# Why don't electrons still have energy after completing a circuit with a battery?

Firstly a battery causes an accumulation of electrons in the negative terminal right? Hence the positive terminal is relatively positive and so an electric field is produced.

This electric field will cause electrons to move from the negative terminal to the positive terminal and on doing so the electrical potential would decrease and the change in electrical potential is what’s called voltage.

Now I always think of electrons moving in a field as a ball being dropped from a height so this potential energy is converted into another kind of energy which is supplied to components.

After an electron supplies energy to a component how does it continue to have no energy? Wouldn’t the electric field accelerate the electron so that the electron would gain more energy even after supplying energy to a component?

• Thank you so much, I didn’t know exactly how to word my question it’s a skill I always lacked Sep 16 '20 at 20:14
• Why do you think electrons that leave components no longer have any energy? Sep 16 '20 at 21:03
• Well that’s atleast how I’ve been taught so far in high school Sep 16 '20 at 21:19

Now I always think of electrons moving in a field as a ball being dropped from a height so this potential energy is converted into another kind of energy which is supplied to components.

Not a bad way to think of it. Normally you don't short-circuit the battery, you put a load across it. Think of this as a resistance to the falling ball. Perhaps rather than air, you drop it in oil and it descends at a constant (but slow) speed.

The ball is losing potential energy as it falls, and the oil is gaining thermal energy. The KE of the ball is constant.

After an electron supplies energy to a component how does it continue to have no energy? Wouldn’t the electric field accelerate the electron so that the electron would gain more energy even after supplying energy to a component?

Here's where the big difference between a circuit and gravity comes. Near the surface of the earth, the gravitational field is approximately constant. But the field in the circuit is not. After the last load is present, there will be a very tiny electric field, just enough to overcome the small resistance in the wires. There is no acceleration after the last load because the field is tiny.

The electric field in steady-state has reconfigured itself such that the current is constant at all points. So the resistors that would otherwise slow down the current have a large field pushing the charge, and the low-resistance wires have very small fields.

• Wow, I’m a high school student and to know all of this occurs is truly humbling. Knowledge is truly a sea. Thank you so much I understood. Sep 16 '20 at 21:37

Wouldn’t the electric field accelerate the electron so that the electron would gain more energy even after supplying energy to a component?

In essence, yes. What you are noticing here is a consequence of the fact that energy is not transported by the charge carriers but by the fields themselves.

The description of energy transport in electromagnetism is called Poynting’s theorem. When you apply it to ordinary circuits you find that most energy transport in a circuit is actually through the fields outside the wires. The only energy transport in the actual wires of a normal circuit is Ohmic losses in the wire itself, and that is energy transport radially into the wire rather than along the wire. All of the useful energy is transported outside the wire.

While it is true that charges in different parts of the circuit have different potential energy, that energy does not ride on the charge to be deposited in a component like a delivery truck dropping off supplies. Again, that potential energy is stored in the resulting field configuration. Hence, as you noticed, the fields may still move the electrons after they exit a component.

The energy in a battery comes from chemical potential. It is the gibbs free energy that is the electrical energy. When the anode and cathode are touching and there is no external circuit the energy is converted to heat not electrical energy. The Nernst Equation relates the Gibbs Free Energy to the Open Circuit Voltage: $$\Delta G=-nFE$$ n is the number of moles of electrons, F is Faradays constant, E is the open circuit potential. This is further related to the concentrations and the equilibrium constant of the reaction. An easy way to guesstimate the voltage is through a standard reduction table which assumes things about Concentration, temperature etc. The standard potential and cell potential is Related to the HOMO and LUMO of the chemical species but is not equal to it.

basically the Nernst equation says: $$\Delta G =-nFE= \mu dN = -RTlnK + RTlnQ$$ mu is the chemical potential and is in between the HOMO and LUMO of the relevant oxidised and reduced species. K is the equilibrium constant and it stands to reason the bigger the equilibrium constant the more negative the gibbs free energy and the larger the cell potential (this reaction likes to form products). Q is the current ratio of concentrations in the cell, when the cell is at equilibrium the voltage is zero as is the Gibbs Free Energy

finally W=QV implies Q=-nF and mu=FV where mu is the chemical potential