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Given Data

1. Current $I = 3.9 \, \text{A}$
2. Square plate length $l = 1.60 \, \text{cm} = 0.016 \, \text{m}$

We need to find the rate of change of the electric field between the plates, expressed in volts per meter per second ($\frac{\text{V}}{\text{m} \cdot \text{s}}$).

Relevant Formulas

1. Displacement Current Density $J_d = \frac{dE}{dt} \times \epsilon_0$
   • Where $\epsilon_0$ (permittivity of free space) is $8.85 \times 10^{-12} \, \text{F/m}$.
   • The displacement current density is linked to the rate of change of the electric field $\frac{dE}{dt}$.
2. Current and Displacement Current Relation
   • Total current $I$ through the capacitor is also equal to the product of displacement current density and area $A$: $I = J_d \cdot A = \epsilon_0 \cdot \frac{dE}{dt} \cdot A$
3. Area of the Plates
   • Since the plates are square: $A = l^2 = (0.016 \, \text{m})^2 = 2.56 \times 10^{-4} \, \text{m}^2$

Solution

Rearrange the equation for $\frac{dE}{dt}$: $\frac{dE}{dt} = \frac{I}{\epsilon_0 \cdot A}$

Substitute the known values: $\frac{dE}{dt} = \frac{3.9}{8.85 \times 10^{-12} \cdot 2.56 \times 10^{-4}}$

Calculating the denominator: $\epsilon_0 \cdot A = 8.85 \times 10^{-12} \times 2.56 \times 10^{-4} = 2.2656 \times 10^{-15}$

Now, divide: $\frac{dE}{dt} = \frac{3.9}{2.2656 \times 10^{-15}} \approx 1.72 \times 10^{15} \, \text{V/m/s}$

Answer

The rate at which the electric field is changing between the plates is approximately: $\frac{dE}{dt} \approx 1.72 \times 10^{15} \, \text{V/m/s}$

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Related Questions

1. How does the permittivity of free space $\epsilon_0$ affect the electric field between capacitor plates?
2. What is the physical meaning of displacement current density?
3. How would the answer change if the plate area was larger?
4. Why is the concept of displacement current necessary in capacitors?
5. How does the geometry of capacitor plates influence electric field changes?

Tip

In physics problems, keep track of units carefully. Ensuring unit consistency is crucial for accurate results.