Your answers were fine as long as you stuck to the narrowest focus of my questions. Where you drifted away from that focus, things started to go a bit wobbly. The activating function approach seems to have given you some false impressions.
Which leg of the Pi is depolarized (has a positive shift of membrane potential)--the one that is attached to the positive terminal of the battery, or the one that is grounded?
The grounded one is depolarized. It's v.m = v.int - v.ext = 1/3V - 0V = 1/3V whereas for the one at 1V its v.m = v.int - v.ext = 2/3V - 1V = -2/3V
Yes, the grounded one is depolarized. Incidentally, the one that is driven by the extracelllar positivity is hyperpolarized by 1/3 volt, not by 2/3 volt.
If you had to generalize, would you say that driving extracellular potential in a positive direction at some location A along a cell will depolarize or hyperpolarize the local membrane?
Extracellular positivity at location A will hyperpolarize membrane potential at location A.
If we have a chain of such pi-elements with a constant gradient of voltage along the chain, the influence on the 'batteries' between the lower contacts on the potentials of adjacent inner nodes compensate each other. Only the first and the last node will be influenced by the applied voltage gradient.
That's what the "activating function" approach predicts, but it's not what would happen in the physical system or in an analytic or numerical solution to the cable equation that describes that physical system. In the steady state, membrane potential will vary continuously with position. For a cylindrical cable in a field with a constant potential gradient, maximum polarization will be at the opposite ends of the cable. Membrane potential will be unchanged midway between the two ends. Derivative of membrane potential relative to distance along the cable will be nonzero everywhere, but steepest near the two ends (this is related to what has been called the "sealed end effect" in other contexts).
In order to have an excitation of all segments, the voltage gradient needs to cange over distance.
First, all segments will not be excited, ever. If an extracellular stimulus depolarizes some part of a neuron, there must be some other part of that neuron that is hyperpolarized by that same stimulus. Remember that positive charge can't accumulate in some part of a cell unless it entered through the membrane in some other part of the cell. And the latter membrane is necessarily hyperpolarized.
Second, the extracellular potential gradient does not have to vary in order for it to affect a cell. Not at all. That misconception comes to you thanks to the activating function approach. The activating function approach predicts that the extracellular potential gradient must have a nonzero second derivative, because one of its key assumptions is that the neurite is so long that effects at the ends can be ignored). But, as many have observed, it's not what you don't know that gets you into trouble--it's what you know for sure that ain't so.
"Well, if the activating function approach is all that bad, what use is it?"
It's most useful in situations in which there is a relatively localized disturbance of extracellular potential, especially when the spatial scale of that disturbance is small compared to the overall length of the neurite that is being stimulated. Good examples would be focal extracellular stimulation of a periperhal nerve many cm away from the origin or termination of that nerve (e.g. in the clinical setting of, say, performing nerve conduction velocity testing of the radial nerve at the wrist, or stimulation of the dorsal columns of the spinal cord, or maybe even deep brain stimulation of a pathway that is a few cm long by a bipolar electrode that is located within a mm or so of the axons). In such cases it can be quite useful.
Up to this point, membrane capacitance has been ignored, and changing extracellular potential at a point affects membrane potential everywhere immediately. What happens if membrane capacitance is not ignored
The time course of the stimulation signal has an impact. Voltage over capacitors can't change abruptly.
True. The time derivative of membrane potential changes abruptly, but charge must accumulate on membrane capacitance, and that process requires the passage of time.