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Figure

In the figure, the circuit is physically isolated from Earth (The black line at the bottom). The red capacitors are parasitic capacitors.

Under switching conditions the node voltages relative to the zero reference swing in such a way that the internal charge in the circuit stays the same. This is understandable because the circuit is isolated from ground and cannot get more charge or lose any charge. So the charges internally rearrange themselves in a way that leads to the node voltages being what they are.

How is this possible. Is this internal rearrangement of the charges a current? Is this current different from the “normal” current of the circuit (The 1mA)?

I think that Electric fields that change and exist between charged bodies will cause something like current flow, called displacement current, such that a change in the electric field on one body affects the electric field on a nearby body. But how does this affect the 1mA current that appears in the circuit?

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Is this internal rearrangement of the charges a current?

Moving charge is a current, so yes, this is a current.

Also, the changing electric field caused by the rearranging charges is a current, called displacement current. Whether a displacement current is "really" a current is a philosophical or semantic question rather than a physics or engineering question. From a physics and engineering point of view, displacement currents must be accounted for when solving the Kirchhoff's Current Law, they contribute to magnetic fields according to Ampere's law, and otherwise affect the rest of the world just as ordinary moving-charge currents do.

If so then is this current different from the “normal” current of the circuit (The 1mA)?

It's a transient current (it falls to zero over time) where the "normal" current in this circuit will continue forever.

Electric fields that change and exist between charged bodies will cause something like current flow, called displacement current, such that a change in the electric field on one body affects the electric field on a nearby body. But how does this affect the 1mA current that appears in the circuit?

If your voltage source has internal resistance (like all real-world voltage source) then the extra current required to charge the parasitic capacitors will cause the source output voltage to drop, reducing the current through the 9 kohm resistor.

This effect will last for only a few nanoseconds or microseconds, until the capacitors are charged, and after that the presence of the parasitic capacitors will not affect the current through the resistor.

Note that in the model drawn, only C3 has its voltage changed when the switch is closed, so only C3 will draw current. The other capacitors won't cause any currents in response to the switch closing or opening.

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  • \$\begingroup\$ Thank you very much, extremely helpful! \$\endgroup\$ – John May 16 at 19:34
  • \$\begingroup\$ If the circuit is grounded, by connecting a point in the circuit to the physical zero reference (Earth). Will the connection to ground affect what is happening in a meaningful way? Will the internal charge distributions and the corresponding currents be different? \$\endgroup\$ – John May 16 at 21:26
  • \$\begingroup\$ @John, Draw the circuit for yourself and think about whether closing or opening the switch changes the voltage across each of the capacitors. \$\endgroup\$ – The Photon May 16 at 22:19
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I'm going to focus on your general question(s) and ignore your specific circuit. I think you only included it to put something down, anyway. There's an underlying question and that's what I'll address.

There is a state that exists before and a state that exists after a source is applied to a circuit. Obviously, there must be a short period in between these two steady state conditions where the circuit transitions between them. This transition period is very fast and most undergrad electronics books don't directly address the physics involved. (But the physics details of this transition period is important to high voltage engineering, for example.)

It's not too complicated to imagine what happens, though. At first, the source will have an excess of positive charges at one node and an excess of negative charges at the other node. These will very, very rapidly impel charges to move (the conductors have a veritable sea of available, conduction band electrons.) These charge motions are, in fact, currents. But many of them are just initial currents needed in order to set up surface charges, leading to a gradient of charge distributions, throughout the circuit. For DC circuits, once these surface charges have been set up, the current required to set them up is no longer required. But these surface charges now act to impel the currents in all the right directions (this way, and that way, at node intersections, etc.) For AC circuits, these surface charge changes are continually changing but they are almost always extremely fast compared to the AC circuit oscillations. So for mental visualization purposes you can just think of each of the infinite number of continuous changes in an AC cycle as being a DC-snapshot. So the DC view is usually sufficient to get the point across.

Imagine bending a wire while a circuit is operating! How is it that the electrons moving along the wire "know" to take the bend? Do they bounce off of things and careen around the bend? No, they don't. Not generally, anyway. What instead happens is that a few of the moving charges will "get stuck" in the surface area of the outside curve of the bend in the wire. This slight excess of charge will be "just enough" to act as a repulsing force that causes the current to accelerate around the bend, as if they just knew to take it.

With that in hand, I recommend that you read a few references. Perhaps a better one I wrote is here. Another is here and only discusses a simple DC battery and one resistor. There's a good video, Surface Charge on a High Voltage Circuit, too. And finally, there's a quantitative discussion: Magnitudes of surface charge distributions associated with electric current flow by Dr. Rosser.

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  • \$\begingroup\$ Thank you very much and yes, I only added the circuit to try help picture the problem that I posed. \$\endgroup\$ – John May 16 at 19:35
  • \$\begingroup\$ If the circuit is grounded, by connecting a point in the circuit to the physical zero reference (Earth). Will the connection to ground affect what is happening in a meaningful way? Will the internal charge distributions and the corresponding currents be different? \$\endgroup\$ – John May 16 at 21:26
  • \$\begingroup\$ @John Suppose a circuit is in an evacuated enclosure in outer space, where it is being continually peppered with protons and electrons. Depending on location and history, it will have accumulated some charge. But too much of a charge will attract the opposite charges and repel same charges, so at some point it reaches equilibium. Now you instantly teleport this to Earth's surface and ground it into Earth (which itself has a different charge balance.) What do you think should happen (and for how long?) Will it matter for the circuit? Think for yourself for a moment before I try and answer. \$\endgroup\$ – jonk May 16 at 21:54
  • \$\begingroup\$ @John Also, if you haven't already done so, you really should allow yourself to enjoy Feynman's Lecture series on physics. It's freely available on the web, now. And it is a treasure and very readable. (Due, in my opinion, mostly to the two co-authors who partnered in order to finally write it.) For example, try Chapter 9, Volume 1. In particular, pay some attention to the Earth's surface vs the Earth's upper atmosphere and how charges are continually transported between them. \$\endgroup\$ – jonk May 16 at 22:05
  • \$\begingroup\$ When a point in the circuit is connected to the Earth, the potential at that point becomes the reference potential of 0V (as assumed for the Earth). Simultaneously, potentials in other points of the circuit also change. But, potential difference between the points still remains same. So, internal charge distribution and the current in the circuit remain unchanged. Only change is in the potentials at the points of the circuit. Please let me know if my answer is correct \$\endgroup\$ – John May 18 at 9:31

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