Question

Two identical 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

Two capacitors are connected in parallel. Hence, the potential on each of them remains the same. So, the charge on each capacitor is

Q_A = Q_B = CV

Formula for energy stored = \(\cfrac12\) CV² = \(\cfrac12\)\(\cfrac{Q^2}{C}\)

Net capacitance with switch S closed = C + C = 2C

\(\therefore\) Energy stored = \(\cfrac12\) x 2C x V² = CV²

After the switch S is opened, capacitance of each capacitor = KC

In this case, voltage only across A remains the same.

The voltage across B changes to V' = \(\cfrac{Q}{C'}=\cfrac{Q}{KC}\)

\(\therefore\) Energy stored in capacitor A = \(\cfrac12\)KCV²

Energy stored in capacitor B = 

 \(\therefore\) Total Energy stored

Required ratio