Given Data
- Plate area, \( A = 5 \, \text{cm}^2 = 5 \times 10^{-4} \, \text{m}^2 \)
- Plate separation, \( d = 3 \, \text{mm} = 3 \times 10^{-3} \, \text{m} \)
- Voltage, \( V(t) = 50 \sin(1000t) \, \text{V} \)
- Permittivity, \( \varepsilon = 2\varepsilon_0 = 2 \times 8.85 \times 10^{-12} \, \text{F/m} \)
Displacement Current Density Calculation
1. Capacitance Calculation
The capacitance \( C \) of the capacitor is given by:
\[ C = \frac{\varepsilon A}{d} \]
Substituting the values:
\[ C = \frac{(2 \times 8.85 \times 10^{-12}) \times (5 \times 10^{-4})}{3 \times 10^{-3}} \]
\[ C \approx 2.943 \times 10^{-12} \, \text{F} \]
2. Voltage Derivative
The displacement current density \( J_d \) is related to the rate of change of electric field:
\[ J_d = \varepsilon \frac{dE}{dt} \]
Where \( E = \frac{V}{d} \).
Thus,
\[ E(t) = \frac{50 \sin(1000t)}{3 \times 10^{-3}} \]
The derivative is:
\[ \frac{dE}{dt} = \frac{50 \times 1000 \cos(1000t)}{3 \times 10^{-3}} \]
3. Displacement Current Density
Now, substituting \( \frac{dE}{dt} \) into the displacement current density formula:
\[ J_d = \varepsilon \frac{dE}{dt} = 2 \varepsilon_0 \frac{dE}{dt} \]
\[ J_d = 2(8.85 \times 10^{-12}) \times \frac{50 \times 1000 \cos(1000t)}{3 \times 10^{-3}} \]
\[ J_d \approx 2 \times 8.85 \times 10^{-12} \times \frac{50000 \cos(1000t)}{3 \times 10^{-3}} \]
\[ J_d \approx 2.95 \times 10^{-3} \cos(1000t) \, \text{A/m}^2 \]
Conclusion
The displacement current density for the given parallel plate capacitor is approximately:
\[ J_d \approx 2.95 \times 10^{-3} \cos(1000t) \, \text{A/m}^2 \]