A parallel plate capacitor of capacitances C is connected to a battery to charge it to a potential V. Similarly, another capacitor of capacitance 2C is charged to a potential 2V. Now the batteries are removed, and the two capacitors are connected in parallel by joining the positive plate to one with the negative plate of the other. Find out the final energy of the system.

Explanation

Charge on the first capacitor of capacitaance C and connected to the battery of voltage V is, `Q_1 = CV`
Similarly charge on the second capacitor,
`Q_2 = (2C)xx(2V)= 4CV`
Since the plates of opposite polarity are connected together, the common potential is,
`V' = (Q_1-Q_1)/(C_1+C_2)= (4CV -CV)/(C+2C) = V`
Now equivalent capacitance, `C'= C+2C= 3C`
So final energy of the configuration,
`U_f= (1)/(2)C'(V')^2= (1)/(2)xx3CxxV^2= (3)/(2)CV^2`.