Question

A parallel plate capacitor has air between disk-shaped plates of radius \(4.00 \mathrm{~mm}\) that are coaxial and \(1.00 \mathrm{~mm}\) apart. Charge is being accumulated on the plates of the capacitor. What is the displacement current between the plates at an instant when the rate of charge accumulation on the plates is \(10.0 \mu \mathrm{C} / \mathrm{s} ?\)

Short answer

Answer: The displacement current between the plates of the capacitor is approximately 0.625/π A/m².

Step-by-step solution

Calculate the area of the plates

Since the plates are disk-shaped, we can calculate the area using the formula for the area of a circle: A = πr^2. In this case, r = 4.00 mm = 4.00 * 10⁻³ m. A = π * (4.00 * 10⁻³ m)² = π * (16 * 10⁻⁶ m²) = 16π * 10⁻⁶ m²

Calculate the rate of change of electric flux

Given the rate of charge accumulation is 10.0 μC/s = 10.0 * 10⁻⁶ C/s, we can calculate the rate of change of electric flux using the formula dQ/dt = ε₀ * (dΦ_E/dt). First, let's convert the rate of charge accumulation to C/s: 10.0 μC/s = 10.0 * 10⁻⁶ C/s Now, we can calculate the rate of change of electric flux: dΦ_E/dt = (10.0 * 10⁻⁶ C/s) / (ε₀ * A)

Calculate the displacement current

We can now find the displacement current using the formula: I_d = ε₀ * (dΦ_E/dt) First, let's plug in the values for ε₀ and A: I_d = (8.85 * 10⁻¹² F/m) * (dΦ_E/dt) Now, substituting the rate of change of electric flux calculated in Step 2, we have: I_d = (8.85 * 10⁻¹² F/m) * ((10.0 * 10⁻⁶ C/s) / (ε₀ * A)) I_d = (8.85 * 10⁻¹² F/m) * ((10.0 * 10⁻⁶ C/s) / ((8.85 * 10⁻¹² F/m) * (16π * 10⁻⁶ m²))) I_d = (10.0 * 10⁻⁶ C/s) / (16π * 10⁻⁶ m²) I_d = 0.625 / π A/m² The displacement current between the plates is approximately 0.625/π A/m².