The plates of an air capacitor have an area of 0.050 m² and are 2.0 mm apart…

Step 1. The capacitance. $C = \dfrac{\varepsilon_0A}{d} = \dfrac{8.85 \times 10^{-12}\ \text{C}^2/(\text{N}\,\text{m}^2) \times 0.050\ \text{m}^2}{2.0 \times 10^{-3}\ \text{m}} = 2.21 \times 10^{-10}\ \text{F} = 221\ \text{pF}$.

Step 2. The charge and the energy. $Q = CV = 2.21 \times 10^{-10}\ \text{F} \times 500\ \text{V} = 1.11 \times 10^{-7}\ \text{C}$. The energy is $U = \tfrac{1}{2}CV^2 = \tfrac{1}{2} \times 2.21 \times 10^{-10}\ \text{F} \times (500\ \text{V})^2 = 2.77 \times 10^{-5}\ \text{J} = 27.7\ \mu\text{J}$.

Step 3. The force. The field is $E = \dfrac{V}{d} = \dfrac{500\ \text{V}}{2.0 \times 10^{-3}\ \text{m}} = 2.5 \times 10^5\ \text{V/m}$. So $F = \tfrac{1}{2}QE = \tfrac{1}{2} \times 1.11 \times 10^{-7}\ \text{C} \times 2.5 \times 10^5\ \text{V/m} = 1.38 \times 10^{-2}\ \text{N} = 13.8\ \text{mN}$.

Step 4. The work. The charge stays the same, and the force $\dfrac{Q^2}{2\varepsilon_0A}$ does not depend on the gap. So $W = F\,\Delta d = 13.8 \times 10^{-3}\ \text{N} \times 1.0 \times 10^{-3}\ \text{m} = 13.8\ \mu\text{J}$.

What to learn. The force per unit area is $\tfrac{1}{2}\varepsilon_0E^2$, which is the energy density. Here, $0.277\ \text{J/m}^3 \times 0.050\ \text{m}^2 = 13.8\ \text{mN}$.