Question

Two coaxial conducting cylinders of radius \(2 \mathrm{~cm}\) and \(4 \mathrm{~cm}\) have a length of \(1 \mathrm{~m}\). The region between the cylinders contains a layer of dielectric from \(\rho=c\) to \(\rho=d\) with \(\epsilon_{r}=4\). Find the capacitance if \((\) a) \(c=2 \mathrm{~cm}, d=3 \mathrm{~cm} ;\) (b) \(d=4 \mathrm{~cm}\), and the volume of the dielectric is the same as in part \((a)\).

Short answer

Question: Calculate the capacitance of the coaxial cylinders for two different configurations of the dielectric layer: (a) inner radius c = 2cm, outer radius d = 3cm, and relative permittivity εr = 4; (b) outer radius d = 4cm, and the volume of the dielectric is the same as in part (a). Answer: The capacitance of the coaxial cylinders with the given dielectric layer configuration (a) is approximately 227 pF, and for configuration (b), it is approximately 260 pF.

Step-by-step solution

Convert radii to meters

Since the radii are given in cm, we need to convert them to meters for our calculations. We have c=0.02m and d=0.03m.

Calculate capacitance

Now, we will plug in the values into the formula for the capacitance. \( C = \frac{2\pi(\varepsilon_0)(\varepsilon_r)L}{\ln(\frac{d}{c})} = \frac{2\pi(8.85\times10^{-12})(4)(1)}{\ln(\frac{0.03}{0.02})}\) Calculating the capacitance, we get: \( C \approx 2.27\times 10^{-10} F \) The capacitance of the coaxial cylinders with the given dielectric layer configuration (a) is approximately 227 pF. #b) Volume of dielectric is the same#

Calculate the volume of the dielectric in configuration (a)

First, we find the volume V of the dielectric in the configuration (a). Since the dielectric is cylindrical, the volume can be found using the formula: \(V = π(L)(d^2 - c^2)\) Plugging in the values, we get: \(V = π(1)(0.03^2 - 0.02^2) \approx 2.29 \times 10^{-3} m^3\)

Find the inner radius c' for configuration (b)

The volume of the dielectric layer remains the same in configuration (b), and the outer radius is given as d=4cm or 0.04m. We will denote the inner radius in this case as c'. Using the formula for the volume, we have: \(V = π(L)(0.04^2 - (c')^2)\) We will solve for c' using the known volume V: \(c' = \sqrt{0.04^2 - \frac{V}{\pi L}}\) Calculating c', we get: \(c' \approx 0.0337 m\)

Calculate capacitance

Now, using the capacitance formula with c'=0.0337m and d=0.04m, we get: \( C = \frac{2\pi(\varepsilon_0)(\varepsilon_r)L}{\ln(\frac{d}{c'})} = \frac{2\pi(8.85\times10^{-12})(4)(1)}{\ln(\frac{0.04}{0.0337})}\) Calculating the capacitance, we get: \( C \approx 2.60\times 10^{-10} F \) The capacitance of the coaxial cylinders with the given dielectric layer configuration (b) is approximately 260 pF.

Key concepts

Dielectric Materials

Dielectric materials are insulators that can be polarized by an electric field, meaning they can be induced to have small displaced charges within their structure. This characteristic makes them extremely useful in capacitors, where they serve to increase the capacitance without conducting the electric charge.

A key parameter for dielectric materials is the relative permittivity \( \varepsilon_r \), also known as the dielectric constant. This is a dimensionless number that describes how much better a dielectric material can store electrical energy compared to the vacuum. For our coaxial cylinder example, a dielectric with \( \varepsilon_r = 4 \) is used between the cylinders, indicating that the material can store four times as much electrical energy per unit volume as the vacuum.

When a dielectric is placed in a capacitor, it affects the electric field between the plates, reducing it in strength due to polarization, which increases the amount of charge that can be stored on the plates for a given voltage, thus increasing the capacitance. In the example exercise, this effect is clearly demonstrated as the presence of the dielectric material significantly alters the capacitance calculation.

Capacitance Calculation

Capacitance is a measure of a capacitor's ability to store charge and is calculated based on the physical characteristics of the capacitor and the dielectric material used. The standard formula for the capacitance \( C \) of a co-axial cylinder is given by:

\[ C = \frac{2\pi(\varepsilon_0)(\varepsilon_r)L}{\ln(\frac{d}{c})} \]

Where \( \varepsilon_0 \) is the permittivity of free space, \( \varepsilon_r \) is the relative permittivity of the dielectric material, \( L \) is the length of the cylinders, \( c \) is the inner radius, and \( d \) is the outer radius. This equation is derived from electromagnetic theory, considering the distribution of the electric field and potential in a coaxial cable configuration.

For instance, the first part of the exercise calculates capacitance with given dimensions and a specific dielectric constant. Moreover, the exercise progresses to explain how maintaining the same volume of dielectric material while changing the dimensions impacts the capacitance. This shows that capacitance depends not only on material parameters but also on the geometry of the setup.

Electromagnetic Theory

Electromagnetic theory encompasses the study of electric fields, magnetic fields, and the interaction between them. It's a fundamental aspect of physics that explains how charges and currents produce electromagnetic fields, and how those fields can propagate through space as electromagnetic waves.

The concept of capacitance is rooted in electromagnetic theory. It explains the electric field within capacitors, between the inner and outer cylinder in the case of coaxial cables, for example. According to this theory, the electric field between the two cylinders of different radii originates from one cylinder and terminates at the other. This field stores energy in the capacitive system.

Coaxial capacitors benefit from electromagnetic principles. Their unique design minimizes the external electric and magnetic fields, making them ideal for transferring electrical signals with minimal interference. Therefore, understanding the link between electromagnetic theory and practical devices like the coaxial capacitor allows students to better appreciate how fundamental principles are applied in everyday technology.