Electric field due to polarized object

[VVS2000]

TL;DR

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

 

[kuruman]

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

View attachment 292998View attachment 292999

So you are. Is there a question you wish to ask?

 

[VVS2000]

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

 

[ergospherical]

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 $\rho_{\mathrm{b}} \equiv - \nabla_{\mathbf{R}} \cdot \mathbf{P}$.

 

[VVS2000]

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