Imagine a straight piece of wire, not connected to anything. Parallel to one of it's ends is a section of circuit with direct current in it. They're placed in such a way, that only about 1/4 (or less) of disconnected wire is near a powered circuit. Question: will electric field in circuit cause at least some of electrons in disconnected wire to move to other end (upper end on picture)? Will this happen, if circuit is a toroidal solenoid, and end of disconnected wire in inserted in hole in the middle of said toroid?
Electric Field and Electrons – Will Electric Field Cause Electrons to Move in Disconnected Wire?
electric-circuitselectric-fields
Related Solutions
There is no effect on the electrons. It is not like a pipe connected to tap and stopped at the other end so that even before unstopping the other end the molecules are being squashed inside. No repulsion as you are thinking of. In the battery case as soon as you connect the + end the current starts to flow. Think of it as not a mechanical force that drives the electrons but the potential difference. so at least two terminals need to be there.
In the following situation
you have the voltage source ensuring a potential difference V = 60 Volts between its terminals. The source's upper terminal is connected to the switch's upper terminal, so they have the same electric potential.
The switch's lower terminal is connected to the resistor's upper terminal, so they also have the same electrical potential. As you correctly stated, there is no current flowing through the resistor, so by Ohm's Law the voltage difference across the resistor's terminals is 0. Therefore, the resistor's upper terminal has the same electric potential as the resistor's lower terminal.
However, the resistor's lower terminal is also the source's lower terminal, which has a potential difference of 60 Volts with the switch's upper terminal.
Therefore, the potential difference across the switch is 60 Volts, even though there is no current flowing through it.
An open switch can be modelled as a resistor with infinite resistance, so if you apply Ohm's Law directly to it, you can have a potential difference even though the current flowing through it is zero.
In the following situation
you have that the ideal voltage source always assures the 60 Volts potential difference between its terminals, regardless if the switch is open or closed. Therefore, there will always be a current $I = \frac{V}{R}$ flowing through the resistor, and if the switch is closed, there will be an infinite current flowing through it (assuming the switch's resistance as zero).
In practice, what would happen is that the current flowing through the switch would be very large, and the wire would melt; I've seen it happen a few times when my students accidentally short-circuit the source in my Circuits Lab class.
Best Answer
If you had an electric field, charge separation would build up in the disconnected conductor such that the electric field inside the conductor was zero. This movement of charges would constitute a current that very rapidly decayed to zero.
However, neither of the diagrams you show appear have a changing magnetic field because the current in the blue wire appears to be depicted as constant. Therefore no electric field is induced and no current flows in either orange wire.