# Why does high voltage imply a high “pushing force” on electrons?

Another word for potential difference, is apparently, "voltage drop", although I've heard potential difference be interchangeable with just voltage, as well, a bit confusingly. However, if I understand it properly, the battery has an electric field. Things closest to it are being pushed with the greatest force.

However, my understanding of potential difference needs a bit of refining.

Then the difference in voltage between any two points, connections or junctions (called nodes) in a circuit is known as the Potential Difference, ( p.d. ) commonly called the Voltage Drop. The Potential difference between two points is measured in Volts with the circuit symbol V, or lowercase “v“, although Energy, E lowercase “e” is sometimes used to indicate a generated emf (electromotive force). Then the greater the voltage, the greater is the pressure (or pushing force) and the greater is the capacity to do work.

The greater the voltage drop, the greater the capacity to do work? Does this have some similarity to closely spaced contour lines? If there's a big voltage drop in a small distance, this implies we are in close proximity to the battery? Because the electrostatic force obeys the inverse square law, so the electrostatic force will be exponential with distance, and thus a high voltage drop implies high force difference in short distance, implying a close spatial proximity to the battery? Is my understanding correct? If not, could someone help refine it?

• The electric field is a potential field (curl = 0), and it's potential is the voltage. Equivalently, the gradient of voltage is the electric field. – τεκ Feb 26 '18 at 2:09

What exerts a force on charged particles is the electric field, which is proportional to the gradient of the potential.

So when the voltage (or potential) is changing rapidly between nearby points in space, that's when you have a strong force exerted on charges.

Things closest to it [a battery] are being pushed with the greatest force.

This isn't necessarily true.

I could connect two wires to the two terminals of the battery and run them a long distance away, and then place the ends of the two wires very close to each other (but not touching). Then )very nearly) the full potential difference of the battery would be found across the small gap between the two wires, and the field strength between the wire ends would be much stronger than the field strength near the actual battery terminals.

Schematically, here's the situation I'm trying to describe:

simulate this circuit – Schematic created using CircuitLab

(Imagine the battery symbols represent several cells within a battery)

• Why would it be stronger? – sangstar Feb 26 '18 at 2:51
• Because it's the same potential difference over a much smaller distance. So the gradient will be larger. – The Photon Feb 26 '18 at 2:51
• Because there's hardly any current flowing through the wires, you know from ohm's law that the potential at the two ends of the wire is the same. – The Photon Feb 26 '18 at 2:57
• So you’re saying since there’s hardly any current flowing through the wires, and this implies there’s hardly any voltage (this would make sense?) this would allow me to infer there’s the same amount of potential at either end? And why is there hardly any current flowing through the wires in the first place? Oh, because the circuit isn’t closed, and thus no current (or hardly any) nor voltage. But then why does this lead us to say the same potential at either end? – sangstar Feb 26 '18 at 3:07
• Yes, because the circuit isn't closed. For each individual wire, if there's no current flowing through it, the potential will be the same at all points within the wire, even if we consider two points at opposite ends of the wire. We know the potential is equal, because if it wasn't then there would be current flowing in the wire. – The Photon Feb 26 '18 at 3:09