I am confused about part d to the following question (defining the reference point $\displaystyle\lim_{x\to\infty}\phi(x)=0$). My instructor has told me that $\displaystyle\lim_{x\to-\infty}\phi(x)=0$ but this does not make sense to me. I include my work on the problem (linked below), which concludes that this potential would usually be non-zero, depending on the values for $R$, $d$, and $\rho$ (only if $d=0$ do my workings predict a potential at negative infinity of $0$). Is the potential actually zero at negative infinity, necessarily? If so, please explain.

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My work typed up on LaTex: link

  • $\begingroup$ Well, I didn't go through all your calculations, but the integral at the beginning of page 2 is written as if you entered the first sphere at $x=R$, while the sphere actually begins at $x=R+d/2$. Similarly, the second sphere actually ends at $x=-R-d/2$, not at $-R$. $\endgroup$ – HicHaecHoc Nov 7 '19 at 22:21
  • $\begingroup$ Also, your calculations are more complicated than necessary. Just calculate the potential $\psi (x)$ of only one positive sphere centered in the origin. Then you can find your solution as the difference between two appropriately translated $\psi$'s. $\endgroup$ – HicHaecHoc Nov 7 '19 at 22:32
  • $\begingroup$ @HicHaecHoc Ah yes, thanks for catching that. With my updated formula, the potential is indeed 0. And that simplification makes sense; it probably would have made my process much easier (now that I have finished integrating the total field there's no point in redoing it though). Link to my new work: drive.google.com/file/d/1qF-XM2L4OJjTBxaRTHmY1AJQ6oBtTvTO/… $\endgroup$ – Murey Tasroc Nov 8 '19 at 0:02

Keep a few things in mind and you won't have to do much integration. The law of superposition will help. So will the consideration of equivalent charge distributions.

The field of a uniformly charged sphere outside of that sphere is the same as it would be had all the charge was located at the center of the sphere.

For A), we are outside of both spheres, so we just calculate the field for to point charges.

For B), we pretend we only have one sphere and calculate the field from the center. Using Gauss' Law, $\vec{E}4\pi r^2=\rho(4\pi r^3)/3\epsilon_0\implies \vec{E}=\frac{\rho r}{3\epsilon_0}\hat{r}$. Keeping in mind $\vec{r}$ is vector pointing from the center of the sphere to the field point. Now, for a given field point, we figure out the vectors to the respective centers, and adjust this electric field formula appropriately, summing the electric field contributed by both spheres.

For C), we combine the approaches of A) and B), i.e., since we are outside of one of the spheres, we treat it as if all of its charge was located at its center, for the part contributed by the other sphere, which the field point is in, we use the radially dependent formula above. Keeping symmetry considerations in mind, a reflection and change in charge sign gives us the field in the complementary region.

For D), we integrate from $x=+\infty$ to the beginning of the sphere on the right. Since the field is effectively the field of two point charges, our integral is elementary. We evaluate the boundary points of the integration asserting the potential is zero at infinity. Once we are in one of the spheres, the integral becomes somewhat more complicated. We have two field contributions, one field term is proportional to r, the other is proportional to $\approx1/r^2$ (adjusting r to represent distance to off-center placement of the center of the sphere. In short, our potential will be functions of $r^2$ and $1/r$ in the spherical regions that don't overlap. Where they do overlap the contribution is only from terms proportional to $r^2$. Finally, given symmetry considerations, we can just use some reflections principles to calculate the field for the rest of the x axis.


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