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.