Topic: Electromagnetics
A parallel plate air-filled capacitor has plate area of $10^{-4} m^{2}$ and plate separation of $10^{-3} m$. It is connected to a 2 V, 1.8 GHz source. The magnitude of the displacement current is $(\epsilon_{0}=1/36\pi\times10^{-9}F/m)$
- A. 200 mA
- B. 20 mA
- C. 20 A
- D. 2 mA
Answer: B. 20 mA
Explanation: Displacement current $I_d = C \frac{dV}{dt}$. Capacitance $C = \frac{\epsilon_0 A}{d}$. Since $V = V_0 \sin(\omega t)$, the magnitude of the displacement current is $|I_d| = \omega C V_0$. So, $|I_d| = 2\pi(1.8 \times 10^9) \left[\frac{(1/36\pi \times 10^{-9})(10^{-4})}{10^{-3}}\right](2) = (3.6\pi \times 10^9) \left(\frac{10^{-13}}{36\pi \times 10^{-3}}\right) (2) = 20 \times 10^{-3} \text{ A} = 20 \text{ mA}$.