Problem 34 - University Physics with Modern Physics

Question

A diclectric of permitivity \(3.5 \times 10^{-11} \mathrm{F} / \mathrm{m}\) completely fills the volume between two capacitor plates. For \(t > 0\) the electric flux through the dielectric is \(\left(8.0 \times 10^{3} \mathrm{V} \cdot \mathrm{m} / \mathrm{s}^{3}\right) t^{3}\) . The dielectric is ideal and nonmagnetic; the conduction current in the dielectric is zero. At what time does the displacement current in the dielectric equal 21\(\mu \mathrm{A} ?\)

Step-by-Step Solution

Answer

The displacement current equals 21μA at 5 seconds.

Step 1: Identify given values and understand the concepts

We are given the permittivity \( \varepsilon = 3.5 \times 10^{-11} \, \text{F/m} \), and the electric flux \( \Phi_E = (8.0 \times 10^{3} \, \text{V} \cdot \text{m/s}^3) \cdot t^3 \). The problem requires us to find the time where the displacement current equals \( 21 \mu \text{A} \). Displacement current \( I_d \) is given by \( I_d = \varepsilon \frac{d\Phi_E}{dt} \).

Step 2: Differentiate the electric flux with respect to time

The displacement current depends on the rate of change of the electric flux. We need to compute the derivative of \( \Phi_E \) with respect to \( t \):\[\frac{d\Phi_E}{dt} = \frac{d}{dt} \left( (8.0 \times 10^{3}) \cdot t^3 \right) = 3 \times (8.0 \times 10^{3}) \cdot t^2 = 2.4 \times 10^{4} \cdot t^2.\]

Step 3: Calculate the displacement current

Using the formula \( I_d = \varepsilon \frac{d\Phi_E}{dt} \), we substitute the permittivity and the derivative we found:\[I_d = (3.5 \times 10^{-11}) \cdot (2.4 \times 10^{4} \cdot t^2) = (8.4 \times 10^{-7}) \cdot t^2.\]

Step 4: Set displacement current equal to 21μA and solve for time

Set the expression for the displacement current equal to \( 21 \times 10^{-6} \, \text{A} \):\[8.4 \times 10^{-7} \cdot t^2 = 21 \times 10^{-6}.\]Solve for \( t^2 \):\[t^2 = \frac{21 \times 10^{-6}}{8.4 \times 10^{-7}} = 25.\]Thus, \( t = \sqrt{25} = 5 \, \text{seconds} \).

Key Concepts

Dielectric

A dielectric is a special type of insulating material that can be polarized by an electric field. This means that when you apply an external electric field, the charges within the dielectric will align in such a way as to reduce the field inside the material.
Dielectrics are crucial in capacitors because they increase the capacitor's ability to store electrical energy. When a dielectric is placed between the plates of a capacitor, it increases the capacitor's capacitance by a factor equal to the dielectric constant (more formally known as relative permittivity) of the material.
Important features of dielectrics include:

• Insulation property—dielectrics do not conduct electricity.
• Polarization—alignment of internal charges.
• Relative permittivity—measure of the dielectric's effect on capacitance.

Electric Flux

Electric flux is an important concept that helps us understand electric fields in various configurations. It measures how much electric field is passing through a given area. Imagine it as the number of electric field lines crossing a surface; more lines represent a higher flux.
The formula for electric flux across a surface is given by:\[\Phi_E = \int \vec{E} \cdot d\vec{A}\]Here, \(\vec{E}\) is the electric field, and \(\vec{A}\) is the area vector. When the electric field is uniform, this simplifies to:\[\Phi_E = E \cdot A \cdot \cos(\theta)\]where \(\theta\) is the angle between the field and the normal to the surface.
In our context, electric flux changes over time, and its rate of change influences the displacement current through the dielectric.

Permittivity

Permittivity is a fundamental property of all materials that is a measure of how much they can resist the electric field. The permittivity of a dielectric material determines how effectively it can increase the capacitance when used between capacitor plates.
There are two types of permittivity:

• Absolute permittivity, usually denoted by \(\varepsilon\), is the measure of permittivity in mediums like vacuum, air, or specific materials. The vacuum permittivity (\(\varepsilon_0\)) is a constant, approximately \(8.85 \times 10^{-12} \, \text{F/m}\).
• Relative permittivity, or dielectric constant (\(\varepsilon_r\)), is the ratio of the permittivity of a medium to the permittivity in a vacuum, given by \(\varepsilon = \varepsilon_r \cdot \varepsilon_0\).

Permittivity affects how electric fields interact with the medium and is crucial in calculating displacement currents.

Capacitor Plates

Capacitor plates form one of the most fundamental components of capacitors, which are used to store and manage electric charge. These plates are typically made of conductive materials, often metals, and are separated by a gap that can be filled with air or a dielectric material.
When a voltage is applied across the plates, one plate accumulates positive charge, while the other accumulates negative charge, forming an electric field between them. The presence of a dielectric increases capacitance by allowing more charge to be stored at the same voltage.
Key points about capacitor plates include:

• Material and spacing significantly affect a capacitor's charge capacity and efficiency.
• Dielectric materials between plates enhance the charge storing capability by increasing capacitance.
• Capacitors are essential in circuits for filtering, tuning, and energy storage applications.

Understanding capacitor plates and their function is crucial, not just for solving problems like displacement current, but for a comprehensive grasp of electrical circuits and electronics.