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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.
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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`
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