then why is there no potential difference between the two capacitors
It's not quite clear what you mean here but do understand that charged capacitors are electrically neutral.
When a capacitor is "charged", it is not electrically charged, it is energy charged in the same sense as when we say a battery is charged.
There is nothing mysterious about two series connected circuit elements having different voltage drops. Think of two series connected resistors with different resistor values.
I would have thought that as plate B is positively charged and plate C
is negatively charged, there would be a potential difference between
the two
You're forgetting something fundamental: The plates B and C along with the wire that connects them are conductors. But, for an ideal conductor, charge distributes itself so that there is no (static) potential difference across the conductor.
The voltage between the bottom plate of C1 and the top plate of C2 is zero precisely because a conductor connects the two plates.
Toby, I agree that this is really counter intuitive and I was also quite surprised as well when I first saw this very demonstration. I am an undergraduate TA and this is how I explained it in my lab section. I hope this helps. I see two parts to a full explanation: (1) Why is the electric field constant and (2) why does the potential difference (or voltage) increase?
Why is the electric field constant as the plates are separated?
The reason why the electric field is a constant is the same reason why an infinite charged plate’s field is a constant. Imagine yourself as a point charge looking at the positively charge plate. Your field-of-view will enclose a fixed density of field lines. As you move away from the circular plate, your field-of-view increases in size and simultaneously there is also an increase in the number of field lines such that the density of field lines remains constant. That is, the electric field remains constant. However, as you continue moving away your field-of-view will larger than the finite size of the circular plates. That is, the density of field lines decreases and therefore, the electric field decreases as well as the potential field.
To show this mathematically, the easiest way to show this for E = constant is using the relation between the electric and potential fields:
$$E = -\frac{\Delta V}{\Delta d} \longrightarrow \Delta V =-E \Delta d$$
I would expect the voltage to increase linearly as long as the field is constant. When the electric field starts decreasing, the voltage also decreases and the fields behave as finite charged plates. Although I’ve only talked about one plate, this idea immediately applies to two plates as well.
Why does the work increase the electrical potential energy of the plates?
One way to interpret why the voltage increases is to view the electric potential (not the electrical potential energy) in a completely different manner. I think of the potential function as representing the “landscape” that the source (of the field) sets up. Let me explain what the gravitational potential acts like when a ball is thrown upwards (of course, you know what happens in terms of the force of gravity or in the conservation of energy scenario). I claim that the potential function is related to the “gravitational landscape” that the earth sets up, which is derived from the potential energy and is equal to the potential energy per mass:
$$ {\Delta U = mg\Delta y} \longrightarrow \frac{\Delta U}{m} = \Delta V = g\Delta y$$
Plotting these functions, the constant gravity field sets up a gravitational potential ramp (linear behavior) that looks like
In terms of energy, the ball moves up this gravitational ramp were the ball is converting its kinetic energy into potential energy until the ball reaches it maximum height. However, the gravitational ramp exists whether the ball is thrown up or not. That is, gravity sets up a gravitational ramp (the landscape) and this is what the ball “sees” before it is thrown up.
If we now apply the above thinking to a constant electric field between the parallel plates, the electric potential function is derived in a similar manner:
$$ {\Delta U = qE\Delta r} \longrightarrow \frac{\Delta U}{q} = \Delta V = E\Delta r$$
If we look at the electric potential of the negative plate (it’s easier than the positive plate), it has a negative electrical ramp that starts at 0V.
So as your TA pulls the plates apart, the work she does moves the positive plate up the electrical ramp and increases the potential of the positive plate. So this interpretation of the electric potential is what you intuitively already think about in terms of mechanical situations like riding your bike up a hill. There is no difference in the electrical situation.
Best Answer
There is in fact a mistake in your argument. This is an error which I believe is made by inquisitive students and one which helps the student learn electrostatics better. I'm glad that you've asked this. V = kq/d isn't applicable everywhere. It's only applicable to a static point charge or a outside a uniform sphere. But it's not true for all situations. An example would be a system of parallel plates.
What is instead true is that V is the integral of electric field with respect to small distance. This is always true, as it's the definition of V. In simpler words, V is the work required to bring a unit positive charge from infinity to the position it's in presently.
Now if you have a system of capacitors, Q = CV. This comes from the very definition of capacitance. Now for a parallel plate capacitor, C is proportional to the area of the plates and inversely proportional to distance between the plates. When you pull two plates apart, the area of the plates stays constant. Since there's nowhere the charge can go , the charge on each plate stays constant too. The only thing that changes is the distance. As V = Q/C, this would mean that as 1/C is proportional to distance, V is proportional to the distance between the plates too. So when you increase the distance, you in fact are increasing the potential difference. I hope I managed to explain it. Have fun learning !