A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the combined system to that stored initially in the single capacitor.

Hint: Use the fact the voltage gets divided into two equal halves for 2 capacitors connected in parallel of equal capacitance, Now use Energy stored formula for both the cases and take ratio.

Complete step-by-step answer:
The energy stored initially,
$ {U_i} = \dfrac{1}{2}C{V^2}$

After connecting to the other unchanged capacitor, charge is disturbed equally.
Since, Q=CV
$ {V_f} = \dfrac{V}{2} $
Hence, energy stored finally in two capacitor is given by:
$
  {U_f} = 2 \times \dfrac{1}{2}CV_f^2 \\
  {U_f} = \dfrac{{C{V^2}}}{4} \\
$
therefore, the ratio of energy stored in combined system to that stored initially is:
$ \dfrac{{{U_f}}}{{{U_i}}} = \dfrac{1}{2} $
The parallel plate capacitor is said to be the simplest form of capacitor. It can be constructed using two metal or metallised foil plates at a distance parallel to each other. In a parallel plate capacitor the current flowing through the circuit gradually becomes less and then zero till the voltage of the capacitor is exactly equal but opposite to the voltage of the battery. Moreover, the capacitor gets charged when it is connected across a d.c battery.

Note:If in case you want to increase the capacitance of parallel plate capacitor then we have to first increase the surface area, then reduce the separation between the plates and use a dielectric material in between the plate which have higher dielectric breakdown strength (dielectric between them act as separator for the plates).