Answer
$N=9090$ capacitors
Work Step by Step
We know that:
$N=\frac{C_{eq}}{C}.......eq(1)$
We also know that:
$C_{eq}=\frac{q}{V}$
We plug in the known values to obtain:
$C_{eq}=\frac{1.00}{110}=9090\times10^{-6}=9090\mu F$
We plug in values in eq(1) to obtain:
$N=\frac{9090}{1.00}=9090$ capacitors
