Electric field inside a resistor

[SparkimusPrime]

A resistor (20 ohm) is made of a very thin piece of metal wire, length = 3mm and diameter = .1mm. Given that it has a potential of 8 volts and .4 amps of current running through it, what is the electric field inside the wire?

I know there's no electrostatic equilibrium here, so I can't just take the easy answer and say there is no electric field internal to a cylinder. It seems I must use the equation that relates voltage to electric field:

V = integral(E * ds)

But, as this seems to be based upon Gauss's law (which I haven't the firmest hold) I hate to assume that I can use the equation when the charge is in motion (current).

 

[Doc Al]

Your equation is almost right: ΔV = - ∫ E ds. This is not based on Gauss's law, but on the relationship between field and potential.

In your example, the magnitude of the field is E = ΔV/L.

 

[M98Ranger]

You are correct in your thinking that the equation V = integral(E * ds) is based on Gauss's law, which states that the electric flux through a closed surface is equal to the charge enclosed by that surface. However, this equation can still be used to find the electric field inside a resistor, even when there is a current running through it.

In this case, the resistor is a thin wire with a length of 3mm and a diameter of 0.1mm. We can assume that the wire is cylindrical in shape, and the electric field will be uniform inside the wire. Using the equation V = integral(E * ds), we can solve for the electric field inside the wire by first calculating the voltage drop across the wire.

The voltage drop across the wire can be found using Ohm's law, V = IR, where V is the voltage, I is the current, and R is the resistance. In this case, the resistance of the wire is 20 ohms, so the voltage drop across the wire is 8 volts.

Next, we can plug this value into the equation V = integral(E * ds) and solve for the electric field, E. Since the wire is cylindrical, the integral becomes V = E * integral(ds), which simplifies to V = E * length. Plugging in the values we know, we get 8 volts = E * 3mm. Solving for E, we get an electric field of 2.67 volts per millimeter inside the wire.

It is important to note that this electric field is only present while there is a current running through the wire. Once the current is turned off, the wire will return to electrostatic equilibrium and there will be no electric field inside the wire.