Question

Figure (31-E26) 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.
Figure

Answer 1

The ratio of the initial total energy stored in the capacitors = 3

Explanation:

When switch s is closed, total energy E = ½ $CV_2$, in this stage we understood that the circuit will acts as closed loop and allows the passage of the currents. Hence the capacitance remain same in the overall circuit.

By considering the other case, When Switch is open the capacitance in the circuit varies which causes variation in the energy.

$E=\frac{1}{2} \times \frac{q^{2}}{c}$

$=\frac{1}{2} \times \frac{c^{2} v^{2}}{3 c}$  $=\frac{c v^{2}}{6}$

In case of 'A"

$C_{\mathrm{eff}}=3 c$

$E=\frac{1}{2} \times C_{\mathrm{eff}} V^{2}$

$=\frac{1}{2} \times 3 c \times v^{2}=\frac{3}{2} c v^{2}$

Hence total final energy is equal to the ratio of initial energy and final energy.

$\text { Total final energy }=\frac{c v^{2}}{6}+\frac{3 c v^{2}}{2}=\frac{10 c v^{2}}{6}$

$\text { Now, } \frac{\text { Initial Energy }}{\text { Final Energy }}=\frac{\mathrm{cv}^{2}}{\frac{10 \mathrm{cv}^{2}}{6}}=3$