Mdthbs

Can someone please explain, intuitively (without any formula, I understand the formulas), why the equivalent capacitance of capacitors in series is less than the any individual capacitor's capacitance?

PS: Let's take a simple case. Say we have 2 capacitors with Capacitance 2 (ignoring units), and we place them in series. A voltage $V$ develops in both, and a charge $+Q$ acculumates on one of their plates. Using the capacitance formulas, the equivalent capacitance is $1/2$ the original. Indeed we get $Q/2V$, where $Q/V$ is the original capacitance. But why? aren't we in total acculumating a charge of $2Q$ over a potential difference of $2V$? Why just $1Q$? (again, I'm speaking intuitively)

Can someone please explain, intuitively (without any formula, I
understand the formulas), why the equivalent capacitance of capacitors
in series is less than the any individual capacitor's capacitance?

I assume you know that the larger the capacitor plates are, the greater the capacitance, all other things being equal. Also I assume you know the greater the separation of the plates (the thicker the dielectric between the plates) the less the capacitance all other things being equal. Given these assumptions, consider the diagrams below.

The top diagram to the left shows two capacitors in parallel. It is equivalent to the digram to the top right. If two or more capacitors are connected in parallel, the overall effect is that of a single (equivalent) capacitor having a total plate area equal to the sum of the plate areas of the individual capacitors. Thus for parallel capacitors the equivalent capacitance is the sum of the capacitances.

The bottom diagram to the left shows two capacitors in series. It is equivalent to the diagram to the bottom right. If two or more capacitors are connected in series, the overall effect is that of a single (equivalent) capacitor having the sum total of the plate spacings of the individual capacitors. Thus for series capacitors the equivalent capacitor is less than the individual capacitors.

For reference the diagram includes an equation for capacitance based on the physical parameters of plate area $A$, plate separation $d$, and electrical permittivity $e$ of the dielectric.

Hope this helps.

The voltage change across a capacitor, or any region of space, is the negative path integral of field dot path. So if two capacitors are placed in series, you have twice the distance for the path integral, compared to one capacitor by itself. Now, since the distance is doubled, to get a final voltage equal to the battery that charges it, the electric field only needs to be 1/2 the electric field of the single capacitor. Since the electric field is proportional to the charge density, you only need 1/2 the charge density that would be on the single capacitor, so C = (Qsingle/2)/V.

why the equivalent capacitance of capacitors in series is less than
the any individual capacitor's capacitance?

To help build your intuition, place in series with another capacitor an ideal parallel plate capacitor (vacuum dielectric for simplicity), with plate spacing $d$. Now consider the limit as the distance $d$ goes to zero (assume zero initial charge).

Intuitively, the capacitance of the ideal parallel plate capacitor increases without bound and what's left in the limit is, effectively, an ideal short circuit when $d = 0$ (the two plates touch).

That is, the series combination of two capacitors has become a capacitor in series with an ideal short circuit. Clearly, the combined capacitance is just the capacitance of the remaining capacitor.

Now, hold the spacing $d$ constant and let the plate area $A$ go to zero instead. Intuitively, the capacitance of the ideal plate capacitor goes to zero and what's left in the limit is, effectively, an ideal open circuit.

That is, the series combination of two capacitors has become a capacitor in series with an ideal open circuit. Clearly, the combined capacitance is zero.

It follows that the total capacitance for two series capacitors with finite capacitance is less than the smaller of the two capacitances