Question

Two identified parallel plate capacitors $A$ and $B$ are connected to a battery of $V$ volts with the switch $S$ closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant $K$. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.

Answer 1

Initially charge on either capacitor is $Q_A=Q_B=CV$ .
After dielectric is introduced, new capacitance of either capacitor =KC
After opening switch potential across capacitor A is V volts.
Let potential across capacitor B be $V_1$
Therefore $Q_B=CV=C_1{V}_1=KC{V}_1$
$V_1=\frac{V}{K}$
Initial energy in both capacitors =$\frac{C{V}^2}{2}+\frac{C{V}^2}{2}=C{V}^2$
Final energy of capacitor A =$\frac{KC{V}^2}{2}$
Final energy of capacitor B =$\frac{KC{V}^2}{2K^2}=\frac{C{V}^2}{2K}$
Total final energy of both capacitors = $\frac{KC{V}^2}{2}+\frac{C{V}^2}{2K}=\left(\frac{K^2+1}{2K}\right)C{V}^2$
Ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric= $\frac{C{V}^2}{\left(\frac{K^2+1}{2K}\right)C{V}^2}=\frac{2K}{K^2+1}$