A 0.1F capacitor is charged by a 10V battery. After disconnecting the battery, this charged capacitor is connected with an uncharged capacitor. If the charge is equally shared between the two, then what will be the energy stored in the two capacitors? Compare this energy with the energy stored initially in the first capacitor.

Explanation

Let `C_1" and "C_2` be the capacitances of the capacitors and their voltage be `V_1" and "V_2` respectively.
Given `C_1=0.1F`,initially `V_1 = 10V " and "V_2=0`.
Initial energy stored in the two capacitors is,
`U_i= (1)/(2)C_1V_1^2+(1)/(2)C_2V_2^2`
`=(1)/(2)xx0.1xx(10)^2+0=5J`
It Q is the initial charge on capacitor `C_1`, its initial energy is,
`U_i= (Q^2)/(2C_1)`
When the two capacitors are connected together, the charge is distributed equally, so the charge on each capacitor is `(Q)/(2)`.
Since the potential difference across the two capacitors is also the same, it follows that their capacitances are equal.
Thus, `C_1= C_2 (" say ")`
Also, `Q_1=Q_2=(Q)/(2)`
Final energy stored in the two capacitors is,
`U_f= (Q_1^2)/(2C_1)+(Q_2^2)/(2C_2)= ((Q/2)^2)/(2C)+((Q/2)^2)/(2C) = (Q^2)/(4C)`
But `U_i= (Q^2)/(2C)`, hence `U_f= (U_i)/(2)= 2.5`J
`Thus, `(U_f)/(U_i)= (2.5)/(5)= (1)/(2)`.