Zero capacitances vs. infinite capacitances

Question

Imagine a rc circuit with a capacitor with movable plates that have changeable area connected to a cell of voltage $V$

1. what if i pushed the plates towards each other till they touched? will this part of the circuit act like a connecting wire? will this 'capacitor' have infinite capcitance (using $C = \varepsilon\frac{A}{d}$)? or if after this made the area of the plates zero(somehow) can i use $C = \varepsilon\frac{A}{d}$ now?

2. what if I decreased the area of the plates first (the plates are at a fixed distance).wont it act like a broken circuit? zero capacitance?

3. initial dist=$d$; initial area=$A$. Imagine a scenario where one of the plates is moving towards the other with a speed $v$ the area is also decreasing at the rate $\frac{A}{d}v$. what will the effective capacitance at $t=\frac{d}{v}$ be?

Answer 1

what if i pushed the plates towards each other till they touched?

If you consider the ideal case where the plates can be brought arbitrarily close together without touching and without dielectric breakdown, you'll find that the work done by the (ideal) cell goes to infinity as the plates are brought together.

Stipulate that the distance $d$ between the plates is initially $d_0$ and approaches zero asymptotically, e.g.,

$$d(t) = d_0e^{-t/\tau}$$

Since the capacitance $C$ is given by $C = \frac{\varepsilon A}{d}$, the charge $Q$ on the more positive plate is given by

$$Q(t) = C(t)\cdot V= \frac{\varepsilon AV}{d(t)} = \frac{\varepsilon AV}{d_0}e^{t/\tau}$$

and it's straightforward to show that the work done by the cell as a function of time

$$W(t) = \frac{\varepsilon AV^2}{d_0}e^{t/\tau}$$

which shows that work done by the cell diverges as the distance $d \rightarrow 0$.

Thus, you must give up the idea that you can push the plates until they touch (requires the cell do infinite work) or give up the idea that the voltage across can be kept at $V$ as $d \rightarrow 0$.

Answer 2

If the plates touch each other, then the plates would simply act as a connecting wire. Even if the plates come extremely close to each other, it would cause an electric breakdown and ionize the air between the plates which would complete the circuit.

You cant make the area of plates zero. If you make it very small, then it would still act as a capacitor but would have negligible capacitance. It would act similar to a broken circuit.

If the area of the plates keeps on decreasing continuously, then the capacitance will also keep on decreasing proportionately.

Answer 3

There are many classical limits for which the expression and simplification can works.

A comparable example is when we treat planets or particle as point object, when they get closer, is there attraction went to infinity because of $\frac{1}{r^2}$?

It's same here, the expression is nonlonger suitable to describe the situations when it got too close.