A capacitor of capacitance `C_(o)` is charged to a potential `V` and then isolated. A small capacitor `C` is then charged from `C_(o)`, discharged and charged again, the process being repeated `n` times. Due to this, potential of the large capacitor is decreased to `V`. Find the capacitance of the small capacitor :

Explanation

When key is closed, common potential `V_(1)=(C_(o)V_(o))/(C_(o)+C)`
charge left on large capacitor after after sharing of charges `Q_(o)^(1) = C_(o)V_(1)`
common potential after second sharing of charges in `V_(2)=(C_(0))/(C_(0)+C)V_(1) , V_(2)=(C_(o)^(2)V_(0))/((C_(o)+C)^(2))`
after `n^(th)` charges `V_(n)=((C_(o))/(C_(o)+C))^(n)V_(0)`
But `V_(n)=V ,V=((C_(0))/(C_(0)+C))^(n) V_(0) , :. C=C_(o)[((V_(o))/(V))^(1//n)-1]`.