Question

A parallel plate air capacitor of capacitance \[C\] is connected to a cell of emf \[V\] and then disconnected from it. A dielectric slab of dielectric constant \[K\], which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect?
$A)$ The change in energy stored is \[\dfrac{1}{2}C{{V}^{2}}\left( \dfrac{1}{K}-1 \right)\].
$B)$ The charge on the capacitor is not conserved.
$C)$ The potential difference between the plates decreases \[K\] times.
$D)$ The energy stored in the capacitor decreases \[K\] times.

Answer 1

Hint: When a capacitor with air between its plates is connected to a voltage source, the amount of charge stored on the capacitor proportional to the capacitance of the capacitor as well as the potential difference between the plates. At the same time, energy stored on this capacitor is dependent on the area of plates as well as distance between the plates. When air between the plates is replaced by a dielectric slab, all these properties of capacitor change according to the dielectric constant of the slab, except one property, which is the prime utility of a capacitor.

Formula used: \[1)Q=CV\]
\[2)C=\dfrac{{{\varepsilon }_{0}}A}{d}\]
\[3)E=\dfrac{1}{2}C{{V}^{2}}\]

Complete step by step answer:
When a capacitor filled with air between the plates is connected to a cell of emf \[V\], then, the charge on it given by
\[Q=CV\]
where
\[C\] is the capacitance of capacitor
\[Q\] is the charge on capacitor
\[V\] is the potential difference between the plates of capacitor
Let this be equation 1.
Here, we know that capacitance of the capacitor is directly proportional to the area of plates and inversely proportional to the distance between the plates. Therefore, equation 1 can be rearranged as
\[C=\dfrac{Q}{V}=\dfrac{{{\varepsilon }_{0}}A}{d}\]
where
\[{{\varepsilon }_{0}}\] is the permeability of free space
\[A\] is the area of plates of capacitor
\[d\] is the separation between plates of capacitor
Let this be equation 2.
Now, when the cell is removed, the charge \[Q\] gets conserved on the plates of the capacitor.
At this point, if we insert a dielectric slab of dielectric constant \[K\] in between the plates, using equation 2, the new capacitance of the capacitor is given by
\[{{C}_{K}}=\dfrac{K{{\varepsilon }_{0}}A}{d}\]
where
\[{{C}_{K}}\] is the capacitance of the capacitor when a dielectric slab is inserted between its plates
$K$ is the dielectric constant of the slab
Let this be equation 3.
From equation 2 and equation 3, we have
\[{{C}_{K}}=KC\]
Let this be equation 4.
Using equation 1 and equation 4, the new potential difference between plates may be written as
\[{{V}_{K}}=\dfrac{Q}{{{C}_{K}}}=\dfrac{Q}{KC}=\dfrac{V}{K}\]
where
\[{{V}_{K}}\] is the potential difference between the plates of the capacitor when a dielectric plate is inserted
$V$ is the potential difference between the plates of the capacitor with air in between the plates
$K$ is the dielectric constant of the slab inserted between the plates
Let this be equation 5.
So, from equation 5, we can conclude that option \[C\] (The potential difference between the plates decreases K times) is correct.

Now, we know that energy stored in the capacitor in the air-filled condition is given by
\[E=\dfrac{1}{2}C{{V}^{2}}\]
where
$E$ is the energy stored in capacitor with air between its plates
$C$ is the capacitance of this capacitor
$V$ is the potential difference across this capacitor
Let this be equation 6.
Similarly, energy stored in capacitor when a dielectric slab is inserted is given by
\[{{E}_{K}}=\dfrac{1}{2}{{C}_{K}}{{V}_{K}}^{2}\]
where
${{E}_{K}}$ is the energy stored in capacitor with dielectric filled between its plates
${{C}_{K}}$ is the capacitance of this capacitor
${{V}_{K}}$ is the potential difference across this capacitor
Let this be equation 7.
Subtracting equation 6 from 7, we have
\[{{E}_{K}}-E=\dfrac{1}{2}{{C}_{K}}{{V}_{K}}^{2}-\dfrac{1}{2}C{{V}^{2}}\]
where
${{E}_{K}}-E$ is the change in energy stored in capacitor
Let this be equation 8.
Putting for the values for \[{{C}_{K}}\] and \[{{V}_{{{K}_{{}}}}}\] from equation 4 and equation 5 in equation 8, we have
\[{{E}_{K}}-E=\dfrac{1}{2}KC{{\left( \dfrac{V}{K} \right)}^{2}}-\dfrac{1}{2}C{{V}^{2}}\Rightarrow {{E}_{K}}-E=\dfrac{1}{2}C{{V}^{2}}\left( \dfrac{1}{K}-1 \right)\]
Let this be equation 9.
From equation 9, it is clear that option \[A\] is also correct.
Now, using 6, equation 9 may also be written as
\[{{E}_{K}}-E=E\left( \dfrac{1}{K}-1 \right)\Rightarrow {{E}_{K}}-E=\left( \dfrac{E}{K}-E \right)\Rightarrow {{E}_{K}}=\dfrac{E}{K}\]
Let this be equation 10.
Hence, from equation 10, we can further conclude that option \[D\] is also correct.

So, the correct answer is “Option B”.

Note: The basic functioning of the given capacitor is as follows. When the capacitor with air in between the plates is fed with voltage, charge gets stored on it. When this voltage is disconnected, no more charge gets stored on it, but the existing charge remains conserved, even when the air between its plates is replaced with a dielectric slab of dielectric constant $K$.