A parallel-plate capacitor with a capacitance of 5 is connected to...

Initial capacitance of the parallel-plate capacitor is given by:

$C_i = \dfrac{\epsilon_0 A}{d} = 5 \mu\text{F}$

The initial charge on the capacitor is:

$Q_i = C_i V = 5 \times 6 = 30 \mu\text{C}$

When a dielectric slab of thickness $t$ and dielectric constant $K$ is introduced into the gap, the new capacitance is:

$C_f = \dfrac{\epsilon_0 A}{d - t + \dfrac{t}{K}}$

Given $d = 2 \text{ mm}$, $t = 1 \text{ mm}$, and $K = 5$, we have:

$C_f = \dfrac{\epsilon_0 A}{2 - 1 + \dfrac{1}{5}} = \dfrac{\epsilon_0 A}{1.2} = \dfrac{2}{1.2} \times \dfrac{\epsilon_0 A}{2} = \dfrac{2}{1.2} \times 5 = \dfrac{25}{3} \mu\text{F}$

The final charge on the capacitor is:

$Q_f = C_f V = \dfrac{25}{3} \times 6 = 50 \mu\text{C}$

The amount of charge that flows through the battery is the difference between the final and initial charges:

$\Delta Q = Q_f - Q_i = 50 - 30 = 20 \mu\text{C}$