What is a series connection of capacitors? Obtain the formula for the effective capacitance in the series combination of two different capacitors.

(N/A) series connection of capacitors is an arrangement where capacitors are connected end-to-end such that the same amount of charge flows through each capacitor.
The figure shows capacitors $C_{1}$ and $C_{2}$ connected in series.
The left plate of $C_{1}$ and the right plate of $C_{2}$ are connected to the terminals of a battery,resulting in charges $+Q$ and $-Q$ on the outer plates.
Due to electrostatic induction,the right plate of $C_{1}$ acquires a charge of $-Q$,and the left plate of $C_{2}$ acquires a charge of $+Q$.
Thus,every capacitor in the series combination carries the same magnitude of charge $Q$,regardless of their individual capacitance values.
Let $V_{1}$ and $V_{2}$ be the potential differences across capacitors $C_{1}$ and $C_{2}$ respectively. If $V$ is the total potential difference across the combination,then:
$V = V_{1} + V_{2} \quad \dots (1)$
Using the relation $V = \frac{Q}{C}$,we can write:
$V = \frac{Q}{C_{1}} + \frac{Q}{C_{2}}$
Dividing by $Q$:
$\frac{V}{Q} = \frac{1}{C_{1}} + \frac{1}{C_{2}} \quad \dots (2)$
For the equivalent capacitor with effective capacitance $C_{eq}$,we have $V = \frac{Q}{C_{eq}}$,which implies:
$\frac{V}{Q} = \frac{1}{C_{eq}} \quad \dots (3)$
Comparing equations $(2)$ and $(3)$,we get:
$\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_{2}}$