Answer

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**Hint**: The charge of a disconnected capacitor remains the same before and after the dielectric has been placed. The voltage drops during after the dielectric has been placed.

Formula used: In this solution we will be using the following formulae;

\[Q = CV\]where \[Q\] is the charge on the capacitor, \[C\] is the capacitance of the capacitor and \[V\] is the voltage across its plate.

\[C = \dfrac{{K\varepsilon A}}{d}\] where \[K\] is the dielectric constant for the material between the plate, \[\varepsilon \] is the permittivity of free space, \[A\] is the area of the capacitor plates, and \[d\] is the distance between the plates.

\[U = \dfrac{1}{2}C{V^2}\] where \[U\] is the potential energy (or energy) possessed by a capacitor.

\[W = - \Delta U\] where \[W\] is the work done, and \[\Delta \] signifies change in a quantity (in this case, \[U\])

**Complete Step-by-Step solution:**

To solve, we note that the charge before and after the dielectric has been placed are the same, since the capacitor was disconnected before it was done.

Generally,

\[Q = CV\] where \[Q\] is the charge on the capacitor, \[C\] is the capacitance of the capacitor and \[V\] is the voltage across its plate.

Capacitance is

\[C = \dfrac{{K\varepsilon A}}{d}\] where \[K\] is the dielectric constant for the material between the plate, \[\varepsilon \] is the permittivity of free space, \[A\] is the area of the capacitor plates, and \[d\] is the distance between the plates.

The initial capacitance is

\[{C_0} = \dfrac{{\varepsilon A}}{d}\] (since \[K = 1\] for air)

Hence, charge is,

\[Q = \dfrac{{\varepsilon A}}{d}V\]

The final capacitance

\[C = \dfrac{{K\varepsilon A}}{d}\]

Hence, charge is also

\[Q = \dfrac{{K\varepsilon A}}{d}{V_f}\]

Hence, by equating, we have

\[Q = \dfrac{{K\varepsilon A}}{d}{V_f} = \dfrac{{\varepsilon A}}{d}V\]

Then, by simplification, we have

\[{V_f} = \dfrac{V}{K}\].

Now the potential energy is given as

\[U = \dfrac{1}{2}C{V^2}\] where \[U\] is the potential energy (or energy) possessed by a capacitor.

Hence, the initial and final energy are respectively

\[{U_i} = \dfrac{1}{2}\left( {\dfrac{{\varepsilon A}}{d}} \right){V^2}\]

\[{U_f} = \dfrac{1}{2}\left( {\dfrac{{K\varepsilon A}}{d}} \right)V_f^2 = \dfrac{1}{2}\left( {\dfrac{{K\varepsilon A}}{d}} \right)\dfrac{{{V^2}}}{{{K^2}}} = \dfrac{1}{2}\left( {\dfrac{{\varepsilon A}}{d}} \right)\dfrac{{{V^2}}}{K}\]

The work done is

\[W = - \Delta U\] where \[W\] is the work done, and \[\Delta \] signifies change in a quantity (in this case, \[U\])

Hence,

\[W = - \left[ {\dfrac{1}{2}\left( {\dfrac{{\varepsilon A}}{d}} \right)\dfrac{{{V^2}}}{K} - \dfrac{1}{2}\left( {\dfrac{{\varepsilon A}}{d}} \right){V^2}} \right]\]

\[ \Rightarrow W = - \dfrac{1}{2}\left( {\dfrac{{\varepsilon A}}{d}} \right){V^2}\left[ {\dfrac{1}{K} - 1} \right]\]

By rearranging, we have

\[W = \dfrac{{\varepsilon A}}{{2d}}{V^2}\left[ {1 - \dfrac{1}{K}} \right]\]

**Hence, the correct option is A**

**Note:**Alternatively, without lengthy calculations and using some reasoning, we can get the answer. We can reason as follows: first work done is a difference between different energy states, hence the answer cannot be C or D, also, energy is \[\dfrac{1}{2}C{V^2}\], hence, the answer cannot be B either. Hence, the answer is A.

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