## and ##\rho_{\mathrm{b}} \equiv - \nabla_{\mathbf{R}} \cdot \mathbf{P}##.
Wow, now I understood
Thanks
 
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