The misleading thing about this picture of capacitor If my understanding of capacitance is correct, capacitance $C$ is defined as the ratio $\frac{Q}{V_{ab}}$ where $Q$ is the magnitude of charge on each conductor for two oppositely charged conductors, and $V_{ab}$ is the potential difference across the two conductors. The two pictures I include here (img-1, img-2) implies that the potential difference must be calculated across the conductors. However, if capacitance is defined the way I think it is, isn't this third image somehow misleading? Shouldn't $C_1$ be equal to $\frac{Q}{V_{pq}}$ and $C_2$ to $\frac{Q}{V_{rs}}$? But if that is the case, $V_{pq}$ + $V_{rs}$ wouldn't add upto $V_{ab}$, would it? The path $ab$ is slightly longer ... In any case, what am I missing here?
 A: In the lumped-model analysis of circuits, each line (or "wire") connecting components in the circuit has a single potential: i.e. the potential does not change along the line. This means that
$$V_{ap}=V_{qr}=V_{sb}=0.$$
So you and the third image are both correct.
Implicit in this "single potential" approximation are the following assumptions:

*

*The wires modeled by these lines have negligible resistance and inductance.

*The wires are electrically short, i.e. shorter than wavelengths associated with signals in the circuit. In other words, voltages and currents in the circuit don't change very fast.

These assumptions are valid in a vast variety of circuits, certainly in this case if $V_{ab}$ doesn't change very fast. For example, at DC, you know that the potential within a conducting wire with zero resistance must be constant, because the E-field within the conductor must be zero.
A: The third image isn't misleading. $C_1=\dfrac{Q}{V_{pq}}$ and $C_2=\dfrac{Q}{V_{rs}}$ are correct and also don't contradict $V_{ab}=V_{pq}+V_{rs}$ as follow
$$\dfrac{1}{C_{eq}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}$$
$$\dfrac{V_{ab}}{Q}=\dfrac{V_{pq}}{Q}+\dfrac{V_{rs}}{Q}$$
$$\therefore V_{ab}=V_{pq}+V_{rs}$$
