Atomic hydrogen contains \(5.5 \times 10^{25}\) atoms \(/ \mathrm{m}^{3}\) at a certain temperature and pressure. When an electric field of \(4 \mathrm{kV} / \mathrm{m}\) is applied, each dipole formed by the electron and positive nucleus has an effective length of \(7.1 \times 10^{-19} \mathrm{~m}\). (a) Find \(P\). (b) Find \(\epsilon_{r}\).

Short Answer

Expert verified
Answer: The electric polarization P is 62.48 pC/m², and the relative permittivity ε_r is approximately 1.

Step by step solution

01

Part (a): Find Electric Polarization P

1. Calculate Dipole Moment (p): The dipole moment can be calculated using the formula p = q * d, where q is the charge of the electron or proton and d is the effective length of each dipole formed. Given: d = 7.1 * 10^-19 m Charge of electron, q = -e = -1.6 * 10^-19 C Charge of proton, q = +e = 1.6 * 10^-19 C Since the charges are equal in magnitude but opposite in direction, we will only consider one of the charges in the calculation. p = q * d p = (1.6 * 10^-19) * (7.1 * 10^-19) p = 11.36 * 10^-38 C m 2. Calculate Electric Polarization P: Electric polarization can be calculated using the formula P = n * p, where n is the number of atoms per unit volume. Given: n = 5.5 * 10^25 atoms/m³ P = n * p P = (5.5 * 10^25) * (11.36 * 10^-38) P = 62.48 * 10^-12 C/m² Thus, the electric polarization P is 62.48 pC/m².
02

Part (b): Find Relative Permittivity εr

1. Calculate Electric Susceptibility χ_e: The electric susceptibility can be calculated using the formula χ_e = P / E, where E is the given electric field. Given: E = 4 kV/m = 4 * 10^3 V/m χ_e = P / E χ_e = (62.48 * 10^-12) / (4 * 10^3) χ_e = 15.62 * 10^-15 2. Calculate Relative Permittivity ε_r: The relative permittivity can be calculated using the formula ε_r = 1 + χ_e. ε_r = 1 + 15.62 * 10^-15 ε_r ≈ 1 Thus, the relative permittivity ε_r for atomic hydrogen is approximately 1.

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Key Concepts

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

Dipole Moment
Imagine a pair of equal and opposite charges separated by a small distance. This arrangement is known as an electric dipole. The dipole moment is a vector quantity that represents the strength and direction of this dipole; it's essentially a measure of the separation of positive and negative charges. It's calculated as the product of the magnitude of one of the charges and the distance between them.

In our case, the atoms of hydrogen get polarized under an electric field, forming diples. Using the formula p = q * d, where q is the charge (of an electron or a proton) and d is the distance between the charges (or the effective length), we can compute the dipole moment for a single hydrogen atom. With a charge of approximately 1.6 * 10-19 C and a distance of 7.1 * 10-19 m, each hydrogen dipole has a moment calculated as 11.36 * 10-38 C*m.

The concept of dipole moment is important as it determines how a dipole aligns with an electric field and the extent to which the material will be polarized.
Electric Field
The electric field is a fundamental concept in electromagnetism, representing the force per unit charge exerted on a positive test charge placed within the field. It is a vector field, meaning it has both a magnitude and a direction at every point in space. For our example, the electric field is provided as 4 kV/m (4 * 103 V/m), which means that a positive charge would feel a force of 4000 newtons for every coulomb of charge.

An electric field is instrumental in creating electric polarization within materials. It causes the charges in atoms and molecules to shift slightly, inducing dipoles. Understanding the nature and strength of the electric field allows us to anticipate the polarization behavior of materials, which is critical in designing electronics, understanding material properties, and studying electrostatic phenomena.
Relative Permittivity
The relative permittivity, often symbolized by εr, is a dimensionless quantity that describes how an electric field affects, and is affected by, a dielectric (insulating) medium. It is also known as the dielectric constant. The relative permittivity of a medium determines how much the medium will reduce the electric field within it compared to the field in a vacuum. It is given by the relation εr = 1 + χe, where χe is the electric susceptibility.

In practical terms, materials with higher relative permittivity can store more electric energy for a given electric field. This concept is essential for understanding how materials influence capacitive behavior in circuits. From the solution provided, for atomic hydrogen under the given conditions, the relative permittivity (εr) is approximately 1, indicating that the material behaves similarly to a vacuum in terms of its ability to permit an electric field.
Electric Susceptibility
The term electric susceptibility, denoted by χe, quantifies how easily a material can be polarized by applying an electric field. It relates the induced polarization P in the material to the electric field E causing it, with the formula P = χe * E. Higher susceptibility indicates a material will become more polarized in response to an electric field.

In the context of our exercise, we calculated the electric susceptibility by dividing the electric polarization by the electric field (χe = P / E). The calculation gives us a small value, suggesting that atomic hydrogen exhibits minimal polarization response under the applied electric field, with its susceptibility essentially being 0 for all practical purposes. This concept helps us understand and quantify the polarizing effects within different materials when subjected to external electric fields, which is especially important in the field of material science and electronic engineering.

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Most popular questions from this chapter

A hollow cylindrical tube with a rectangular cross section has external dimensions of \(0.5 \mathrm{in}\). by 1 in. and a wall thickness of \(0.05 \mathrm{in}\). Assume that the material is brass, for which \(\sigma=1.5 \times 10^{7} \mathrm{~S} / \mathrm{m}\). A current of \(200 \mathrm{~A} \mathrm{dc}\) is flowing down the tube. \((a)\) What voltage drop is present across a \(1 \mathrm{~m}\) length of the tube? (b) Find the voltage drop if the interior of the tube is filled with conducting material for which \(\sigma=1.5 \times 10^{5} \mathrm{~S} / \mathrm{m}\).

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Let the surface \(y=0\) be a perfect conductor in free space. Two uniform infinite line charges of \(30 \mathrm{nC} / \mathrm{m}\) each are located at \(x=0, y=1\), and \(x=0, y=2 \cdot(a)\) Let \(V=0\) at the plane \(y=0\), and find \(V\) at \(P(1,2,0)\). (b) Find \(\mathbf{E}\) at \(P\).

Given the potential field \(V=100 x z /\left(x^{2}+4\right) \mathrm{V}\) in free space: \((a)\) Find \(\mathrm{D}\) at the surface \(z=0 .(b)\) Show that the \(z=0\) surface is an equipotential surface. ( \(c\) ) Assume that the \(z=0\) surface is a conductor and find the total charge on that portion of the conductor defined by \(0

Two identical conducting plates, each having area \(A\), are located at \(z=0\) and \(z=d\). The region between plates is filled with a material having \(z\) -dependent conductivity, \(\sigma(z)=\sigma_{0} e^{-z / d}\), where \(\sigma_{0}\) is a constant. Voltage \(V_{0}\) is applied to the plate at \(z=d ;\) the plate at \(z=0\) is at zero potential. Find, in terms of the given parameters, \((a)\) the resistance of the material; \((b)\) the total current flowing between plates; ( \(c\) ) the electric field intensity \(\mathbf{E}\) within the material.

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