Electric field in a current carrying conductor

Thread Starter


Joined Nov 26, 2014
We all know the ohm's law which states that V = IR. Now lets take a simple circuit where a battery is connected to a resistor via copper conductors. If the battery is of 'V' volts then the drop in the resistor is sightly less but approximately equal to V volts. And a small tiny fraction of voltage is lost along the copper conductor too since it too has some resistance. Now, from electrostatics we know that E, Electric field = dV/dx. Since there is a huge voltage drop in the resistor and it is also a voltage gradient along its length, there exist a uniform electric field along the resistor since E=dV/dx. Also, since there is a very weak voltage gradient in the copper conductor, a very very weak electric field exist in the copper conductor. What I need to know is, how a weak electric field in the conductor can abruptly become into a strong electric field in the resistor.


Joined Feb 5, 2010
are you confusing a static (non moving) collection of electrons and a flowing stream of moving charge carriers? it has been many decades since college, but your presumption of conditions seems flawed in its application. someone with a better memory will be along shortly to correct us both. :)


Joined Oct 2, 2009
Good question. I think the proper answer is you do not know the value of dx which is a physical distance.

Instead of a resistor and a wire, let us imagine 1k - 1 - 1kΩ, i.e. a small value resistance in series between two larger value resistances.
Since the resistors are in series, the current I is the same for all resistors. The voltage drop across each resistor is proportional to its resistance.

But what is the electric field between the ends of the resistor? That we don't know because dx is unknown.
Hence you are confusing electric field and potential difference.

Maybe someone else has a better answer.


Joined Sep 17, 2013
What I need to know is, how a weak electric field in the conductor can abruptly become into a strong electric field in the resistor.
Are you expecting a gradual change? If so, why? The high resistivity material inside the resistor can't 'flow' into and blend with the low resistivity wire. If you connected, say, an AA cell in series with a car battery the electric field inside each would be unaltered.


Joined Apr 2, 2015
Voltage is caused by excess charge on the outer shell of the conductor (remember, extra charge always moves to the outside in a conductor). So it is essentially the difference in charge density that gives you voltage.

Now electric fields have superposition, so let's look at the field acting on a positive charge towards the middle of your conductor. Since the charge density on either side of it is roughly the same, the electric fields will equal and cancel out resulting in a net ~0 E field on the test charge.

At the beginning of the resistor however, the charge density is higher than at the other end of the resistor, so those fields won't completey cancel out, and you'll have a net nonzero E field through the resistor (the reason why the voltage drops across the resistor is a different matter).

So the keyword is NET E field. Your deriving the E field from the voltage difference across two points(dx). Each charge has its own E field, but depending on where your at in the circuit, they cancel out to different magnitudes.


Joined Sep 22, 2013
"What I need to know is, how a weak electric field in the conductor can abruptly become into a strong electric field in the resistor."

It doesn't. As Kermit2 said, this is a moving field.

In your DC set up the electric field forms 2 consecutive and smaller sized tapered cylinders and a cone. All connected together.

An electric field has a rotating, perpendicular,(to the conductor) radius(x).

The radius that you use is arbitrary, because in reality the radius propagates without end.

We will pick a 3" radius to start. This gives us a 6" field diameter around the conductor at the battery terminal. At 3" out from the conductor the field strength will be a certain level dependent on source voltage. This reference "E" at 3" will represents maximum field size. Note that in reality the field grows infinitely large and infinitely weak. Even after circuit off.

When the field rotates down the first conductor...The radius(voltage) will shorten a little because of resistance. At the end of the conductor and at the beginning of resistor the radius might be 2.9". The cylinder radius might decreased by 1/10 " across the first conductor. This would be a greater change than normal, but I use this to show resistance.

So next we have a 2.9" radius in and a 1/10" radius out of the resistor(cone shape(field, not resistor)). This is the load which is a big radius(voltage) drop.

From the bottom of resistor to negative terminal of battery we have a radius of 1/10" to a radius of zero(origin) at negative terminal.

3 electric field tapered cylinders in series.

If the circuit was AC.....There would be another set of negative cylinders, going in the opposite direction that alternate with the positive.

And with an AC circuit the cylinders in the circuit MAY become larger than the source cylinders.

Also in an AC circuit, the walls of the cylinder may become non linear.

This will allow the cylinder to expand and contract, like a spring.

The amount of charge moved by this field will be limited by resistance.

The velocity and rotation rate of the field is limited by reactance.

One other critical property of the field is the field velocity.

This propagation delay changes with the media. This is very important for the way nature works and some strange things can happen when the media(field velocity) changes quickly.

You can measure voltage across a circuit, but voltage moves down thru a circuit, similar to current.

Study charge and you will find nature.


Joined Mar 31, 2012
Remove the resistor and conductor completely from the picture and just have the two terminals of the battery, having voltage V, separated by some distance, D. The electric field along the path from between the two terminals is going to be E = V/D, right? Now put a resistor of length L (L<D) midway between the two terminals, but not connected to either. You will now have a E field of zero within the resistor (otherwise current would be flowing) and the E field between each end of the battery and the end of the resistor will be E=V/(D-L), which will be higher than it was before. How is this possible? Simple, when you put the resistor into the existing E field you DID get a tiny, transient current flow that resulted in a sufficient charge separation within the resistor to create an E-Field that exactly cancels out the external E-field. This charge separation results in charges at the ends of the resistor creating an equivalent voltage potential across the smaller gap resulting in a higher E-field.

Repeat this process with a conductor and the same thing happens. Repeat this process with a conductor and a resistor and the same thing happens. Repeat this process and allow contact between some, but not all, of the gaps and the same thing happens. Eventually, you allow all of the gaps to close so that you have steady state current flow and it is pretty easy to see why the E-field is concentrated and diminished the way that it is.