A parallel-plate capacitor with a plate area of 20 cm^2 and a...

The initial capacitance of the parallel-plate capacitor is given by:

$C_1 = \dfrac{\varepsilon_0 A}{d_1}$

The final capacitance after the plates are pulled apart is:

$C_2 = \dfrac{\varepsilon_0 A}{d_2}$

Since the battery remains connected, the potential difference $V$ across the capacitor is constant. The initial and final charges on the capacitor are $Q_1 = C_1 V$ and $Q_2 = C_2 V$ respectively.

As the separation increases, the capacitance decreases, causing charge to flow back into the battery. The charge that flows back is:

$\Delta Q = Q_1 - Q_2 = (C_1 - C_2) V$

The energy absorbed by the battery is the work done on it by the returning charge:

$W = \Delta Q \times V = (C_1 - C_2) V^2$

Substituting the given values $A = 20 \times 10^{-4} \text{ m}^2$, $d_1 = 1.00 \times 10^{-3} \text{ m}$, $d_2 = 2.0 \times 10^{-3} \text{ m}$, and $V = 12.0 \text{ V}$:

$C_1 = \dfrac{8.85 \times 10^{-12} \times 20 \times 10^{-4}}{1.00 \times 10^{-3}} = 17.7 \times 10^{-12} \text{ F}$

$C_2 = \dfrac{8.85 \times 10^{-12} \times 20 \times 10^{-4}}{2.0 \times 10^{-3}} = 8.85 \times 10^{-12} \text{ F}$

The difference in capacitance is:

$C_1 - C_2 = 17.7 \times 10^{-12} - 8.85 \times 10^{-12} = 8.85 \times 10^{-12} \text{ F}$

The energy absorbed by the battery is:

$W = (8.85 \times 10^{-12}) \times (12.0)^2$

$W = 8.85 \times 10^{-12} \times 144 = 1274.4 \times 10^{-12} \text{ J}$

$W = 12.7 \times 10^{-10} \text{ J}$