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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.

A
9 : 16
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B
5 : 9
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C
2 : 3
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D
3 : 5
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Solution

The correct option is D 3 : 5
Initial condition when switch S is closed
Ei=EAi+EBi; EAi=12CV2; EBi=12CV2
Ei=12CV2+12CV2
Ei=CV2

Now, in second case the switch is disconnected and dielectric of constant 3 is inserted in both capacitors

Ef=EAf+EBf
EAf=12×3×C×V2=32CV2

In B the potential changes as dilectric is inserted to a free charged capacitor, which is given as VK.


EBf=12×3C×(V3)2=12×3CV29EBf=12×13CV2

Ef=32CV2+16CV2=12CV2[3+13]Ef=12×103×CV2

Therefore, the ratio of energies is given as:
EiEf=CV212×103×CV2=3:5

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