Question

A parallel plate capacitor with a dielectric slab of dielectric constant 3, filling the space between the plates, is charged to a potential V. The battery is then disconnected, and the dielectric slab is withdrawn, it is then replaced by another dielectric slab of dielectric constant 2. If the energies stored in the capacitor before and after the dielectric slab is changed are ${{E}_{1}}$ and ${{E}_{2}}$, then ${{E}_{1}}$/${{E}_{2}}$ is:
A. $\dfrac{4}{9}$
B. $\dfrac{2}{3}$
C. $\dfrac{3}{2}$
D. $\dfrac{9}{5}$

Answer 1

First, we will find the charge that the plates of the capacitor will have and then use the expression of energy with charge and capacitance. We will define how the capacitance changes on changing the dielectric slab and after that how the energy stored in the capacitor changes on changing the dielectric slab.
Formula used:
The capacitance of a capacitor
$C=\dfrac{\varepsilon A}{d}=\dfrac{k{{\varepsilon }_{o}}A}{d}$
The relation between capacitance-voltage across the capacitor and charge stored in the capacitor
$Q=CV$
Energy stored in the capacitor
$E=\dfrac{C{{V}^{2}}}{2}=\dfrac{QV}{2}=\dfrac{{{Q}^{2}}}{2C}$

_Complete step-by-step solution: _
The capacitance of the capacitor will be given by the following formula
$C=\dfrac{\varepsilon A}{d}=\dfrac{k{{\varepsilon }_{o}}A}{d}$
Here k is the dielectric constant, A is the area of plates of the capacitor, and d is the distance between the plates of the capacitor. ${{\varepsilon }_{o}}$ is the permittivity of free space. All the values except for ‘k’ remains constant throughout the process so let us define the capacitance as
$C=k{{C}_{o}}$
The charge on the plates of capacitors will also remain the same once the battery is disconnected. The charge they will have is
$Q=kV{{C}_{o}}$
Initially, it will be $Q=3V{{C}_{o}}$as the dielectric slab of constant three has been inserted inside.
With the new dielectric slab, the charge will remain the same, but capacitance will change. So, we will use the formula for energy as
$E=\dfrac{{{Q}^{2}}}{2C}$ and the energy will be inversely proportional to the capacitance of the capacitor which in turn is directly proportional to the dielectric constant of the slab inside the capacitor. So, the energy will be inversely proportional to the dielectric constant.
$\therefore \dfrac{{{E}_{1}}}{{{E}_{2}}}=\dfrac{{{k}_{2}}}{{{k}_{1}}}=\dfrac{2}{3}$

Hence the correct option is B. i.e., $\dfrac{2}{3}$.

Note: There is no electrical work being done on the capacitor but the energy increases. This happens due to the physical force needed to remove the dielectric slab from the capacitor. Similarly, energy is lost when a dielectric slab is inserted in the capacitor.