A capaitor of capacitance `C_(1) = 1.0 muF` withstands teh maximum voltage `V_(1) = 6.0 kV` while a capacitor of capacitance `C_(s) = 2.0 muF`, the maximum voltage `V_(s) = 4.0 kV`. What voltage will the system of these two capacitors withsatand if they are connected in sereis ?

Explanation

Amount of charge, that the capacitor of capacitance`C_(1)` can withstand, `q_(1) = C_(1) V_(1)` and similarly the charge, that the capacitor of capacitance `C_(2)` can withstand , `q_(2) = C_(2) V_(2)`. But in series combination, charge on both the capacitors will be same, so `q_(max)`, that the conservation can withstand `= C_(1) V_(1)`,
as `C_(1) V_(1) lt C_(2) V_(2)`, from the numberical data, given.
Now, net capacitance of the system,
`C_(0) = (C_(1) C_(2))/(C_(1) + C_(2))`
and hence, `V_(max) = (q_(max))/(C_(0)) = (C_(1) V_(1))/(C_(1) C_(2)//C_(1) + C_(2)) = V_(1) (1 + (C_(1))/(C_(2))) = 9 kV`