# Electric potential vs gravitational potential

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#1
I found that, when considering the gravitational potential energy a particle has (on a universal scale - not mgh, but -Gm1m2/r), it is necessary to consider one direction to be positive and the other to be negative (resolving forces) before converting from force terms to GPE terms.

How does one do something similar for electric fields? The problem is this sort of thing suggests that, the electric field strength on a line between a positive and negative charge never being 0, when you resolve the forces they will not be 0 at any point either, so the electric potential energy will never be 0 and thus the electric potential (which is just EPE per unit charge) will not be 0. This thinking must be wrong because I have various problems where EPE is 0 on just such a line.
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8 years ago
#2
I found that, when considering the gravitational potential energy a particle has (on a universal scale - not mgh, but -Gm1m2/r), it is necessary to consider one direction to be positive and the other to be negative (resolving forces) before converting from force terms to GPE terms.

How does one do something similar for electric fields? The problem is this sort of thing suggests that, the electric field strength on a line between a positive and negative charge never being 0, when you resolve the forces they will not be 0 at any point either,
correct
so the electric potential energy will never be 0
Why? There is no logical reason for this based on your previous correct statement.

Electric field strength is found from potential gradient. Not the actual value of the potential.
The potential can be zero at the mid point between two opposite charges because potential is a scalar and at that point one potential value is positive and the one due to the other charge is negative by an equal amount.
And if you are about to ask how can the potential be zero but the potential gradient (field) is not zero.
Think of any straight line function going through the origin of a graph at some angle, say, 45 degs.
The gradient at 0,0 is 45 degs (0.5) but the value of the function at 0.0 is of course zero.
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#3
(Original post by Stonebridge)
Why? There is no logical reason for this based on your previous correct statement.

Electric field strength is found from potential gradient. Not the actual value of the potential.
The potential can be zero at the mid point between two opposite charges because potential is a scalar and at that point one potential value is positive and the one due to the other charge is negative by an equal amount.
And if you are about to ask how can the potential be zero but the potential gradient (field) is not zero.
Think of any straight line function going through the origin of a graph at some angle, say, 45 degs.
The gradient at 0,0 is 45 degs (0.5) but the value of the function at 0.0 is of course zero.
I am beginning to see, but why is the methodology so different between electric and gravitational fields (when it comes to the calculation of potential energies and potentials)?

If I chose a point between two massive objects affecting a gravitational field, that point could have 0 potential energy for any mass there and thus 0 potential. Mass is essentially positive, so I would expect the same result for two positively- or negatively-charged particles producing an electric field, but no - instead, the potential energy and potential between the two positive particles is additive (above 0 if positive etc.), but between oppositely charged particles can be 0.

I'm not understanding how to think about calculating the electric potential energy and potential for a given point in a field with some number (we could say 2) of point charges? I suppose the question of why it's different for gravitational fields can wait till I at least know what I'm talking about with electric fields...
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8 years ago
#4
I've just written answers to this today.

http://www.thestudentroom.co.uk/show....php?t=2696361

There is a fundamental difference between gravitational and electrical fields when it comes to the positive and negative signs.
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8 years ago
#5
I found that, when considering the gravitational potential energy a particle has (on a universal scale - not mgh, but -Gm1m2/r), it is necessary to consider one direction to be positive and the other to be negative (resolving forces) before converting from force terms to GPE terms.

How does one do something similar for electric fields? The problem is this sort of thing suggests that, the electric field strength on a line between a positive and negative charge never being 0, when you resolve the forces they will not be 0 at any point either, so the electric potential energy will never be 0 and thus the electric potential (which is just EPE per unit charge) will not be 0. This thinking must be wrong because I have various problems where EPE is 0 on just such a line.

*cough*
r^2
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#6
(Original post by Joinedup)
*cough*
r^2
No
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#7
(Original post by Stonebridge)
I've just written answers to this today.

http://www.thestudentroom.co.uk/show....php?t=2696361

There is a fundamental difference between gravitational and electrical fields when it comes to the positive and negative signs.
Thanks for the link, it helps a lot.

Let me try and get this straight then. So the calculation of force and field strength is the same for the two fields (just resolve forces, being aware of what is attractive and repulsive; and if you want field strength at that point, consider the test mass 1 kg).

But the calculation of potential energy and potential is fundamentally different: for gravitational fields, you consider one direction to be positive and the other negative and then use the magnitudes of the GPEs of the interactions with each mass, to find GPE of a particle at that point (and then consider the particle being a test mass 1 kg at that point, for potential).
But for electric fields, you find EPE by essentially summing up contributions from all charges (which would thus be negative for a negative charge acting on a positive test charge - but would also be negative for a positive charge acting on a negative test charge), and then to find electric potential, divide by charge of the particle being considered (i.e. call it test charge 1C)?

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8 years ago
#8
Thanks for the link, it helps a lot.

Let me try and get this straight then. So the calculation of force and field strength is the same for the two fields (just resolve forces, being aware of what is attractive and repulsive; and if you want field strength at that point, consider the test mass 1 kg).

But the calculation of potential energy and potential is fundamentally different: for gravitational fields, you consider one direction to be positive and the other negative and then use the magnitudes of the GPEs of the interactions with each mass, to find GPE of a particle at that point (and then consider the particle being a test mass 1 kg at that point, for potential).
But for electric fields, you find EPE by essentially summing up contributions from all charges (which would thus be negative for a negative charge acting on a positive test charge - but would also be negative for a positive charge acting on a negative test charge), and then to find electric potential, divide by charge of the particle being considered (i.e. call it test charge 1C)?

Yes for the electric potential you find the potential at the point in question due to the individual charges, taking into account whether they are positive (giving a positive potential at the point) or negative, giving a negative potential at that point.
In the case of gravitational potential, as there are no "positive and negative" masses, and gravitation is always attractive, the potential at any point due to a mass is always negative, by the convention that an attractive force will require work to be done to separate the two masses, meaning that the potential is negative if it is taken as zero at infinity.
So potential can never be zero except at infinity for gravitation, because all masses in the system contribute negative potential at a point. Of course, you can find points where the field (force) is negative if the resultant is zero. For example, half way between two equal masses. The potential is not zero there, though. It's negative, being the sum of the negative potentials due to the two masses.
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#9
(Original post by Stonebridge)
Yes for the electric potential you find the potential at the point in question due to the individual charges, taking into account whether they are positive (giving a positive potential at the point) or negative, giving a negative potential at that point.
I see, thanks. And it's a linear sum? You just add up the magnitude of potential due to each charge, with a - sign for negative charges and + for positive, and for the r term do you have to take components in some way, or is r just the straight-line distance between the particle and charge?

And for electric potential energy you simply find potential at the position of the particle, as above, and then multiply by the charge of the particle for which you want to find EPE? (or vice versa you could find EPE first then divide by its charge to find electric potential of the field at that point - a test charge 1 C is just this dividing by 1 to give electric potential)
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8 years ago
#10
I see, thanks. And it's a linear sum? You just add up the magnitude of potential due to each charge, with a - sign for negative charges and + for positive, and for the r term do you have to take components in some way, or is r just the straight-line distance between the particle and charge?

And for electric potential energy you simply find potential at the position of the particle, as above, and then multiply by the charge of the particle for which you want to find EPE? (or vice versa you could find EPE first then divide by its charge to find electric potential of the field at that point - a test charge 1 C is just this dividing by 1 to give electric potential)
The r value is just the straight line distance from the charge producing the field to the point you want the potential at. Potential is scalar so they just add magnitudes. No vectors or components.
Yes. Potential energy E of a charge q at a point is just the value of that charge times the potential V at that point.
E = Vq

Or V = E/q if you wish to find V given the other two.
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8 years ago
#11
No
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