Why doesn't an electric charge use up all its potential energy travelling from positive to negative terminal?

Question

When an electric charge flows from high potential (positive terminal of battery) to low potential (negative terminal of battery), why doesn't the charge use up all of its potential energy at the start to move to the negative terminal? Why is there still enough energy to power a light bulb? E.g. A ball which is lifted above the ground has potential energy and all the energy is converted to kinetic energy to move to the ground when released. Why isn't all the electric potential energy of the charge converted to energy to move to the negative terminal as well? Where does extra energy to power a light bulb come from?

Answer 1

Speed of electrons

You are still struggling with the concept of charge and current raised in your [previous question] (If an electron has 0 electric potential after passing through a resistor, how does it flow to the other terminal). Let's deal with your question first.

E.g. A ball which is lifted above the ground has potential energy and all the energy is converted to kinetic energy to move to the ground when released.

This isn't a good analogy for a current. At the instant the ball is released it still has all the potential energy and zero kinetic energy. As the ball accelerates due to gravity the potential energy decreases (due to loss of height) and the kinetic energy increases (due to increased velocity). An electrical analogy of this would be an electron beam in a vacuum tube where the effect on a single electron and its acceleration from cathode to anode can be considered.

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. Electron current in a thermionic valve / tube.

Have a look at the Wikipedia article on drift velocity. It shows the calculations for a current I = 3 amperes, and a wire of 1 mm diameter and comes up with the result that in this wire the electrons are flowing at the rate of −0.0000028 m/s. At this rate it would take 99 hours for an electron to travel 1 m along that wire! If we double the current the time taken is halved.

As the electrons move through the resistive material they interact with the atoms in that material and that's what limits the current.

Speed of electricity

With reference to Wikipedia's Speed of electricity:

The speed at which energy or signals travel down a cable is actually the speed of the electromagnetic wave, not the movement of electrons. Electromagnetic wave propagation is fast and depends on the dielectric constant of the material. In a vacuum the wave travels at the speed of light and almost that fast in air.

The speed of the electric current will be somewhere between 50% and 99% of c, the speed of light (\$3 \cdot 10^8~m/s\$). See Wikipedia's Velocity factor for more information.

The end result

The electrical current is a very slowly moving jostling of electrons through the conductor. The propagation of the wave is very fast. Think of a line of marbles in a pipe. If we push a marble in one end a marble will pop out the other end immediately despite the fact that each marble has only moved one marble diameter.

Stop thinking of charge. Start thinking of current.

Answer 2

In your ball-drop analogy, not all of the potential energy of the ball is converted permanently into kinetic energy, moving the ball to the ground. Actually, virtually none of it is.

1. Ball is released, potential energy begins converting into kinetic energy of ball.
2. Ball strikes air molecules on way to floor/ground, some kinetic energy of ball transfers to kinetic energy of air, & some converts to heat.
3. Ball strikes floor/ground, some kinetic energy of ball transfers to floor/ground kinetic energy, some converts back into opposite-vector kinetic energy in ball, some converts to heat, & some converts to sound pressure wave.
4. Ball bouncess off floor/ground, repeat step 2, converts remaining kinetic energy into potential energy (peak of bounce), then repeats all steps until ball energy is fully discharged.

Now, with electrons in a closed circuit, the transfers are usually voltage (potential energy) to current (kinetic energy), and current to heat/light/work output (through a resistance, whether that be wire, light, transistor, motor, etc.), with the ball-bounce energy actually being (normally in DC circuits) avoided, as the last remaining energy in the electron is spent heating the battery as it passes through the battery's ESR (effective series resistance).

Answer 3

You're mixing up conventions here. If you assume the charge carriers are negative (electrons), then the negative terminal of the battery is the high-potential terminal. Electric charge has a sign, not just a magnitude. Gravity only has one kind of "charge", but electromagnetism has two! Electric fields push positive charges in one direction and negative charges in the opposite direction.

In most areas of electrical engineering, we pretend that the moving charges in a circuit are always positive, and that current flows out of the positive terminal of a battery. This is mainly done for historical reasons. In most cases, it's really the negative electrons that are moving, but you get the same results from the math as long as you're consistent.