A capacitence of `C_(0)` is charged to a potential `V_(0)` and then isolated. `A` small cpacitor `C` is then charged from `C_(0)`, discharged and charged again, the process bing repeated `n` times. Due to this, potential of the large capacitor is decreased to `V`. find the capacitance of the small capacitor. is it possible to remove charge on `C_(0)` completely in this way?

Explanation

First time:
common potential :
`V_(1)=(C_(0)V_(0)+Cxx0)/(C_(0)+C)=((C_(0))/(C_(0)+C))V_(0)`
Final charge on (i),
`Q'_(1)=C_(0)V_(1)=(C_(0)^(2)V_(0))/(C_(0)+C)`
second time:
Common potential :
`V_(2)=(Q'_(1)+0)/(C_(0)+C)=((c_(0))/(C_(0)+C))^(2)V_(0)`
`n^(th) "time" : V_(n)=((C_(0))/(C_(0)+C))^(n)V_(0)`
`V=((C_(0))/(C_(0)+C))^(n)V_(0)`
`(C_(0))/(C_(0)+C)=((V)/V_(0))^((1)/(n))implies(C+C_(0))/(C_(0))=((V_(0))/(V))^(1//n)`
`C=C_(0)[((V_(0))/(V))^((1)/(n))-1]`
The charge on `C_(0)` cannot be removed completely in this way.