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.

Question

Philoid · Accepted Answer

Initially the switch is closed and the capacitors are fully charged.

When the switch is closed, both capacitors are in parallel as shown in fig,

Hence the total energy stored by the capacitor when switch is closed is –

where C is the capacitance and V is the applied voltage.

When the switch is opened and dielectric is induced, the capacitance is

Given dielectric constant as 3

The total energy stored by the capacitor when switch is closed is –

For capacitor at AB

And E_BC is –

Total energyE_f)

Now, the ratio of the initial total energy stored in the capacitors to the final total energy stored –

=3/5