- #1
SparkimusPrime
- 35
- 0
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).
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).