Question

Let the internal dimensions of a coaxial capacitor be \(a=1.2 \mathrm{~cm}, b=4 \mathrm{~cm}\), and \(l=40 \mathrm{~cm}\). The homogeneous material inside the capacitor has the parameters \(\epsilon=10^{-11} \mathrm{~F} / \mathrm{m}, \mu=10^{-5} \mathrm{H} / \mathrm{m}\), and \(\sigma=10^{-5} \mathrm{~S} / \mathrm{m}\). If the electric field intensity is \(\mathbf{E}=\left(10^{6} / \rho\right) \cos 10^{5} t \mathbf{a}_{\rho} \mathrm{V} / \mathrm{m}\), find \((a) \mathbf{J} ;(b)\) the total conduction current \(I_{c}\) through the capacitor; \((c)\) the total displacement current \(I_{d}\) through the capacitor; \((d)\) the ratio of the amplitude of \(I_{d}\) to that of \(I_{c}\), the quality factor of the capacitor.

Short answer

Given a coaxial capacitor with internal dimensions, parameters, and electric field intensity, we have found: a) The current density, J: \(\mathbf{J} = 10 \cos{10^5 t} \mathbf{a}_{\rho} A / m^2\) b) The total conduction current, \(I_c\): \(I_c \approx 3.87\cos{10^5 t} A\) c) The total displacement current, \(I_d\): \(I_d \approx -12.32 \cdot 10^5 \sin{10^5 t} A\) d) The quality factor of the capacitor: Q ≈ \(3.18 \cdot 10^5\)

Step-by-step solution

Write down the given quantities and equations

We are given the following quantities: Inner radius: \(a = 1.2 cm\) Outer radius: \(b = 4 cm\) Length of the capacitor: \(l = 40 cm\) Electric field intensity: \(\mathbf{E}=\left(10^{6} / \rho\right) \cos 10^{5} t \mathbf{a}_{\rho} V / m\) Parameters of the homogeneous material: \(\epsilon = 10^{-11} F / m\) \(\mu = 10^{-5} H / m\) \(\sigma = 10^{-5} S / m\)

Calculate the current density J

The current density, J, can be calculated using Ohm's Law for current density: \(\mathbf{J} = \sigma \mathbf{E}\) Substitute the given values for \(\sigma\) and \(\mathbf{E}\): \(\mathbf{J} = (10^{-5} S / m) \cdot \left(10^6 / \rho \right) \cos{10^5 t} \mathbf{a}_{\rho} V / m\) Simplifying, we have: \(\mathbf{J} = 10 \cos{10^5 t} \mathbf{a}_{\rho} A / m^2\)

Calculate the total conduction current, \(I_{c}\)

The total conduction current, \(I_{c}\), can be found by integrating the current density over the surface of the capacitor: \(I_{c} = \int \mathbf{J} \cdot d\mathbf{S}\) With cylindrical coordinates, we have \(d\mathbf{S} = \rho dl d\phi \mathbf{a}_{\rho}\), so the integral becomes: \(I_{c} = \int_{0}^{2\pi}\int_{0}^{0.4}\int_{0.012}^{0.04} 10\cos{10^5 t} \cdot \rho dl d\phi d\rho\) With the simplified integral: \(I_{c} = 10\cos{10^5 t} \cdot \int_{0}^{2\pi} d\phi \int_{0}^{0.4} dl \int_{0.012}^{0.04} \rho d\rho\) Compute the integral, we get: \(I_{c} = 10\cos{10^5 t} \cdot 2\pi \cdot 0.4 \cdot \left(\frac{0.04^2 - 0.012^2}{2}\right)\) After solving, we have: \(I_{c} = 10\cos{10^5 t} \cdot \pi (0.1232)\) \(I_{c} \approx 3.87\cos{10^5 t} A\)

Calculate the total displacement current, \(I_{d}\)

The total displacement current, \(I_{d}\), can be found using the formula: \(I_{d} = \epsilon \frac{d\Phi}{dt}\), where \(\Phi\) is the electric flux. Electric flux, \(\Phi = \int \mathbf{E} \cdot d\mathbf{S} = EA\), where A is the area of the surface. Differentiating E with respect to time, we have: \(\frac{d\mathbf{E}}{dt} = -10^5 \cdot 10^6 \cdot \sin{10^5 t} \mathbf{a}_{\rho} V/m^2\) Now, we find the area A: \(A = \pi(b^2 - a^2)\) Substitute the given values and solve for A: \(A = \pi((0.04)^2 - (0.012)^2) = \pi(0.1232)\) Now, we find the displacement current: \(I_{d} = \epsilon \frac{d\Phi}{dt} = \epsilon A \frac{d \mathbf{E}}{dt}\) \(I_{d} = (10^{-11} F/m) \cdot \pi (0.1232) \cdot (-10^5 \cdot 10^6 \sin{10^5 t} V/m^2)\) Simplifying, we have: \(I_{d} \approx -12.32 \cdot 10^5 \sin{10^5 t} A\)

Calculate the ratio of the amplitude of \(I_{d}\) to that of \(I_{c}\), the quality factor of the capacitor

The ratio of the amplitude of the displacement current \(I_{d}\) to that of the conduction current \(I_{c}\) gives us the quality factor, Q. Here, the ratio can be represented as: \(Q = \frac{12.32 \cdot 10^5}{3.87} \approx 3.18 \cdot 10^5\) Thus, the quality factor for the given coaxial capacitor is approximately \(3.18 \cdot 10^5\).

Key concepts

Current Density

Current density is an essential concept in electromagnetism describing the electric current per unit area of cross-section. It's mathematically depicted by the symbol \textbf{J} and quantifies how much charge is flowing through a given area. The formula for current density is:
\[\begin{equation} \textbf{J} = \frac{I}{A} \text{, where \(I\) is the total current and \(A\) is the cross-sectional area.} \text{In the context of a coaxial capacitor, we consider the movement of charges between the inner and outer conductors.} \end{equation}\] Intuitively, if you increase the area through which current can pass, the current density decreases, and vice versa. This concept contributes to understanding how effectively current can be transferred through different mediums.

Conduction Current

Conduction current refers to the movement of electric charge that is facilitated through a material due to the movement of electrons or ions. This is the kind of current we typically measure with an ammeter in a circuit. For a coaxial capacitor, the conduction current flows along the conductive material between the two cylindrical conductors.
In equations, we can express the conduction current \textbf{I} as an integral of the current density \textbf{J} over the conductor's cross-sectional area \[\begin{equation} I = \text{\int }\mathbf{J} \cdot d\mathbf{S} \end{equation}\] where \textbf{S} is the area vector. In simple terms, this tells us how much current is passing through the entire cross-section of the conductor.

Displacement Current

Displacement current is a bit more abstract than conduction current. It doesn't involve actual charges moving through a conductor but rather a time-varying electric field. This concept was introduced by James Clerk Maxwell to explain the continuity of current in scenarios where no physical charge flow exists, such as the region in a capacitor.
\[\begin{equation} I_d = \epsilon \frac{d\Phi_E}{dt} \end{equation}\] Here, \(\epsilon\) is the permittivity of the material and \(\Phi_E\) is the electric flux linked with the area. For a coaxial capacitor, this displacement current plays a role when dealing with an alternating electric field, as it ensures that the laws of electromagnetism remain consistent in situations where the electric field changes over time.

Quality Factor

The quality factor, often denoted as Q, is a dimensionless parameter that compares the displacement current's amplitude to the conduction current's amplitude in a coaxial capacitor. A higher quality factor indicates that a capacitor is more reactive (stores more energy) than resistive (dissipates energy).
\[\begin{equation} Q = \frac{|I_d|}{|I_c|} \end{equation}\] In our case, the quality factor gives us an understanding of how the capacitor behaves under an alternating current scenario and signifies its efficiency at storing and releasing energy. The computed high value of the quality factor suggests that the considered coaxial capacitor is particularly efficient in its capacitive properties, meaning it has a superior capacity to store charge before significant losses due to conduction occur.

Electric Flux

Electric flux, denoted by \(\Phi_E\), is a measure that describes the quantity of electric field lines passing through a given area. In a physical sense, it helps to characterize the amount of 'electric field' present over an area. For a uniform electric field and flat area, the electric flux can be found by multiplying the electric field strength, E, by the area, A, that the field lines are passing through.
\[\begin{equation} \Phi_E = \mathbf{E} \cdot \mathbf{A} \end{equation}\] In the context of our coaxial capacitor, when evaluating the displacement current, we first calculate the electric flux through the surface area between the inner and outer conductors. This step is crucial for determining how the changing electric field influences the displacement current within the capacitor.

Ohm's Law for Current Density

Ohm's Law is well understood when it comes to voltage, current, and resistance. However, when dealing with continuous media, we adapt the law to relate electric field and current density—a form often referred to as Ohm's Law for current density. The relationship is given by:
\[\begin{equation} \mathbf{J} = \sigma \mathbf{E} \end{equation}\] Here, \textbf{J} is current density, \(\sigma\) is the electrical conductivity of the material, and \textbf{E} is the electric field intensity. This equation suggests that current density in a material is directly proportional to the electric field applied, assuming the conductivity is constant. In a coaxial capacitor, by knowing the electric field and the material's conductivity, we can determine how dense the current will be that flows through this material, which is crucial for understanding the capacitor's behavior under different electrical conditions.