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Displacement Current in a Dielectric. Suppose that the parallel plates in Fig. 29.23 have an area of \(3.00 \mathrm{~cm}^{2}\) 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 \(120 \mathrm{~V}\) and the conduction current \(i_{\mathrm{C}}\) equals \(6.00 \mathrm{~mA} .\) 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?

Short Answer

Expert verified
The capacitance of the capacitor is approximately \(0.0532 \, \mu F\), the charge on each plate is approximately \(6.38 \, \mu C\), their rate of change equals the conduction current \(6.00 \, mA\), and the displacement current inside the dielectric is the same, \(6.00 \, mA\).

Step by step solution

01

Calculate the capacitance

One should first calculate the capacitance \(C\) of the capacitor using the formula for a plate capacitor: \[C = \varepsilon_0 \cdot \varepsilon_r \cdot \frac{A}{d}\] where \(\varepsilon_0 = 8.854 \times 10^{-12} \, Fm^{-1}\) is the permittivity of free space, \(\varepsilon_r\) is the relative permittivity or dielectric constant of the material between the plates, \(A\) is the area of one of the plates, and \(d\) is the separation between the plates. Here, \(\varepsilon_r = 4.70\), \(A = 3.00 \times 10^{-4} \, m^2\), and \(d = 2.50 \times 10^{-3} \, m\).
02

Calculate the charge on each plate

Use the formula \(Q = C \cdot V\) to calculate the charge \(Q\) on each plate. \(V = 120 \, V\) is the potential difference between the plates. Substitute \(C\) from step 1 to get \(Q\).
03

Determine the rate of change of charge

The rate of change of charge on the plates equals the conduction current \(i_C\), which is \(6.00 \times 10^{-3} \, A\). This fact is due to the principle of conservation of charge.
04

Calculate the displacement current

According to the Maxwell-Ampère law, in a capacitive system the displacement current \(i_D\) inside the dielectric equals the rate of change of charge on the plates (or equivalently the conduction current \(i_C\)). Hence, \(i_D = i_C\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dielectric Constant
The dielectric constant, denoted as \( \varepsilon_r \), is a crucial parameter in the study of capacitors. It represents how well a material can store electrical energy in an electric field. When a dielectric material is placed between the plates of a capacitor, it increases the capacitor's ability to hold charge.
  • An increased dielectric constant implies that the material allows the capacitor to store more charge at the same voltage.
  • Mathematically, the dielectric constant is expressed as the ratio of the permittivity of the dielectric material \( \varepsilon \) to the permittivity of free space \( \varepsilon_0 \).
  • The formula is \( \varepsilon_r = \frac{\varepsilon}{\varepsilon_0} \).
In our exercise, the dielectric constant is given as \(4.70\), indicating that the material within the capacitor can store 4.7 times more electrical energy compared to a vacuum. This enhancement is vital in practical applications such as improving the efficiency of electronic circuits.
Capacitance
Capacitance is a measure of a capacitor's ability to store charge per unit voltage. It plays a central role in determining the charge on the plates and is given by the formula:
\[ C = \varepsilon_0 \cdot \varepsilon_r \cdot \frac{A}{d} \]
  • \( C \) is the capacitance measured in farads (F).
  • \( \varepsilon_0 \) is the permittivity of free space, approximately \(8.854 \times 10^{-12} \, F/m \).
  • \( A \) is the area of one plate, and \( d \) is the separation between the plates.
  • The dielectric constant \( \varepsilon_r \) modifies this capacitance, increasing it as the constant increases.
In the given exercise, we calculate the capacitance to determine the charge stored when a 120 V potential difference is applied across the plates. By manipulating the capacitance equation and substituting the given values, you find that an increased dielectric constant leads to a higher capacitance, and thus, the ability to store more charge.
Maxwell-Ampère Law
The Maxwell-Ampère law is a pivotal principle in electromagnetism. It extends Ampère's law to include the effects of changing electric fields, particularly in capacitors where displacement current is relevant. This law is expressed with the equation:
\[ \oint B \cdot \mathrm{d}l = \mu_0 (i_C + i_D) \]
where:
  • \( B \) is the magnetic field.
  • \( \mu_0 \) is the permeability of free space.
  • \( i_C \) is the conduction current, and \( i_D \) is the displacement current.
In systems involving capacitors, like in our problem, the displacement current \( i_D \) effectively allows the continuation of current within a dielectric when the physical movement of charge (conduction current) does not pass through. The law shows that displacement current helps maintain magnetic fields in the dielectric without the need for physical charge carriers. By equating \( i_D \) to the rate of change of electric displacement field, this concept bridges the gap between static fields and dynamic systems, showing that the displacement current is as crucial as the conduction current in this context.

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