Question

A capacitor of capacitance `C_(0)` is charged to a patential `V_(0)` and then isolated. A small capacitor C is then charged from `C_(0)`, discharged and charge again, the process being is decreased to V. Find the value of C.

Answer 1

Correct Answer - `C_(0)[(V_(0)/V)^(1/n)-1)]`
`q_(0)=C_(0)V_(0)`
After charging, `q_(1)=q_(0)(C_(0)/(C_(0)+C))`
After discharging and again charging first time,
`q_(0_(1))=(C_(0)V_(0))C_(0)/((C_(0)+C))=(C_(0)^(2)V_(0))/((C_(0)+C))`
i.e. `V_(0_(1))=(C_(0)V_(0))/((C_(0)+C))`
`q_(0_(2))=(q_(0_(1)))(C_(0))/(C_(0)+C)=(C_(0)^(3)V_(0))/((C_(0)+C)^(2))`
`V_(0_(2))=(C_(0)/C_(0)+C)^(2)V_(0)`
After nth chargeing
`V_(0_(n))=V=(C_(0)/(C+C_(0)))^(n)V_(0)`
or `(C_(0)/(C+C_(0)))^(n)=(V/V_(0))`
`C/C_(0)+1=(V_(0)/V)^(1//n)`
or `C=C_(0)(V_(0)/V)^(1//n)-C_(0)=C_(0)[(V_(0)/V)^(1//n)-1]`.