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:

$C_1 = \dfrac{\epsilon_0 A}{d_1}$

Substituting the given values $A = 20 \text{ cm}^2 = 20 \times 10^{-4} \text{ m}^2$ and $d_1 = 1.00 \text{ mm} = 10^{-3} \text{ m}$:

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

The initial energy stored in the capacitor is:

$U_1 = \dfrac{1}{2} C_1 V^2$

$U_1 = \dfrac{1}{2} \times 17.7 \times 10^{-12} \times (12.0)^2 = 1274.4 \times 10^{-12} \text{ J} \approx 12.7 \times 10^{-10} \text{ J}$

When the plates are pulled apart to a new separation $d_2 = 2.0 \text{ mm}$, the new capacitance becomes:

$C_2 = \dfrac{\epsilon_0 A}{d_2} = \dfrac{C_1}{2}$

Since the battery remains connected, the potential difference across the plates remains constant at $V = 12.0 \text{ V}$. The final energy stored in the capacitor is:

$U_2 = \dfrac{1}{2} C_2 V^2 = \dfrac{1}{2} \left( \dfrac{C_1}{2} \right) V^2 = \dfrac{U_1}{2}$

$U_2 = \dfrac{1274.4 \times 10^{-12}}{2} = 637.2 \times 10^{-12} \text{ J} \approx 6.37 \times 10^{-10} \text{ J}$