In fig. `C_(1) = 20 muF, C_(2) = 30 muF and C_(3) = 15 muF` and the insulated plate of `C_(1)` is at a potential of 90 V, one plate of `C_(3)` being earthed. What is the potential difference between the plates of `C_(2)` three capacitors being connected in series ?

Explanation

Here `C_(1) = 20 muF, C_(2) = 30 muF`.
C_(3) = 15 muF, V = 90V

Equivalent capacitance C' is given by
`(1)/(C ) = (1)/(C_(1)) + (1)/(C_(2)) + (1)/(C_(3)) = (1)/(20) + (1)/(30) + (1)/(15)`
`C = (20)/(3) muF = (20)/(3) xx 10^(-6) F`
As potential difference 'V' = 90V
Charge, `q = CV = (20)/(3) xx10^(-6)xx90 = 600xx10^(-6)C`
P.d between the plates of capacitor `C_(2)`
`V_(2) = (q)/(C_(2)) = (600xx10^(-6))/(30xx10^(-6)) = 20V`