Does path independence still hold if permittivity is non-uniform?

[eyeweyew]

TL;DR Summary

Does path independence still hold if permittivity is non-uniform

Consider a scenario in the picture where one half of space consists of a material with permittivity ϵ1 and the other half consists of a material with permittivity ϵ2, where ϵ1 > ϵ2. A unit positive charge is fixed at the interface between the two materials. Path1 is entirely within the material with permittivity ϵ1 and Path2 is entirely within the material with permittivity ϵ2. The paths are reflectively symmetric, with the interface acting as the mirror line. Just for simplicity, also assume the arc part of the both paths is perpendicular to the electric field comes from the fixed charge so they don't contribute to the work. Does it still require the same amount of work to move a unit positive charge from point A to point B along Path1 and Path2?

$$W=\int_{A}^{B}\overrightarrow{E}\cdot\overrightarrow{dl}$$

where $$\overrightarrow{E} = \frac{1}{4\pi\epsilon r^2}\widehat{r}$$

 

[Alex Schaller]

TL;DR Summary: Does path independence still hold if permittivity is non-uniform

View attachment 347696
Consider a scenario in the picture where one half of space consists of a material with permittivity ϵ1 and the other half consists of a material with permittivity ϵ2, where ϵ1 > ϵ2. A unit positive charge is fixed at the interface between the two materials. Path1 is entirely within the material with permittivity ϵ1 and Path2 is entirely within the material with permittivity ϵ2. The paths are reflectively symmetric, with the interface acting as the mirror line. Just for simplicity, also assume the arc part of the both paths is perpendicular to the electric field comes from the fixed charge so they don't contribute to the work. Does it still require the same amount of work to move a unit positive charge from point A to point B along Path1 and Path2?

$$W=\int_{A}^{B}\overrightarrow{E}\cdot\overrightarrow{dl}$$

where $$\overrightarrow{E} = \frac{1}{4\pi\epsilon r^2}\widehat{r}$$

TL;DR Summary: Does path independence still hold if permittivity is non-uniform

View attachment 347696
Consider a scenario in the picture where one half of space consists of a material with permittivity ϵ1 and the other half consists of a material with permittivity ϵ2, where ϵ1 > ϵ2. A unit positive charge is fixed at the interface between the two materials. Path1 is entirely within the material with permittivity ϵ1 and Path2 is entirely within the material with permittivity ϵ2. The paths are reflectively symmetric, with the interface acting as the mirror line. Just for simplicity, also assume the arc part of the both paths is perpendicular to the electric field comes from the fixed charge so they don't contribute to the work. Does it still require the same amount of work to move a unit positive charge from point A to point B along Path1 and Path2?

$$W=\int_{A}^{B}\overrightarrow{E}\cdot\overrightarrow{dl}$$

where $$\overrightarrow{E} = \frac{1}{4\pi\epsilon r^2}\widehat{r}$$

I think that work would be different for both sides.

 

[TSny]

The work is the same for both paths. It turns out that the electric field is given by the same expression in both dielectrics: $$\vec E = \frac{q}{2\pi(\epsilon_1+\epsilon_2) r^2} \hat r,$$ where $r$ is the distance from the charge $q$ and $\hat r$ is the unit radial vector pointing away from $q$.

 

[renormalize]

TL;DR Summary: Does path independence still hold if permittivity is non-uniform

Yes, it can be easily shown that path independence holds for this particular case. Label the half space containing $\varepsilon_{1}$ as Region 1 and that of $\varepsilon_{2}$ as Region 2. Erect a spherical coordinate system $r,\theta,\phi$ centered on the charge $Q$ and aligned such that the interface between regions 1 and 2 is the $x,y$ plane, i.e., where $\theta=\pi/2$. In both regions, the electric field satisfies Coulomb's Law: $$\nabla\cdot\overrightarrow{E}=0\Rightarrow\overrightarrow{E}\left(r,\theta,\varphi\right)\propto\frac{\hat{r}}{r^{2}}\Rightarrow\overrightarrow{E}_{1}\left(r,\theta,\varphi\right)=\frac{k_{1}}{r^{2}}\hat{r},\quad\overrightarrow{E}_{2}\left(r,\theta,\varphi\right)=\frac{k_{2}}{r^{2}}\hat{r}$$In addition, the tangential component of $\overrightarrow{E}$ at the dielectric interface must be continuous:$$\overrightarrow{E}_{1}\left(r,\frac{\pi}{2},\varphi\right)=\overrightarrow{E}_{2}\left(r,\frac{\pi}{2},\varphi\right)\Rightarrow k_{1}=k_{2}\equiv k$$In other words, the electric field $\overrightarrow{E}_{1}=\overrightarrow{E}_{2}=\overrightarrow{E}$ has the single functional form $\overrightarrow{E}=k\,\hat{r}/r^2$ throughout all space. The constant $k$ can easily be found by applying Gauss's law to a spherical surface surrounding $Q$, one hemisphere in Region 1 and the other in Region 2, with the result:$$k=\frac{Q}{4\pi\overline{\varepsilon}}\Rightarrow\overrightarrow{E}=\frac{Q}{4\pi\overline{\varepsilon}r^{2}}\hat{r}\quad\text{where}\;\overline{\varepsilon}\equiv\frac{1}{2}\left(\varepsilon_{1}+\varepsilon_{2}\right)$$Thus, a charge $Q$ located at the interface between two dielectrics gives rise to a uniform coulomb field of exactly the same form as if the charge were embedded in a single medium with the average dielectric. Hence, the work done on transporting a test charge is path-independent.

 

[tech99]

If the charge is a point, how can it be simultaneously in both dielectrics?