# Voltage and electric field in a closed circuit confusion

I have a doubt about the intuition behind Electric field and Potential difference. Let's imagine a closed circuit with a battery, a resistance and a lightbulb. We observe that the lightbulb is on. What I infer from this is that electrons are slowly going around, and thus there must be an electric field. It seems intuitive to think that to keep the electrons moving, the electric field must act everywhere in the circuit (1).

However, if I measure with a voltmeter the voltage difference at two points in an ideal wire, without no components in between, the voltage difference is zero, which by the definition of electric field and voltage implies that the electric field is also zero (2).

Point 1 and point 2 are contradicting one another. What is wrong here?

• Measuring the voltage in an ide wire wire a battery attached to it, will not be zero. May 9 at 19:27

What is wrong here?

You aren't thinking about what the "ideal wire" model is simplifying.

The real wire has a small resistance. Maybe a few milliohms. And so there is a small electric field along the wire to produce the current.

When we use the ideal wire approximation we ignore these small potential differences and small electric field components in order to simplify the circuit analysis, knowing it won't affect the important result significantly. But that doesn't mean there is really no electric field and no potential difference in a real circuit built with real wire instead of ideal wire.

It seems intuitive to think that to keep the electrons moving, the electric field must act everywhere in the circuit (1).

You have to realize this intuition is like the intuition that in order for the Moon to orbit theEarth, there has to be some force pushing it around the Earth.

In other words, your intuition is based on experience where motion is opposed by the environment (via some kind of friction force) in a visible way.

But imagine there is no resistance to motion in the wire, just as we imagine there is no resistance to celestial motions of Moon or planets around the Sun. Then, the motion can be self-perpetuating and there is no need for any force to keep things moving. In ideal conductor, force needed to keep current going is, by definition of the ideal conductor, zero.

• OP mentioned a battery is attached to this circuit. Therefore current is not constant inside this wire for the situation described. Applying Ohms law in this sense is consensicle. May 9 at 19:11
• Abstractions and applications of ohms law, about zero resistance wires lead to so many people actually believing that when a battery is connected to a superconducting wire, that E=0. For OP, imagine a loop and a battery, if E=0 in the wire, then the field cannot be Conservative. May 9 at 19:13
• Paull, your example is making me develop the right intuition. Then if I only have a 5V battery and a wire, will the potential field decrease linearly along the wire from 5V to 0V (measured from the neg. terminal of the battery)? May 10 at 7:19

An ideal wire with a voltage source attached to it Would obviously have a potential difference across it. Resistance or not. V=IR Is a steady state solution required for a constant current. When R=0, V=0, for a constant current. Aka, no potential is applied ( as no work needed to maintain that current as no resistance)

In this situation, current is definitely not constant in this superconducting wire.

Ohms law is one of the most universally misunderstood laws out there despite its simplicity, since no one actually goes through the derivation from the drude model.

There IS an electric field across a zero resistance wire with a battery attached to it,and thus there IS a potential difference. Any answer other than that, is obsurd.

The work done by the field in accelerating these charges in the wire is VERY small, compared to the work done against the resistance in the components, so is treated as zero.

If you don't believe me. Imagine a battery with a loop of superconducting wire.

If E=0 in the wire, $$\oint -\vec{E} \cdot \vec{dl} = V_{0}$$

Which violates the fact that E is conservative.