Displacement Current in a Dielectric. Suppose that the parallel plates in Fig. 29.23...

The area of the capacitor plate is,

$$ \begin{aligned} A &=3.00 \mathrm{~cm}^{2} \\ &=3.00 \mathrm{~cm}^{2}\left(\frac{10^{-4} \mathrm{~m}^{2}}{\mathrm{~cm}^{2}}\right) \\ &=3 \times 10^{-4} \mathrm{~m}^{2} \end{aligned} $$

The thickness of dielectric sheet or distance between the capacitor plates is,

$$ \begin{aligned} d &=2.50 \mathrm{~mm} \\ &=2.50 \mathrm{~mm}\left(\frac{10^{-3} \mathrm{~m}}{\mathrm{~mm}}\right) \\ &=2.5 \times 10^{-3} \mathrm{~m} \end{aligned} $$

The given conduction of the current is,

$$ \begin{aligned} i_{c} &=6.00 \mathrm{~mA} \\ &=6.00 \mathrm{~mA}\left(\frac{10^{-3} \mathrm{~A}}{\mathrm{~mA}}\right) \\ &=6.00 \times 10^{-3} \mathrm{~A} \end{aligned} $$

(a)

Calculate the charge on each plate.

The capacitance of the parallel plate capacitor having dielectric is,

$$ C=k \frac{A \varepsilon_{0}}{d} $$

Here, \(k\) is the dielectric constant, \(A\) is the area of the plate capacitor, \(\varepsilon_{0}\) is the permittivity of the free space and \(d\) is the separation between the capacitors.

The charge on the each plate is,

$$ q=C V $$

Here, \(C\) is the capacitance of the parallel plate capacitor and \(V\) is the potential difference between the plates.

Substitute \(k \frac{A \varepsilon_{0}}{d}\) for \(C\) in the above equation.

$$ q=\left(k \frac{A \varepsilon_{0}}{d}\right) V $$

Substitute \(4.7\) for \(k, 3 \times 10^{-4} \mathrm{~m}^{2}\) for \(A, 8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{Nm}^{2}, 2.5 \times 10^{-3} \mathrm{~m}\) for \(d\) and \(120 \mathrm{~V}\) for \(V\) in the above equation.

$$ \begin{aligned} q &=\left((4.7) \frac{\left(3 \times 10^{-4} \mathrm{~m}^{2}\right)\left(8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{Nm}^{2}\right)}{2.5 \times 10^{-3} \mathrm{~m}}\right)(120 \mathrm{~V}) \\ &=5.99 \times 10^{-13} \mathrm{C} \end{aligned} $$

Therefore, the charge on each plate is \(5.99 \times 10^{-13} \mathrm{C}\)

(b)

Calculate the rate of change of the charge on the plates.

The rate of change of charge on the plates is equal the conduction current.

$$ \frac{d q}{d t}=i $$

The conduction current is given as,

$$ \begin{aligned} i_{e} &=6 \mathrm{~mA}\left(\frac{10^{-3} \mathrm{~A}}{1 \mathrm{~mA}}\right) \\ &=6 \times 10^{-3} \mathrm{~A} \end{aligned} $$

Substitute \(6 \times 10^{-3} \mathrm{~A}\) for \(i_{c}\) in the equation \(\frac{d q}{d t}=i_{c}\).

$$ \frac{d q}{d t}=6 \times 10^{-3} \mathrm{~A} $$

Therefore, the rate of change of charge on the plates is \(6 \times 10^{-3} \mathrm{~A}\).

(c)

Calculate the displacement current in the dielectric.

The displacement current \(I_{D}\) is equal to the conduction current \(i_{e}\) in the in the dielectric. Then,

$$ i_{D}=i_{c} $$

Substitute \(6 \times 10^{-3} \mathrm{~A}\) for \(i_{c}\) in the above equation.

$$ i_{D}=6 \times 10^{-3} \mathrm{~A} $$

Therefore, the displacement current in the dielectric is \(6 \times 10^{-3} \mathrm{~A}\)