A 0.134-A current is charging a capacitor that has square plates 6.00 cm on each side. The plate separation is 4.00 mm. (a) Find the time rate of change of electric flux between the plates. V·m/s (b) Find the displacement current between the plates. A

Final answer:

To find the time rate of change of electric flux between the plates of the capacitor, use the formula \(\frac{d\phi_E}{dt} = \frac{I}{A_\text{plate}}\). The displacement current between the plates can be found using the formula \(I_d = \varepsilon_0 \kappa \frac{d\phi_E}{dt}\).

Explanation:

To find the time rate of change of electric flux between the plates of the capacitor, we can use the formula: \(\frac{d\phi_E}{dt} = \frac{I}{A_\text{plate}}\), where \(\frac{d\phi_E}{dt}\) is the time rate of change of electric flux, \(I\) is the current, and \(A_\text{plate}\) is the area of one plate. In this case, the area of each plate is \((0.06 \,\text{m})^2\) and the current is 0.134 A. Thus, the time rate of change of electric flux is \(\frac{0.134 \,\text{A}}{(0.06 \,\text{m})^2}\) V·m/s.

The displacement current between the plates of a capacitor can be found using the formula: \(I_d = \varepsilon_0 \kappa \frac{d\phi_E}{dt}\), where \(I_d\) is the displacement current, \(\varepsilon_0\) is the vacuum permittivity, \(\kappa\) is the dielectric constant, and \(\frac{d\phi_E}{dt}\) is the time rate of change of electric flux. In this case, \(\varepsilon_0\) is a constant, \(\kappa\) depends on the material between the plates (not provided), and we found \(\frac{d\phi_E}{dt}\) to be \(\frac{0.134 \,\text{A}}{(0.06 \,\text{m})^2}\) V·m/s. So the displacement current can be calculated once these values are known.

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Final answer:

The time rate of change of electric flux can be found using I/ε₀A. The displacement current can be found using ε₀A(dE/dt)

Explanation:

The time rate of change of electric flux between the plates of the capacitor can be found using the formula:

dΦ/dt = I/ε₀A

Where dΦ/dt is the time rate of change of electric flux, I is the current, ε₀ is the permittivity of free space, and A is the area of one of the plates.

We are given that the current is 0.134 A and the area of each plate is (0.060 m)² = 0.0036 m². Plugging these values into the equation, we get: the time rate of change of electric flux between the plates is 37.22 V·m/s.

Similarly, the displacement current between the plates can be found using the formula:

Id = ε₀A(dE/dt)

Where Id is the displacement current, ε₀ is the permittivity of free space, A is the area of one of the plates, and dE/dt is the time rate of change of electric field intensity between the plates.

We are given that ε₀ is 8.854 × 10⁻¹² F/m and dΦ E/ dt is 37.22 V·m/s. Plugging these values into the equation, we get:

Id = (8.854 × 10⁻¹² F/m)(37.22 V·m/s) = 3.29 × 10⁻¹⁰ A

Therefore, the displacement current between the plates is 3.29 × 10⁻¹⁰ A.

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