A number of capacitors, each of capacitance 1μF and each one of which gets punctured a potential difference just exceeding 500 volt is applied . Then an arrangement suitable for giving a capacitor of capacitance  3 μFacross which 2000 V may be applied requires at least

to increase the potential from 500 V to 2000 V four capacitor has to be connected in series 2000500but in this process capacitance becomes 14μF  Cs=C4 so to get a capacitance of 3μF   =nx14μF  n =12so toatal no of capacitors  used as 4x12=48   ( each row four are in series and twelve such rows are in parallel)