Question

The initial and final energy stored in the capacitor, ${{U}_{i}}$ and ${{U}_{f}}$ ?

Answer 1

Hint : In order to solve the question, first know what the capacitors are and the capacitors formula for the capacitance and the charge and when the potential is applied across the plates, the value of capacitance changes and then, the energies for the initial and final states are calculated.
Formula for the capacitance is given by
$C=\dfrac{A{{\varepsilon }_{0}}}{d}$
And that for charge stored in capacitor,
$Q=CV=\dfrac{A{{\varepsilon }_{0}}V}{d}$
Where $A$ is the area of the plates and $d$ is the distance between the plates.
The energy stored in a capacitor is given by the formula:
$U=\dfrac{1}{2}C{{V}^{2}}$ .

Complete Step By Step Answer:
A parallel plate capacitor consists of two metal plates separated by a distance, then, the electric field is applied across the plates to store the energy.
Let us consider a parallel plate capacitor, with the distance between the plates, $d$ and area of the plates $A$ , let the medium between the plates be air
Therefore, the initial plate capacitance is given by
${{C}_{i}}=\dfrac{A{{\varepsilon }_{0}}}{d}$
And the charge is ${{Q}_{i}}={{C}_{i}}{{V}_{i}}=\dfrac{A{{\varepsilon }_{0}}V}{d}$
The final parallel plate capacitance is ${{C}_{f}}=\dfrac{A{{\varepsilon }_{0}}}{2d}$
And the final charge is given as
${{Q}_{f}}={{C}_{f}}{{V}_{f}}=\dfrac{A{{\varepsilon }_{0}}{{V}_{f}}}{2d}$
As the battery is disconnected so the charge will remain the same
Therefore, ${{Q}_{i}}={{Q}_{f}}$
Now putting the values of both the charges
$\begin{aligned} & \Rightarrow \dfrac{A{{\varepsilon }_{0}}V}{d}=\dfrac{A{{\varepsilon }_{0}}{{V}_{f}}}{2d} \\\ & \Rightarrow {{V}_{f}}=2V \\\ \end{aligned}$
The energy stored in a capacitor is given by the formula:
$U=\dfrac{1}{2}C{{V}^{2}}$
Thus, the initial energy will be
${{U}_{i}}=\dfrac{1}{2}{{C}_{i}}{{V}_{i}}^{2}=\dfrac{1}{2}\times \dfrac{A{{\varepsilon }_{0}}}{d}\times {{V}^{2}}=\dfrac{A{{\varepsilon }_{0}}{{V}^{2}}}{2d}$
And the final energy will be
${{U}_{f}}=\dfrac{1}{2}{{C}_{f}}{{V}_{f}}^{2}=\dfrac{1}{2}\times \dfrac{A{{\varepsilon }_{0}}}{2d}\times 4{{V}^{2}}=\dfrac{A{{\varepsilon }_{0}}{{V}^{2}}}{d}$

Note :
The final capacitance of the capacitor is about half of its value at the initial stage, its charge could become approximately half too but it depends on the voltage also. However, the final energy is twice than that of the initial value. The electric field is applied to store the charge in the capacitor.