Chapter 11: Problem 15
= If the atomic model is taken instead as a nucleus of charge \(+Q\) at the centre of a spherical volume of radius a carrying the total electronic charge \(-Q\). show that the atomic polarizability is again \(4 \pi \varepsilon_{0} a^{3}\).
Short Answer
Expert verified
Answer: The atomic polarizability of this atomic model is 4πε₀a³.
Step by step solution
01
Understanding the concepts involved in this problem
First, recall the definition of the electric displacement field (\(\vec{D}\)) and the relationship to the electric field (\(\vec{E}\)): \(\vec{D} = \varepsilon_{0} \vec{E} + \vec{P}\), where \(\vec{P}\) is the polarization density of the material. The atomic polarizability is related to the polarization and the electric field as \(\vec{P} = \alpha \vec{E}\). Combining these equations, we get \(\vec{D} = \varepsilon_{0} \vec{E} + \alpha \vec{E}\). We will use these equations to analyze the atomic model and find the atomic polarizability.
02
Determine the electric field of the atomic model
The electric field for the atomic model can be found using Gauss's law: \(\oint_{S} \vec{E} \cdot \text{d}\vec{S} = \frac{Q_{\text{enclosed}}}{\varepsilon_{0}}\). Since we have a spherical model, the electric field is symmetric and can be written as \(\vec{E} = E_{r} \hat{r}\), where \(E_{r}\) depends only on the distance r from the center of the atom. For a sphere with radius r < a, the enclosed charge is uniformly distributed throughout the sphere. To find the charge q enclosed within a sphere of radius r, we use the reasoning that the charge density is constant: \(\frac{q}{\frac{4}{3}\pi r^3} = \frac{-Q}{\frac{4}{3}\pi a^3}\). Solving for q, we get \(q(r) = -Q\left(\frac{r^3}{a^3}\right)\). Applying Gauss's law, we find \(E_{r} = \frac{-Q r}{4 \pi \varepsilon_{0} a^{3}}\) for r < a.
03
Determine the electric displacement field of the atomic model
Using the same procedure as for the electric field, we can find the electric displacement field (\(\vec{D}\)) using Gauss's law, considering the total charge enclosed within the sphere: \(\oint_{S} \vec{D} \cdot \text{d}\vec{S} = Q_{\text{enclosed}}\). Therefore, \(D_{r} = \frac{Q}{4 \pi r^{2}}\), where \(D_{r}\) is the radial component of \(\vec{D}\), and \(r\) is the distance from the center of the atom.
04
Apply the relationship between electric field and electric displacement field
We have found both the electric field \(E_{r}\) and the electric displacement field \(D_{r}\) for the atomic model. Applying Gauss's law, we get \(D_{r} = \varepsilon_{0} E_{r} + P_{r}\), where \(P_{r}\) is the radial component of the polarization density. Now, we can use the relationship between the polarization (\(\vec{P}\)) and the electric field (\(\vec{E}\)) through the atomic polarizability \(\alpha\): \(P_{r} = \alpha E_{r}\).
05
Find the atomic polarizability
Combining the equations in step 4: \(\frac{Q}{4 \pi r^{2}} = \varepsilon_{0} \frac{-Q r}{4 \pi \varepsilon_{0} a^{3}} + \alpha \frac{-Q r}{4 \pi \varepsilon_{0} a^{3}}\). Simplifying this equation, we get \(\alpha = \frac{a^3}{r^3}\left(\frac{Q}{4 \pi \varepsilon_{0}a^{2}} - \frac{Q}{4 \pi a^{2}}\right)\). If we evaluate this expression at the radius r = a, we get \(\alpha = 4 \pi \varepsilon_{0} a^3\), which is the desired result for the atomic polarizability of the given atomic model.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Electric Displacement Field
The electric displacement field, often denoted by \( \vec{D} \), is a vector field that appears in Maxwell's equations. It accounts for the effects of free charge and bound charge within a material.
When an electric field is applied to a material, the electrons and nuclei within the atoms may shift slightly from their equilibrium positions, creating a polarization of the material. This shift is indicative of polarization density, \( \vec{P} \), and the electric displacement field is defined as \( \vec{D} = \varepsilon_{0} \vec{E} + \vec{P} \), where \( \varepsilon_{0} \) is the vacuum permittivity and \( \vec{E} \) is the electric field in the material.
This relationship shows how the electric displacement field encompasses both the vacuum response to the electric field and the additional effect introduced by the material's polarization. The concept is crucial to understand because it describes the overall 'displacement' of charge in response to an electric field and applies whether inside or outside the material.
When an electric field is applied to a material, the electrons and nuclei within the atoms may shift slightly from their equilibrium positions, creating a polarization of the material. This shift is indicative of polarization density, \( \vec{P} \), and the electric displacement field is defined as \( \vec{D} = \varepsilon_{0} \vec{E} + \vec{P} \), where \( \varepsilon_{0} \) is the vacuum permittivity and \( \vec{E} \) is the electric field in the material.
This relationship shows how the electric displacement field encompasses both the vacuum response to the electric field and the additional effect introduced by the material's polarization. The concept is crucial to understand because it describes the overall 'displacement' of charge in response to an electric field and applies whether inside or outside the material.
Gauss's Law
Gauss's Law is a fundamental law of electromagnetism that relates the electric fields to the charges that produce them. Mathematically, it is expressed as \( \oint_{S} \vec{E} \cdot \text{d}\vec{S} = \frac{Q_{\text{enclosed}}}{\varepsilon_{0}} \) for an electric field \( \vec{E} \) and a closed surface \( S \) with an enclosed charge \( Q_{\text{enclosed}} \).
In simpler terms, Gauss's Law states that the total electric flux out of a closed surface is proportional to the charge enclosed within the surface. This law has profound implications in calculating electric fields around symmetrical charge distributions, such as spherical or cylindrical objects. For the atomic model in question, which has spherical symmetry, Gauss's Law allows for the straightforward determination of the electric field produced by the spherical volume of charge, laying the groundwork for establishing the relationship between polarizability and electric field.
In simpler terms, Gauss's Law states that the total electric flux out of a closed surface is proportional to the charge enclosed within the surface. This law has profound implications in calculating electric fields around symmetrical charge distributions, such as spherical or cylindrical objects. For the atomic model in question, which has spherical symmetry, Gauss's Law allows for the straightforward determination of the electric field produced by the spherical volume of charge, laying the groundwork for establishing the relationship between polarizability and electric field.
Polarization Density
Polarization density, represented by the symbol \( \vec{P} \), measures the electric dipole moment per unit volume of a dielectric material. It results from the response of a material to an external electric field, which aligns the microscopic dipole moments within the material.
In the context of the atomic polarizability, the polarization density is directly proportional to the electric field \( \vec{E} \) the material is subjected to, according to the relationship \( \vec{P} = \alpha \vec{E} \), where \( \alpha \) is the atomic polarizability. This concept helps to understand how materials influence an electric field, as the polarization density relates to the ability of the material to polarize in response to that field and thus, changes the effective electric field within the material.
The polarization density plays a pivotal role in the electric displacement field and is significant in the analysis of dielectric materials, where the ability of a material to resist an electric field is crucial for applications in capacitors and insulators.
In the context of the atomic polarizability, the polarization density is directly proportional to the electric field \( \vec{E} \) the material is subjected to, according to the relationship \( \vec{P} = \alpha \vec{E} \), where \( \alpha \) is the atomic polarizability. This concept helps to understand how materials influence an electric field, as the polarization density relates to the ability of the material to polarize in response to that field and thus, changes the effective electric field within the material.
The polarization density plays a pivotal role in the electric displacement field and is significant in the analysis of dielectric materials, where the ability of a material to resist an electric field is crucial for applications in capacitors and insulators.