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

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

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eyeweyew said:
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 $$\overrightarrow{E} = \frac{1}{4\pi\epsilon r^2}\widehat{r}$$
eyeweyew said:
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.

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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##.

eyeweyew, nasu and renormalize
eyeweyew said:
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.

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

The two path integrals of E must be equal because the potential at B must be the same for each path.
Your assumption, "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." is WRONG. The lines of E will not be straight lines in either dielectric.

Meir Achuz said:
The lines of E will not be straight lines in either dielectric.
That's wrong. The E-field lines point straight in the radial direction in both dielectrics, as discussed in posts #3 and #4 above.

Meir Achuz said:
#3 and #4 are wrong. ##\nabla\cdot{\bf D}=\nabla\cdot[\epsilon({\bf r)E]}=0.##
OK, please start from this equation and show your detailed derivation of the "solution" that exhibits curved E-field lines. Meantime, I stand by my post #4.

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Meir Achuz said:
I am a bit puzzled. You have shown that ##\nabla\cdot{\bf E}=0## implies
that ##{\bf E}##~##{\bf\hat r}##.
Then, would you agree that ##\nabla\cdot[\epsilon({\bf r) E]}=0## implies
that ##\epsilon({\bf r) E}##~##{\bf\hat r}##.
This shows that ##{\bf E(r)}## cannot ~##{\bf\hat r}##.
I don't understand what you're getting at. If ##\epsilon({\bf r) E}##~##{\bf\hat r}## I can divide by the scalar field ##\epsilon({\bf r)}## to conclude that ##{\bf E(r)}##~##{\bf\hat r}## at any point where ##\epsilon({\bf r)}\neq 0## (which is everywhere in this particular problem). The result of dividing (or multiplying) a vector by any scalar points in exactly the same direction as the original vector.

Meir Achuz said:
This is not even wrong, because the left-hand side is undefined, but it does show that you're guess for E is wrong.
The right-hand side shows that there must be some vector dependence beyond ##{\bf\hat r}##.
\begin{align}
0&=\epsilon\left(\nabla\cdot{\bf E\left(r\right)}\right)+\mathbf{E}\left(\mathbf{r}\right)\cdot\left[\mathbf{\hat{i}}\delta\left(x\right)\left(\epsilon_{\mathrm{2}}-\mathrm{\epsilon}_{\mathrm{1}}\right)\right]\nonumber\\&=\epsilon\left(\nabla\cdot{\bf E}\left(\mathbf{r}\right)\right)+E_{x}\left(x,y,z\right)\delta\left(x\right)\left(\epsilon_{\mathrm{2}}-\mathrm{\epsilon}_{\mathrm{1}}\right)\nonumber\\&=\epsilon\left(\nabla\cdot{\bf E}\left(\mathbf{r}\right)\right)+E_{x}\left(0,y,z\right)\delta\left(x\right)\left(\epsilon_{\mathrm{2}}-\mathrm{\epsilon}_{\mathrm{1}}\right)\nonumber\\&=\epsilon\left(\nabla\cdot{\bf E}\left(\mathbf{r}\right)\right)\nonumber
\end{align}
since:$${\bf E}\left(\mathbf{r}\right)\left|_{x=0}\right.=\left(\frac{k}{r^{2}}\mathbf{\hat{r}}\right)\left|_{x=0}\right.=\frac{k}{y^{2}+z^{2}}\left(\mathbf{\hat{j}}\cos\phi+\mathbf{\hat{k}}\sin\phi\right)\Rightarrow E_{x}\left(0,y,z\right)=0$$The solution ##{\bf E}\left(\mathbf{r}\right)=\\k\,\mathbf{\hat{r}}/r^{2}## is correct.

Meir Achuz said:
Some more points:
1. All textbook derivations I have seen for getting ##\nabla({\bf r}/r^3=4\pi\delta({\bf r})## use the fact that
##{\bf r}/r^3## is spherically symmetric, and not would work if ##\epsilon({\bf r})## is included.
2. If a configuration is not spherically symmetric, then no solution will be spherically symmetric.
3. This problem does have a solution which clearly shows that the electric field is not in the radial direction.
It is the multipole expansion, and requires all even multipoles.
Again, I claim this is entirely wrong.
The following reference: https://unlcms.unl.edu/cas/physics/tsymbal/teaching/EM-913/section4-Electrostatics.pdf solves for the electrostatic potential ##\Phi## due to a point charge ##q## standing-off at a distance ##d## from a planar dielectric interface. It uses the method of images and works in cylindrical coordinates:

The result is:

Our interest is the case where ##q## resides exactly on the interface, so we set ##d=0## to find:$$\Phi_{1}\left(z\right)\left|_{d=0}\right.=\Phi_{2}\left(z\right)\left|_{d=0}\right.=\frac{q}{2\pi\sqrt{s^{2}+z^{2}}\left(\varepsilon_{1}+\varepsilon_{2}\right)}=\frac{1}{4\pi\overline{\varepsilon}}\frac{q}{r}\:\text{where }\overline{\varepsilon}\equiv\frac{\varepsilon_{1}+\varepsilon_{2}}{2}$$Therefore, the electric field is:$$\overrightarrow{E}\left(r\right)=-\nabla\left[\frac{1}{4\pi\overline{\varepsilon}}\frac{q}{r}\right]=\frac{1}{4\pi\overline{\varepsilon}}\frac{q}{r^{2}}\mathbf{\hat{r}}$$which is precisely the solution from my post #4 that exhibits straight E-field lines in the radial direction.
If you still assert otherwise, the burden is on you to post either your own derivation or else a scholarly reference that obtains the solution in terms of multipoles and manifests curved E-field lines.

Last edited:
eyeweyew, Motore, Meir Achuz and 1 other person
Yes, that's a good derivation. In fact, it's so standard I had forgotten about it. I'm sorry to have wasted your time.

eyeweyew, weirdoguy and renormalize
renormalize said:
Again, I claim this is entirely wrong.
The following reference: https://unlcms.unl.edu/cas/physics/tsymbal/teaching/EM-913/section4-Electrostatics.pdf solves for the electrostatic potential ##\Phi## due to a point charge ##q## standing-off at a distance ##d## from a planar dielectric interface. It uses the method of images and works in cylindrical coordinates:
View attachment 349461
The result is:
View attachment 349462
Our interest is the case where ##q## resides exactly on the interface, so we set ##d=0## to find:$$\Phi_{1}\left(z\right)\left|_{d=0}\right.=\Phi_{2}\left(z\right)\left|_{d=0}\right.=\frac{q}{2\pi\sqrt{s^{2}+z^{2}}\left(\varepsilon_{1}+\varepsilon_{2}\right)}=\frac{1}{4\pi\overline{\varepsilon}}\frac{q}{r}\:\text{where }\overline{\varepsilon}\equiv\frac{\varepsilon_{1}+\varepsilon_{2}}{2}$$Therefore, the electric field is:$$\overrightarrow{E}\left(r\right)=-\nabla\left[\frac{1}{4\pi\overline{\varepsilon}}\frac{q}{r}\right]=\frac{1}{4\pi\overline{\varepsilon}}\frac{q}{r^{2}}\mathbf{\hat{r}}$$which is precisely the solution from my post #4 that exhibits straight E-field lines in the radial direction.
If you still assert otherwise, the burden is on you to post either your own derivation or else a scholarly reference that obtains the solution in terms of multipoles and manifests curved E-field lines.
Thanks for posting this reference!

Thank you so much for all the answers here! Just one last thing, I want to confirm other than this particular case, it is true that path independence holds if permittivity is non-uniform in general right? Thanks!

eyeweyew said:
I want to confirm other than this particular case, it is true that path independence holds if permittivity is non-uniform in general right?
Yes, that's right. A time-dependent magnetic field must be present for ##\int_A^B \vec E \cdot \vec{dl}## to be path-dependent. This follows directly from one of Maxwell's equations.

I want to confirm I understand path independence correctly in circuit analysis. In the circuit below, since for loop1, we have $$\oint_{closed}^{}\overrightarrow{E}\cdot{d}\overrightarrow{l}=-V+I_tR_t+I_1R_1=0$$

and for loop2 we have $$\oint_{closed}^{}\overrightarrow{E}\cdot{d}\overrightarrow{l}=-V+I_tR_t+I_2R_2= -L\frac{dI_2}{dt}\neq0$$

so that's why we say it is path dependent right?

eyeweyew said:
I want to confirm I understand path independence correctly in circuit analysis. In the circuit below, since for loop1, we have $$\oint_{closed}^{}\overrightarrow{E}\cdot{d}\overrightarrow{l}=-V+I_tR_t+I_1R_1=0$$

and for loop2 we have $$\oint_{closed}^{}\overrightarrow{E}\cdot{d}\overrightarrow{l}=-V+I_tR_t+I_2R_2= -L\frac{dI_2}{dt}\neq0$$

so that's why we say it is path dependent right?

View attachment 350453
Yes, that's right. The time-dependent B field of the inductor makes ##\oint^{}\overrightarrow{E}\cdot{d}\overrightarrow{l}## path dependent.

eyeweyew
renormalize said:
Again, I claim this is entirely wrong.
The following reference: https://unlcms.unl.edu/cas/physics/tsymbal/teaching/EM-913/section4-Electrostatics.pdf solves for the electrostatic potential ##\Phi## due to a point charge ##q## standing-off at a distance ##d## from a planar dielectric interface. It uses the method of images and works in cylindrical coordinates:
View attachment 349461
The result is:
View attachment 349462
Our interest is the case where ##q## resides exactly on the interface, so we set ##d=0## to find:$$\Phi_{1}\left(z\right)\left|_{d=0}\right.=\Phi_{2}\left(z\right)\left|_{d=0}\right.=\frac{q}{2\pi\sqrt{s^{2}+z^{2}}\left(\varepsilon_{1}+\varepsilon_{2}\right)}=\frac{1}{4\pi\overline{\varepsilon}}\frac{q}{r}\:\text{where }\overline{\varepsilon}\equiv\frac{\varepsilon_{1}+\varepsilon_{2}}{2}$$Therefore, the electric field is:$$\overrightarrow{E}\left(r\right)=-\nabla\left[\frac{1}{4\pi\overline{\varepsilon}}\frac{q}{r}\right]=\frac{1}{4\pi\overline{\varepsilon}}\frac{q}{r^{2}}\mathbf{\hat{r}}$$which is precisely the solution from my post #4 that exhibits straight E-field lines in the radial direction.
If you still assert otherwise, the burden is on you to post either your own derivation or else a scholarly reference that obtains the solution in terms of multipoles and manifests curved E-field lines.
I look into the example 1 in this reference and I find something that bothers me a little bit.

When I try to find potential and electric field at the point 2d distance away from the q charge in both dialectrics ε1 and ε2 on the z-axis(s=0), i.e.: p1 and p2 in the following picture. I got for p1

$$\\\\\Phi_{1}\left(z\right)\left|_{s=0,z>0}\right.=\frac{1}{4\pi\varepsilon_{1}}\left(\frac{q}{d-z}+\frac{\left(\varepsilon_{1}-\varepsilon_{2}\right)}{\left(\varepsilon_{1}+\varepsilon_{2}\right)}\frac{q}{d+z}\right) \\\\E_{1}\left(z\right)\left|_{s=0,z>0}\right.=-\nabla\Phi_{1}\left(z\right)\left|_{s=0,z>0}\right.=-\frac{d\Phi_{1}\left(z\right)\left|_{s=0,z>0}\right.}{dz}=\frac{1}{4\pi\varepsilon_{1}}\left(\frac{\left(\varepsilon_{1}-\varepsilon_{2}\right)}{\left(\varepsilon_{1}+\varepsilon_{2}\right)}\frac{q}{\left( d+z \right)^{2}}-\frac{q}{\left( d-z \right)^{2}}\right) \hat{z}$$

assume q is a positive unit charge so

$$\\\\E_{1}\left(z=3d\right)\left|_{s=0,z>0}\right.=\frac{1}{4\pi\varepsilon_{1}}\left(\frac{\left(\varepsilon_{1}-\varepsilon_{2}\right)}{\left(\varepsilon_{1}+\varepsilon_{2}\right)}\frac{1}{16d^{2}}-\frac{1}{4d^{2}}\right) \hat{z}$$

Since the term inside the bracket is less than 0, the electric field at z=3d is pointing toward the unit positive charge which seems wrong?

and for p2

$$\\\\ \Phi_{2}\left(z\right)\left|_{s=0,z<0}\right.=\frac{1}{2\pi}\frac{1}{\left(\varepsilon_{1}+\varepsilon_{2}\right)}\frac{q}{d-z} \\\\E_{2}\left(z\right)\left|_{s=0,z<0}\right.=-\nabla\Phi_{2}\left(z\right)\left|_{s=0,z<0}\right.=-\frac{d\Phi_{2}\left(z\right)\left|_{s=0,z<0}\right.}{dz}=-\frac{1}{2\pi}\frac{1}{\left(\varepsilon_{1}+\varepsilon_{2}\right)}\frac{q}{\left( d-z \right)^{2}}\hat{z}$$

again assume q is a positive unit charge so

$$\\\\E_{2}\left(z=-d\right)\left|_{s=0,z<0}\right.=-\frac{1}{2\pi}\frac{1}{\left(\varepsilon_{1}+\varepsilon_{2}\right)}\frac{1}{4d^{2}}\hat{z}$$

Electric field at z=-d is pointing away from the unit positive charge so it is correct.

eyeweyew said:
I look into the example 1 in this reference and I find something that bothers me a little bit.

When I try to find potential and electric field at the point 2d distance away from the q charge in both dialectrics ε1 and ε2 on the z-axis(s=0), i.e.: p1 and p2 in the following picture. I got for p1

View attachment 350573

$$\\\\\Phi_{1}\left(z\right)\left|_{s=0,z>0}\right.=\frac{1}{4\pi\varepsilon_{1}}\left(\frac{q}{d-z}+\frac{\left(\varepsilon_{1}-\varepsilon_{2}\right)}{\left(\varepsilon_{1}+\varepsilon_{2}\right)}\frac{q}{d+z}\right)$$
The denominator of the first term should represent the (positive) distance from ##q## to the point on the z-axis where you are evaluating the potential in medium 1.

So the denominator should be written ##|d-z|## instead of ##(d-z)##. In the neighborhood of ##p_1##, ##z > d##. So, ##|d-z| = z-d##.

renormalize

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