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

(N/A) Consider $n$ capacitors with capacitances $C_{1}, C_{2}, \ldots, C_{n}$ connected in parallel.
In a parallel combination,the potential difference $V$ across each capacitor is the same,while the total charge $Q$ is the sum of the charges on individual capacitors.
Let $Q_{1}, Q_{2}, \ldots, Q_{n}$ be the charges on capacitors $C_{1}, C_{2}, \ldots, C_{n}$ respectively.
The total charge $Q$ is given by $Q = Q_{1} + Q_{2} + \ldots + Q_{n}$.
Since $Q_{i} = C_{i}V$ for each capacitor,we have:
$Q = C_{1}V + C_{2}V + \ldots + C_{n}V$
$Q = (C_{1} + C_{2} + \ldots + C_{n})V$
If $C_{p}$ is the effective (equivalent) capacitance of the parallel combination,then $Q = C_{p}V$.
Comparing the two expressions for $Q$,we get:
$C_{p}V = (C_{1} + C_{2} + \ldots + C_{n})V$
$C_{p} = C_{1} + C_{2} + \ldots + C_{n}$
Thus,the effective capacitance of capacitors in parallel is the algebraic sum of the individual capacitances,and it is always greater than the capacitance of any individual capacitor in the combination.