Electric potential

I've been spending some sizable amounts of time on this, but there remains one critical point I can't make heads nor tails of, and I need some help.

What does it means for an environment charge to have an electric potential of X volts?

As far as I can tell, all the electrostatic qualities (force, field, potential energy and potential) depend intimately on the distance between the source (environment charges) and the target (victim charge or point in space).

https://lh5.googleusercontent.com/-s...UNSET%255D.gif

It just does not seem to make sense to talk about an object being charged to a potential of -866V.

Exemplary problem

A thin, spherical, conducting shell of radius R is mounted on an isolating support and charged to a potential of –866 V. An electron is then fired directly toward the center of the shell, from point P at distance r from the center of the shell (r >> R). What initial speed v0 is needed for the electron to just reach the shell before reversing direction?

Isn't it just like talking about it if it was negatively charged (lots of electrons added) and positively charged (lots of electronics removed).

In your -866V case...lots of electronics added to it.

Should be, but an object having potential by itself has no meaning (with regards to my current understanding of electric potential).

In order for there to be potential energy, there needs to be a distance between the two charges, like with a compressed spring, you need to know how much is actually compressed. I have little idea how to put the electron into the picture.

U = qV = (-e)*(-866) = [some value] (J)

at which distance?

I could have sworn this was going to be about a new potential line tier.

*leaves disappointed*

Quote:

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Originally Posted by Kalovale

Should be, but an object having potential by itself has no meaning (with regards to my current understanding of electric potential).

In order for there to be potential energy, there needs to be a distance between the two charges, like with a compressed spring, you need to know how much is actually compressed. I have little idea how to put the electron into the picture.

U = qV = (-e)*(-866) = [some value] (J)

at which distance?

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I think the reference point for potential is at infinity, for both gravitational and electrical potential.

Scratch that, I think I got it.

The potential of an object (or at least a point charge) is the potential difference (caused by its "electricity-ness") between the position which it occupies and infinity.

The first bit seems to have been implied and caused confusion due to the object being spherically-shaped with a non-negligible radius.

O.k. So, you would need to translate that potential difference (V) into the electrostatic force, which is simply dividing the V by R. Now, you have N/C, and have a charge and mass, so can get the acceleration in the opposite dirrection. I didn't think too much about it, but I guess you would integrate the acceleration equation with respect to time, and go from velocity = infi to v = 0.

Sounds like you know what you are doing at any rate. Good luck!

Quote:

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Originally Posted by WayOfTime

O.k. So, you would need to translate that potential difference (V) into the electrostatic force, which is simply dividing the V by R. Now, you have N/C, and have a charge and mass, so can get the acceleration in the opposite dirrection. I didn't think too much about it, but I guess you would integrate the acceleration equation with respect to time, and go from velocity = infi to v = 0.

Sounds like you know what you are doing at any rate. Good luck!

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I was pursuing a more conceptual understanding of potential (essentially just what the heck it is and what does it mean to say something has a specific quantity of potential).
The answer I came up with:

- Electrical object(s) render(s) the space around them "electrical", every point in space where there is "electrical-ness", there is something called "electric potential"
- From Coulomb's law and various other sources, we know that "electrical-ness" gets stronger near the source(s) and dissipates at infinity
- The potential value at every point in space represents the amount of work per unit charge needed to move the test charge from infinity to that position
- Extrapolating that, people refer to "the potential of an object", which isn't supposed to make sense, to refer to the potential value of the point in space which that object occupies.

I still can't be certain with that definition, since it would be impossible to bring any test charge to that point in space, since it is already occupied by the source charge and therefore the work needed to bring a test charge close to it would blow up to infinity.

The whole idea above is elegantly encapsulated in the concept of equipotential
http://www.physics.uc.edu/~bortner/l...s/image007.jpg

Every point in the 2D surface is given a height (potential at that point) and the height AT the source should be infinity in the z-direction. Obviously it is not (as the problem in OP suggests), so I must still be missing something.

Ew, I should've come into this thread earlier.

Just FYI for potential energy (U), it's the negative integral of the force (k Q q over R squared, the electric field) from infinity to r. They decided to define the potential energy at infinity 0.

So the definitive answer to this whole issue is just a lack of clarity in wording:

Charging an object to an electric potential of 866 V
means
Charging an object to a charge Q such that the electric potential on the surface (NOT at the center nor at a point charge) is 866 V.

Yes, assuming the electric potential at infinity distance away is 0.

Quote:

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Originally Posted by Kalovale

So the definitive answer to this whole issue is just a lack of clarity in wording:

Charging an object to an electric potential of 866 V
means
Charging an object to a charge Q such that the electric potential on the surface (NOT at the center nor at a point charge) is 866 V.

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Well, technically the voltage inside the object is the same as at the surface, but that is moot.

Quote:

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Originally Posted by Kalovale

So the definitive answer to this whole issue is just a lack of clarity in wording:

Charging an object to an electric potential of 866 V
means
Charging an object to a charge Q such that the electric potential on the surface (NOT at the center nor at a point charge) is 866 V.

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The problem states that it is a conducting sphere which means that it has a uniform charge density. There is no distinction between the surface charge and the charge at the center. If I understand correctly, you problem actually stems from wanting to understand too much, too fast. Your simple problem doesn't really make a distinction between the charge on the sphere and the charge in it (hence placing the particle at r >> R), because its trivial to the results of the problem and incredibly complicated to define in most cases. Effectively, this idealized sphere is a particle.

There are better definitions around I'm certain, but I would attempt to define electric potential as a measure of the potential resistive or attractive force exerted by an electric field on a charged particle.

To clarify your graph, yes but no. You're delving into the realm of super colliders trying to split particles by running them into each other at relativistic speeds. Here there be complicated pomegranate were people just really shouldn't go without a Ph.D and I won't presume to try and explain effectively. With only the Maxwell's equations at your disposal, I don't believe it's possible to accurately explain it.