Two circular plates of radius 5.0 cm are 0.50 mm apart, with a vacuum between them…

Step 1. The capacitance. $A = \pi R^2 = 7.854 \times 10^{-3}\ \text{m}^2$. Then $\varepsilon_0 A = 8.85 \times 10^{-12} \times 7.854 \times 10^{-3} = 6.95 \times 10^{-14}$ in SI units, and $C = \dfrac{\varepsilon_0 A}{d} = \dfrac{6.95 \times 10^{-14}}{0.50 \times 10^{-3}\ \text{m}} = 139\ \text{pF}$.

Step 2. The peak displacement current. $\omega = 2\pi f = 1.257 \times 10^{7}\ \text{rad/s}$. So $I_d = C\omega V_0 = 139 \times 10^{-12}\ \text{F} \times 1.257 \times 10^{7}\ \text{rad/s} \times 120\ \text{V} = 210 \times 10^{-3}\ \text{A} = 210\ \text{mA}$.

Step 3. The peak rate of change of the field. $E = V/d$, so $\dfrac{dE}{dt} = \dfrac{\omega V_0}{d} = \dfrac{1.257 \times 10^{7}\ \text{rad/s} \times 120\ \text{V}}{0.50 \times 10^{-3}\ \text{m}} = 3.02 \times 10^{12}\ \text{V/(m s)}$.

Step 4. The field at the edge. $B = \dfrac{\mu_0 I_d}{2\pi R} = \dfrac{2 \times 10^{-7}\ \text{T m/A} \times 0.2096\ \text{A}}{0.050\ \text{m}} = 0.839\ \mu\text{T}$.