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To solve this problem, we will use the relationship between the displacement current (\( I_d \)) and the rate of change of the electric flux (\( \frac{d\Phi_E}{dt} \)) in a capacitor. 

### Given Information:
1. **Radius of the capacitor plates** (\( R \)): \( 2.0 \, \text{cm} = 0.02 \, \text{m} \)
2. **Separation distance** (\( d \)): \( 0.60 \, \text{mm} = 0.0006 \, \text{m} \)
3. **Frequency** (\( f \)): \( 75.0 \, \text{Hz} \)
4. **Maximum displacement current** (\( I_d \)): \( 38 \, \mu\text{A} = 38 \times 10^{-6} \, \text{A} \)

### Objective:
Determine the maximum value of \( \frac{d\Phi_E}{dt} \) between the plates, with fringing effects neglected.

### Solution:
1. **Displacement Current and Electric Flux Relationship**:
   The displacement current \( I_d \) is related to the rate of change of electric flux \( \frac{d\Phi_E}{dt} \) by the equation:
   \[
   I_d = \epsilon_0 \frac{d\Phi_E}{dt}
   \]
   where \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{F/m} \).

2. **Solving for \( \frac{d\Phi_E}{dt} \)**:
   Rearranging the equation to solve for \( \frac{d\Phi_E}{dt} \):
   \[
   \frac{d\Phi_E}{dt} = \frac{I_d}{\epsilon_0}
   \]
   
   Substituting the given values:
   \[
   \frac{d\Phi_E}{dt} = \frac{38 \times 10^{-6} \, \text{A}}{8.85 \times 10^{-12} \, \text{F/m}}
   \]

3. **Calculating the Result**:
   \[
   \frac{d\Phi_E}{dt} \approx \frac{38 \times 10^{-6}}{8.85 \times 10^{-12}} = 4.3 \times 10^6 \, \text{V}\cdot\text{m/s}
   \]

### Answer:
The maximum value of \( \frac{d\Phi_E}{dt} \) between the plates is approximately \( 4.3 \times 10^6 \, \text{V}\cdot\text{m/s} \), expressed with two significant figures.

Would you like more details, or do you have any questions? 

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Here are some related questions:
1. How does the displacement current relate to the magnetic field around a capacitor?
2. What would happen to \( \frac{d\Phi_E}{dt} \) if the separation \( d \) between plates increased?
3. How does frequency affect the maximum displacement current in a capacitor?
4. How would the electric field between the plates change if the radius of the plates increased?
5. What assumptions are made by neglecting fringing effects?

**Tip:** When dealing with time-varying electric fields, the displacement current plays a role similar to conduction current in generating a magnetic field.

Suppose an air-gap capacitor has circular plates of radius R = 2.0 cm and separation d = 0.60 mm. A 75.0 Hz emf, E = E0 cos(ωt), is applied to the capacitor. The maximum displacement current is 38 μA. Determine the maximum value of dΦE/dt between the plates. Neglect fringing. Express your answer using two significant figures.

Learn how to determine the maximum rate of change of electric flux (dΦE/dt) in an air-gap capacitor with circular plates. Given the plate radius, separation, frequency, and maximum displacement current, we apply Maxwell's displacement current equation to solve for dΦE/dt. This solution is ideal for undergraduate-level physics students studying electromagnetism.

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