Obtain the formula for the effective capacitance of the series combination of $n$ capacitors.

(N/A) The figure shows $n$ capacitors of capacitance $C_{1}, C_{2}, C_{3}, \ldots, C_{n}$ arranged in series.
The characteristics of a series combination are that the charge on each capacitor is the same,while the potential difference across each capacitor is different.
Suppose the potential differences across capacitors of capacitance $C_{1}, C_{2}, \ldots, C_{n}$ are $V_{1}, V_{2}, \ldots, V_{n}$ respectively.
The total potential difference $V$ of the series combination is given by:
$V = V_{1} + V_{2} + V_{3} + \ldots + V_{n}$
Since $V = \frac{Q}{C}$,we have:
$V = \frac{Q}{C_{1}} + \frac{Q}{C_{2}} + \frac{Q}{C_{3}} + \ldots + \frac{Q}{C_{n}}$
Dividing by $Q$:
$\frac{V}{Q} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}} + \ldots + \frac{1}{C_{n}}$
If $C$ is the effective capacitance of the combination,then $\frac{V}{Q} = \frac{1}{C}$. Therefore:
$\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}} + \ldots + \frac{1}{C_{n}}$
Hence,the effective capacitance is smaller than the smallest individual capacitance in the series combination.
The reciprocal of the effective capacitance of capacitors connected in series is equal to the sum of the reciprocals of the individual capacitances.