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In my AP physics guide the graph for electric potential (for a radial field, i.e. point charge) and gravitational potential (for radial field, i.e. point mass) slightly differ from each other and I want to know the "math" behind this.

The one given for gravitational potential:

enter image description here

Ones for electric potential:

enter image description here

I am not worried about there being two graphs for electric potential because I know this is because electric potential is positive or negative depending on the charge whose electric field is being described. However, I noticed that the gravitational field graph only starts at $R_E$, which is defined in the book as being the radius of the mass we are describing. On the other hand, there is no corresponding $R_E$ for the electric potential graphs. Does that mean that electric potential is also defined inside a charge as well as above its surface? (I assume gravitational potential is not defined inside a mass because the given graph for gravitational potential stops at $R_E$)

Qmechanic
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user45220
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    By mass, do you mean mass distribution? (like a system of particles.), if you do, I'm afraid the graph is incorrect. Mass distributions like the Earth, for example, do have potential due to their fields at a distance (from the center) less than their radius. – Hritik Narayan May 04 '15 at 12:33
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    @HritikNarayan: Sorry, I meant point mass or anything which behaves as if all its mass is concentrated at its center. (Edited) – user45220 May 04 '15 at 12:35
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    The plot you have for electric potential is drawn for a point-like charge. For a finite sized charged particle the field inside the charged particle indeed does not diverge. For example for a hollow sphere with some charge, the potential is constant inside the sphere and outside the sphere follows the behavior shown in your figure. – Echows May 04 '15 at 12:36
  • @Echows: Thanks! I don't think I fully understand how the graph for a point charge and normal charge would differ. Does it mean it's correct as given (if it's for a point charge)? – user45220 May 04 '15 at 12:43
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    Comment to the question (v3): $R_E$ undoubtedly stands for radius of the Earth. See e.g. this and this Phys.SE posts and links therein. – Qmechanic May 04 '15 at 12:43
  • @Qmechanic: Yes, you're right! On one page it gives "radius of earth" and on another it is "radius of the mass." (It uses $R_E$ in both of them) so I wasn't sure which one to put. – user45220 May 04 '15 at 12:45
  • @Qmechanic: I just saw the links and they are really helpful, thanks! – user45220 May 04 '15 at 12:59

1 Answers1

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The difference probably is that the graph for the gravitational potential is the one for a spherical mass distribution (or a sphere with a certain mass if you wish) and the electric one is given for a point charge. You could also draw the gravitational potential for a point mass, then it would look equivalent to your electrical potential, or the other way round.

In both cases you can calculate the electric/gravitational potential inside a charge/matter distribution. Note however, that it would look different than the one you presented here. So what you can not do is simply extend the same graph for the inside of your distribution.

Noldig
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  • Thanks for your answer! If we can't extend the same graph for the inside of the distribution how do we explain the given graph of electric potential? (It looks extended to me because they didn't label the radius of the charge, or does it have no radius because it's a point charge? If it doesn't then can we keep extending the graph forever?) – user45220 May 04 '15 at 12:41
  • I think this is the point, its the graph for a point charge. If you mean by "extending the graph forever" that you can go arbitrarily close to 0 then yes. You can see that the potential/field diverges for d=0. – Noldig May 04 '15 at 12:44
  • Sorry, I'm not sure I understand "diverges". Does it mean it changes when you get to 0 because that's the radius of a point charge? So this generalizes to any charge and the field diverges when you get to its radius? Thanks. – user45220 May 04 '15 at 12:46
  • Diverges means it becomes infinite. This only happens for a point charge, because you can get arbitrarily close to the charge. If the radius is finite, the potential will also have a finite value on the surface of the sphere. – Noldig May 04 '15 at 12:52
  • Thanks, I understand now. So if we can find electric/gravitational potential inside a charge/matter distribution how would the graph continue "below" the surface, and why doesn't the $R_E$ graph in the book show this? Also, just to make sure I know I would really appreciate if you could please confirm if this is correct: 1. For a point charge/mass the potential diverges because it has 0 radius. 2. For a charge/matter distribution that behaves like a point mass (or maybe it doesn't need to behave like a point mass? I'm not sure) the value is finite on the surface like the $R_E$ on the graph. – user45220 May 04 '15 at 12:57