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**Hint:**When a capacitor is charged, it not only stores charge but also energy. The type of energy stored in the capacitors is the electrical potential energy. When the battery is removed, the capacitor dissipates the electrical energy stored in it (until it gets completely discharged) to the electrical device it is connected to and so it can be used like a temporary battery.

**Complete step by step solution:**

When it is connected to a battery, the charge stored in the capacitor is its capacitance times the voltage of the battery. The charge is stored until the potential of the capacitor is equal to the voltage of the battery.

$Q = CV$

Where,

$Q =$ charge stored

$C =$ capacitance of the capacitor

$V =$ voltage of the battery

The capacitors are connected in parallel, therefore their total capacitance is calculated as ${C_{total}} = {C_1} + {C_2}$. Here, ${C_1} = 3\mu F$ and ${C_2} = 4\mu F$.

Calculating the charge on the first capacitor:

$\Rightarrow {Q_1} = {C_1}{V_1}$

$\Rightarrow {Q_1} = \left( {3\mu F} \right)\left( {6V} \right)$

$\Rightarrow {Q_1} = 18\mu C$

Calculating the charge on the second capacitor:

$\Rightarrow {Q_2} = {C_2}{V_2}$

$\Rightarrow {Q_2} = \left( {4\mu F} \right)\left( {6V} \right)$

$\Rightarrow {Q_2} = 24\mu C$

Common potential difference, ${V_{common}}$is equal to the total charge of the capacitors$({Q_{total}})$ divided by their total capacitance $\left( {{C_{total}}} \right)$

i.e., ${V_{common}} = \dfrac{{{Q_{total}}}}{{{C_{total}}}}$

$\Rightarrow {V_{common}} = \dfrac{{24\mu C - 18\mu C}}{{3\mu F + 4\mu F}}$

$\Rightarrow {V_{common}} = \dfrac{6}{7}V$

Energy of a capacitor, ${E_{capacitor}} = \dfrac{1}{2}QV$

Substituting the value of $Q = CV$ in the above equation, we get

$\Rightarrow {E_{capacitor}} = \dfrac{1}{2}C{V^2}$

Hence, the final energy stored $\left( {{U_{final}}} \right)$ is given as:

$\Rightarrow {U_{final}} = \dfrac{1}{2}\left( {{C_1} + {C_2}} \right)V_{common}^2$

$\Rightarrow {U_{final}} = \dfrac{1}{2}\left( {3 \times {{10}^{ - 6}} + 4 \times {{10}^{ - 6}}} \right){\left( {\dfrac{6}{7}} \right)^2}$

$\Rightarrow {U_{final}} = \dfrac{{{{10}^{ - 6}}}}{2}(7)\left( {\dfrac{6}{7}} \right)\left( {\dfrac{6}{7}} \right)$

$\Rightarrow {U_{final}} = \dfrac{{18}}{7} \times {10^{ - 6}}J$

$\therefore {U_{final}} = 2.57 \times {10^{ - 6}}J$

**Therefore, the correct answer [D], $2.57 \times {10^{ - 6}}J$.**

**Note:**The change in electrical potential energy $\Delta P.E{._{electric}}$ is given as: $\Delta P.E{._{electric}} = QV$. But Energy of a capacitor, ${E_{capacitor}} = \dfrac{1}{2}QV$. This is because as the charge stored on the capacitor decreases during discharge of charge, so does the voltage across the capacitor decrease. the sum of the different values of energy at each instant as the voltage across the capacitor drops is equal to ${E_{capacitor}}.$

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