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

When the plates are pulled apart to a separation $d_2$, the final capacitance becomes $C_2 = \dfrac{\varepsilon_0 A}{d_2}$.

Since the battery remains connected, the potential difference $V$ across the capacitor is constant.

The initial charge on the capacitor is $Q_1 = C_1 V and the final charge is$Q_2 = C_2 V$.<br><br>The charge flown through the circuit is the difference between the initial and final charges:<br><br>$\Delta Q = Q_1 - Q_2 = (C_1 - C_2) V$<br><br>$\Delta Q = \varepsilon_0 A V \left( \dfrac{1}{d_1} - \dfrac{1}{d_2} \right)

Substituting the given values $A = 20$\times$10^{-4}$\text{ m}^2$,$V = 12.0 $\text{ V},$d_1 = 1.00 $\times$10^{-3} $\text{ m},$d_2 = 2.0 $\times$10^{-3} $\text{ m}, and$\varepsilon_0 = 8.85 $\times$10^{-12} $\text{ F/m}:<br><br>\Delta$Q = 8.85 $\times$10^{-12} $\times$20 $\times$10^{-4} $\times$12 $\times$\left( \dfrac{1}{10^{-3}} - \dfrac{1}{2 \times 10^{-3}} \right)

\Delta$Q = 8.85$\times$10^{-12}$\times$24$\times$10^{-3}$\times \left(1000 - 500 \right)

\Delta$Q = 8.85 \times10^{-12} \times24 \times10^{-3} \times500$

\Delta$Q = 8.85 \times10^{-12} \times12$

\Delta$Q = 106.2 \times10^{-12}$\text{ C} =$1.06 \times10^{-10}$\text{ C}