#### Question

Figure shows two identical parallel plate capacitors connected to a battery through a switch *S*. Initially, the switch is closed so that the capacitors are completely charged. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant 3. Find the ratio of the initial total energy stored in the capacitors to the final total energy stored.

#### Solution

When the switch is closed, both capacitors are in parallel.

⇒ The total energy of the capacitor when the switch is closed is given by `E_i = 1/2 CV^2 + 1/2 CV^2 = CV^2`

When the switch is opened and the dielectric is induced, the capacitance of the capacitor A becomes

`C^' = KC = 3C`

The energy stored in the capacitor A is given by

`E_A = 1/2 C^'V^2`

`⇒ E_A = 1/2 (3C)V^2 = 3/2 CV^2`

The energy in the capacitor B is given by

`E_B = 1/2 xx C/3 xx V^2`

`therefore` Total final Energy

`E_f = E_A + E_B`

`⇒ E_f = 3/2 CV^2 + 1/6 CV^2`

`⇒ E_f = (9 CV^2 + 1CV^2)/6 = 10/6 CV^2`

Now,

Ratio of the energies, `E_1/E_2 = (CV^2)/(10/6 CV^2) = 3/5`