Answer
385.8k+ views
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.
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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