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.
VVS2000
Messages
150
Reaction score
17
TL;DR 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
16377788745095814392662940793349.jpg
16377789157864281810467250891417.jpg
 
Physics news on Phys.org
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

View attachment 292998View attachment 292999
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}##.
 
Last edited:
  • Like
Likes Astronuc, berkeman, TSny and 1 other person
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
 
  • Like
Likes dlgoff and berkeman
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
I passed a motorcycle on the highway going the opposite direction. I know I was doing 125/km/h. I estimated that the frequency of his motor dropped by an entire octave, so that's a doubling of the wavelength. My intuition is telling me that's extremely unlikely. I can't actually calculate how fast he was going with just that information, can I? It seems to me, I have to know the absolute frequency of one of those tones, either shifted up or down or unshifted, yes? I tried to mimic the...
Back
Top