Hello all. I wasn't exactly sure of the best phrasing for title, so I hope I've attracted the right folks to answer this question. It has to do with the physical description of circuits, and the electric fields within them.
The question is: What is the electric field in an ideal wire, in a circuit, for a wire that is connected to ground?
Here's what I mean: Imagine a circuit with a battery and a resistor. The battery is 1.5V, the resistor is a carbon resistor, nothing special, with value R. We have the (+) terminal of the battery, a wire (called wire A), then the lead of the resistor. The other lead of the resistor is connected to ground, and that wire is called wire B. The (-) terminal of the battery is also ground. Here it is, with electric fields shown:
Note, the electric field within the resistor is not shown. The red field, and the green field, simply distinguish between the field outside of the conductor (red) and the field within the conductor (axial field, green). For sake of argument, all wires and leads are 'superconductors', and the battery is ideal. The electric field of interest in this question is the green electric field.
The first assumption is that there is no voltage drop across the grounded wire from the resistor lead to the (-) terminal of the battery i.e. it is a net with 0V at all points. But this is about the physical description, so the distance and organization matters for the circuit. The second assumption is that a current is flowing through the circuit. The third assumption is that a non-zero electric field must exist in order for a current to flow in a conductor.
Now, the fourth assumption, is that the definition of electric potential is the negative of the line integral along the path from a reference (0V) to the point of interest. Take the reference as the negative side of the battery, and the point of interest is the closest point to the resistor lead on the grounded wire, wire B.
The proposed question comes from the fact that you have current, and therefore an assumed electric field, along the wire's axis, but in order for the electric potential to be 0V, you must have either a) a zero-length path, or b) a zero electric field. The third option would be an electric field that is conservative, i.e. a net electric field of zero that still changes. There seems to be, in all cases, something of a contradiction. If the entirety of the voltage gradient is dropped across the resistor, and there is no voltage dropped across the ground wire, why does current flow?
Now, some solutions to the question that I have considered are: perhaps there is a mechanical aspect, because of course charge is moving through the wire, and if current was not flowing through the ground wire into the battery, then you would start to build up charge and of course that doesn't happen. When I picture a voltage, I imagine naturally the entire circuit at once so considering only the ground wire has been quite confusing for me, as you can probably tell. What resolution is there to my problem?
Any help is greatly appreciated. Thanks in advance.
Sam
The question is: What is the electric field in an ideal wire, in a circuit, for a wire that is connected to ground?
Here's what I mean: Imagine a circuit with a battery and a resistor. The battery is 1.5V, the resistor is a carbon resistor, nothing special, with value R. We have the (+) terminal of the battery, a wire (called wire A), then the lead of the resistor. The other lead of the resistor is connected to ground, and that wire is called wire B. The (-) terminal of the battery is also ground. Here it is, with electric fields shown:
Note, the electric field within the resistor is not shown. The red field, and the green field, simply distinguish between the field outside of the conductor (red) and the field within the conductor (axial field, green). For sake of argument, all wires and leads are 'superconductors', and the battery is ideal. The electric field of interest in this question is the green electric field.
The first assumption is that there is no voltage drop across the grounded wire from the resistor lead to the (-) terminal of the battery i.e. it is a net with 0V at all points. But this is about the physical description, so the distance and organization matters for the circuit. The second assumption is that a current is flowing through the circuit. The third assumption is that a non-zero electric field must exist in order for a current to flow in a conductor.
Now, the fourth assumption, is that the definition of electric potential is the negative of the line integral along the path from a reference (0V) to the point of interest. Take the reference as the negative side of the battery, and the point of interest is the closest point to the resistor lead on the grounded wire, wire B.
The proposed question comes from the fact that you have current, and therefore an assumed electric field, along the wire's axis, but in order for the electric potential to be 0V, you must have either a) a zero-length path, or b) a zero electric field. The third option would be an electric field that is conservative, i.e. a net electric field of zero that still changes. There seems to be, in all cases, something of a contradiction. If the entirety of the voltage gradient is dropped across the resistor, and there is no voltage dropped across the ground wire, why does current flow?
Now, some solutions to the question that I have considered are: perhaps there is a mechanical aspect, because of course charge is moving through the wire, and if current was not flowing through the ground wire into the battery, then you would start to build up charge and of course that doesn't happen. When I picture a voltage, I imagine naturally the entire circuit at once so considering only the ground wire has been quite confusing for me, as you can probably tell. What resolution is there to my problem?
Any help is greatly appreciated. Thanks in advance.
Sam