In a parallel plate capacitor connected to a constant voltage source, a dielectric is introduced such that its surface area is the same as that of plates but the thickness of the dielectric is half the separation between the plates. If dielectric constant of the dielectric between the plates is $6$, then
(A) percentage loss of energy of the capacitor, for introducing the dielectric between the plates is $41.67\% $
(B) the capacitors decreases by about $71\% $
(C) the potential difference across the plates of the capacitors decreases by about $71\% $
(D) the charge of the capacitors remains conserved while introducing the dielectric between the plates.

Hint: To give the answer this problem we must know about the concept of series and parallel connection of capacitor. As in this problem the series connection is given so use the formula of series connection to find equivalent capacitance. By using this capacitance we will answer the question.

Complete step by step answer:
Initially the capacitance in the capacitor was $C = \dfrac{{A{\varepsilon _0}}}{d}$.
Here, $C$ is capacitance, $A$ is area, ${\varepsilon _0}$ is permittivity of free space, and $d$ is the separation.

When dielectric fill the half separation $\left( {\dfrac{d}{2}} \right)$ of plates, there are two capacitors, one is air filled with capacitance $2C$ and another is dielectric filled with capacitance $2kC$ and they are in series.

The equivalent capacitance for series connection is give by,
$\dfrac{1}{{{C_{{\rm{eq}}}}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}}$

By substituting $2C$ for ${C_1}$, and $2kC$ for ${C_2}$, we get,
\[\begin{array}{c}
\dfrac{1}{{{C_{{\rm{eq}}}}}} = \dfrac{1}{{2C}} + \dfrac{1}{{2kC}}\\
 = \dfrac{{2kC + 2C}}{{\left( {2C} \right)\left( {2kC} \right)}}\\
{C_{{\rm{eq}}}} = \dfrac{{\left( {2C} \right)\left( {2kC} \right)}}{{2kC + 2C}}\\
 = \dfrac{{2kC}}{{k + 1}}
\end{array}\]

By substituting $6$ for $k$, we get,
\[\begin{array}{c}
{C_{{\rm{eq}}}} = \dfrac{{2\left( 6 \right)}}{{6 + 1}}C\\
 = \dfrac{{12}}{7}C\\
 = 1.71C
\end{array}\]

Find the percentage of increase in capacitance as follows.
$\left( {1.71 - 1} \right) \times 100\% = 71\% $

So, option (B) is wrong because capacitance is increased by $71\% $ not decreases by about $71\% $.

As the capacitor is connected to voltage source so total potential remain constant and energy stored in capacitors will also increase because the energy stored in capacitor is given by $U = \dfrac{1}{2}{C_{{\rm{eq}}}}{V^2}$. So, the option (A) is also wrong.

In a capacitor charge is always conserved and it will flow to maintain constant potential.

Thus, the correct option is (D).

Note:
The thing which needs to be taken care of is the type of connection because students generally get confused while writing the equivalent capacitance formula for series and parallel connection. We must remember the concept about the charge in the capacitor.