A `20muF` capacitor is charged to a potential of 20V and is then connected to an uncharged `10muF` capacitor. Find out the common potential and the ratio of the energies stored in the two capacitors.

Explanation

Here, capacitance of first capacitor,
`C_1 = 20 muF = 20 xx10^(-6)F`
Capacitance of second capacitor,
`C_2 = 10muF = 10xx10^(-6)F`
Initially, the charge on the first capacitor,
`Q_1= C_1V_1= (20xx10^(-6))xx20= 400xx10^(-6)C`
and the charge on the second capacitor, as it was uncharged, `Q_2=0`.
Net charge, `Q = Q_1+Q_2= 400xx10^(-6)C`
A parallel combination of the capacitors attains a common potential, say V. Equivalent capacitance,
`C = C_1+C_2= (20+10)xx10^(-6)F = 30xx10^(-6)F`
`V= (Q)/(C )= (400xx10^(-6))/(30xx10^(-6))= (40)/(3) = 13.33V`
The ratio of the energies stored in the two capacitors is,
`(E_1)/(E_2)= ((1)/(2)C_1V^2)/((1)/(2)C_2V^2) = (C_1)/(C_2)= 20)/(10)= (2)/(1)`.