I'm wondering about the meaning of voltage in a circuit. Voltage is always described as related to the potential energy of the charge carriers. Often, the analogy is made to gravitational potential energy. This is fine, but what quality of the charge carriers gives rise to or contains this potential energy?

To put it in a different way, imagine I connect a battery with a resistor, so that a current flows around the circuit. On the wire connected to the positive terminal of the battery, we say that we have a "high" potential, while on the wire connected to the negative terminal the potential is low. What is actually different about the charge carriers on these two lengths of wire?

One possibility seems to me to be that the potential energy is related to the Coulomb force between the charge carriers, which could be different if the average distance between charge carriers is (ever so slightly) different on those two lengths of wire. I find this an attractive explanation, but I've never seen it written anywhere so I am skeptical.


2 Answers 2


What is actually different about the charge carriers on these two lengths of wire?

Nothing, they are the same carriers with same properties. What is different is that electric potential at the + terminal is higher than electric potential at the - terminal.

Electric potential is function of position. It depends on distribution of electric charge everywhere via the Coulomb-potential formula

$$ \varphi(\mathbf x) = K\int \frac{\rho(\mathbf x')}{|\mathbf x - \mathbf x'|}~d^3\mathbf x. $$ The closer the point $\mathbf x$ is to positively charged bodies, the higher the potential at $\mathbf x$ is.

The reason the potential is higher at the + terminal is that distribution of electric charge on surface of the conductors is such that there is more positive charge near + terminal and more negative charge near the - terminal. In a circuit, this distribution has to be maintained somehow, either it is maintained by a capacitor or a battery.

In your case, the + terminal of the battery and the connected wire has more positive electric charge distributed on its surface than the - terminal has.

  • $\begingroup$ Thanks, please et me extend your description of the battery: so when the battery is on its own, the potential difference is related to the excess positive charges on the positive terminal (likewise for the neg terminal). The electric potential thus seems related to the electrostatic repulsion acting on the charges at the terminals. So if I connect the battery to a resistor, and current flows, but the wires connected to the terminals maintain a potential difference, does that not imply that there is, e.g. still a relative excess of positive charge on the wire connected to the positive terminal? $\endgroup$
    – user
    Commented Jan 20, 2021 at 16:59
  • 1
    $\begingroup$ @user, that is true because a battery is not a capacitor. The potential is maintained by chemical processes, not the presence of charge. While some charge leaves via the connected circuit, the chemical reaction produces replacement charge, maintaining (approximately) the same potential. $\endgroup$
    – BowlOfRed
    Commented Jan 20, 2021 at 20:17

One difference between electrical and gravitational potential is the role of the dissipative forces.

We are used to deal with gravitational potential problems in a context of conservative forces, neglecting air resistance.

When a DC circuit is connected, that "conservative" part of the problem represents only a short transient, before a constant current flows. When the difference of potential is measured along a resistor, it corresponds to a constant charge velocity, not to an increase of its kinetic energy.

So, the situation can be compared to falling raindrops. Its velocity depends on $g$ and the air drag. $g$ plays the role of the electric field, and the air drag plays the role of the electric resistance. This analogy has limitations, because it requires a constant resistance between the battery poles.


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