Problem

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

Displacement Current in a Dielectric. Suppose that the parallel plates in Fig. 29.23 have an area of 3.00 cm2 and are separated by a 2.50-mm-thick sheet of dielectric that completely fills the volume between the plates. The dielectric has dielectric constant 4.70. (You can ignore fringing effects.) At a certain instant, the potential difference between the plates is 120 V and the conduction current iC equals 6.00 mA. At this instant, what are (a) the charge q on each plate; (b) the rate of change of charge on the plates; (c) the displacement current in the dielectric?

#### Step-by-Step Solution

Solution 1

The equation for the charge on the capacitor is,

$$Q=C V$$

Here, $$C$$ is the capacitance and $$V$$ is the potential difference.

The equation for the capacity of the parallel plate capacitor is,

$$C=\frac{\varepsilon A}{d}$$

Here, $$\varepsilon$$ is the permittivity, $$A$$ is the area of the capacitor plates and $$d$$ is the distance between the two capacitor plates.

The equation for the current density is,

$$J_{D}=\varepsilon \frac{d E}{d t}$$

In the dielectric, $$i_{D}=i_{c}$$

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