• The capacitance $C$ of the capacitor when filled with dielectric is:
  $$C = \frac{Q}{V} = \frac{0.06 \times 10^{-6}\,\text{C}}{60\,\text{V}} = 10^{-9}\,\text{F}$$
• The capacitance $C_0$ in the absence of dielectric (vacuum/air) is:
  $$C_0 = \frac{\varepsilon_0 A}{d} = \frac{(8.854 \times 10^{-12}\,\text{F/m})(200 \times 10^{-4}\,\text{m}^2)}{2.0 \times 10^{-3}\,\text{m}} = \frac{1.7708 \times 10^{-13}}{2.0 \times 10^{-3}} = 8.854 \times 10^{-11}\,\text{F}$$
• The dielectric constant $K$ of the material is:
  $$K = \frac{C}{C_0} = \frac{10^{-9}\,\text{F}}{8.854 \times 10^{-11}\,\text{F}} = \frac{100}{8.854} \approx 11.294 \approx 11.3$$

In a series combination of capacitors, both capacitors acquire the same charge $Q$ ($Q_1 = Q_2 = Q$).

• Potential difference across a capacitor is $V = \frac{Q}{C}$:
  $$\frac{V_1}{V_2} = \frac{Q/C_1}{Q/C_2} = \frac{C_2}{C_1} = \frac{3\,\mu\text{F}}{2\,\mu\text{F}} = \frac{3}{2}$$
• Stored electrostatic energy in terms of charge $Q$ is $U = \frac{Q^2}{2C}$:
  $$\frac{U_1}{U_2} = \frac{Q^2/(2C_1)}{Q^2/(2C_2)} = \frac{C_2}{C_1} = \frac{3\,\mu\text{F}}{2\,\mu\text{F}} = \frac{3}{2}$$

• The capacitance of a parallel plate capacitor with plate separation $d$ in air is:
  $$C_0 = \frac{\varepsilon_0 A}{d}$$
• When a dielectric slab of thickness $t$ and dielectric constant $K$ is introduced, and plate separation is increased by $\Delta d$, the new capacitance is:
  $$C = \frac{\varepsilon_0 A}{(d + \Delta d) - t + \frac{t}{K}} = \frac{\varepsilon_0 A}{d + \left(\Delta d - t\left(1 - \frac{1}{K}\right)\right)}$$
• For the capacitance to remain equal to its original value $C_0$:
  $$\Delta d = t \left(1 - \frac{1}{K}\right)$$
  Given $\Delta d = 3.2\,\text{mm}$ and $t = 4\,\text{mm}$:
  $$3.2 = 4 \left(1 - \frac{1}{K}\right) \implies 1 - \frac{1}{K} = \frac{3.2}{4} = 0.8$$
  $$\frac{1}{K} = 1 - 0.8 = 0.2 \implies K = \frac{1}{0.2} = 5$$