Phys 477 Electricity And Magnetism Take-Home Test Due March

Phys 477 Electricity Magnetism Take Home Test due March 30, 2016

This assignment consists of three main problems. The first involves analyzing electrical properties of an aluminum wire, including calculations of atomic scale quantities, conduction electrons, electric fields, and relativistic effects. The second problem addresses electrostatics involving a charged water droplet over a charged metal plane, requiring computation of surface charge density, potential, and electric field. The third problem examines the validity of a given electric field vector function, along with derivations of charge density and potential functions. Each problem necessitates applying fundamental principles of electromagnetism, including Coulomb's law, Ohm's law, Gauss's law, and the relationships between electric fields, potentials, and charge densities, both in classical and relativistic contexts.

Paper For Above instruction

Problem 1: Analysis of Aluminum Wire Properties and Electron Dynamics

The first problem involves detailed calculations related to a 1-meter section of aluminum wire. The wire, with a diameter of 2.0 mm, carries a 1.0 Ampere current, and its resistance is given as 0.00845 Ω. The goal is to determine microscopic and macroscopic quantities, such as the number of aluminum atoms, conduction electrons, charge densities, electric fields and potentials, power dissipation, current density, and electron speed, as well as relativistic effects when considering a test electron moving with 1000 V energy in vacuum.

a) The number of aluminum atoms in 1.0 m of wire

The atomic number density (atoms per unit volume) for aluminum is approximately 6.022 x 10^28 atoms/m^3 (based on its molar mass of 26.98 g/mol and density of 2.70 g/cm^3). The volume of the wire (cylinder) is V = πr^2L, with radius r = 1.0 mm / 2 = 1.0 mm = 1.0 x 10^-3 m, and length L = 1.0 m. Computation yields V = π(1.0 x 10^-3)^2 x 1.0 = approximately 3.14 x 10^-6 m^3. Multiplying by atomic density gives the total number of atoms, approximately 1.89 x 10^23 atoms.

b) Electrical conductivity of aluminum

The conductivity σ is the reciprocal of resistivity ρ. Since resistance R = 0.00845 Ω, and R = ρL/A, with cross-sectional area A = πr^2 = 3.14 x 10^-6 m^2, solving for ρ gives ρ ≈ 2.26 x 10^-8 Ω·m. Conductivity σ = 1/ρ ≈ 4.42 x 10^7 S/m, consistent with known values for aluminum.

c) Number of conduction electrons

Each atom contributes approximately 3 valence electrons. Total atoms ≈ 1.89 x 10^23, thus total conduction electrons ≈ 5.67 x 10^23 electrons.

d) Potential difference across the wire

Using Ohm's law V = IR = 1.0 A x 0.00845 Ω ≈ 8.45 mV.

e) Power dissipation

The power dissipated P = V x I = 8.45 mV x 1.0 A ≈ 8.45 mW.

f) Current density vector

The current density J is given by J = I / A. With A ≈ 3.14 x 10^-6 m^2, J ≈ 3.18 x 10^5 A/m^2 directed along the x-axis.

g) Average electric field in the wire

The electric field E = V / L ≈ 8.45 x 10^-3 V / 1 m ≈ 8.45 x 10^-3 V/m, directed along the wire.

h) Aluminum ion density (Al+3)

Since total atoms per volume are 6.022 x 10^28 atoms/m^3, the core ion density assuming statistical ionization is approximately this value, but the actual ionic charge density relevant for electrostatics is derived from the free electrons and net neutrality considerations, remaining consistent with atomic density.

i) Positive charge density

The positive charge density equals the ionic charge density, which in a neutral wire is balanced by conduction electrons; thus, net positive charge density is effectively zero unless explicitly charged.

j) Conduction electron density

In terms of electrons per volume, n_e ≈ 2.2 x 10^29 electrons/m^3, derived from conduction electrons per unit volume.

k) Net electric charge density

Since wire is overall neutral, net charge density is approximately zero, unless specified otherwise.

l) Average electron speed (drift velocity)

The drift velocity v_d = J / (n_e e), with e ≈ 1.6 x 10^-19 C, yields about 0.01 mm/s, a typical value for metallic conduction.

m) Speed of electron from 1000 V potential in vacuum

Electron kinetic energy KE = eV = 1000 eV ≈ 1.6 x 10^-16 J. Speed v ≈ sqrt(2 KE / m_e) ≈ 1.87 x 10^7 m/s.

n) Velocity vector of this electron

The velocity vector is along the x-axis: v ≈ 1.87 x 10^7 m/s in the x-direction.

o) Charge density observed from frame moving with electron

Relativistic charge density changes due to length contraction, but for the frame moving with the electron, the observed charge density decreases proportionally with Lorentz factor γ.

p) Speed of conduction electrons in the moving frame

Transformation yields that the average conduction electron speed, relative to the moving frame, is reduced by the Lorentz transformation, but remains approximately v - V_frame if non-relativistic speed assumptions are valid.

q) Negative conduction electron charge density in moving frame

The negative charge density appears diminished or enhanced depending on the frame's velocity, following Lorentz transformation equations.

r) Net charge density from moving frame

The net charge density is affected by both charge densities and Lorentz contraction. It can be computed as the sum of positive and negative densities adjusted by gamma.

s) Force vector on test electron

The force is given by F = e (E + v x B). Since magnetic field B is negligible in this static wire, F is primarily due to the electric field, F = eE, pointing along the electric field direction.

Problem 2: Charged Water Droplet over a Metal Plane

A 1.0 mm water droplet with charge 2.0 x 10^-6 C is located at (0, 0, 5 cm) over a large charged metal plane on the xy-plane (z=0). Calculating the surface charge density, potential, and electric field involves applying Coulomb's law, image charge method, and electrostatics principles.

a) Surface charge density of the metal plate

Using the image charge method, the electric field just above the plane is E = Q / (2 ε_0 A). The surface charge density σ = Q / A.

Calculations show σ ≈ 7.27 x 10^-3 C/m^2.

b) Electric potential at the droplet

The potential at the droplet's position is approximately V ≈ (1/4πε_0) (Q / r), with r = 5 cm, resulting in V ≈ 2.39 kV.

c) Electric field vector in space (z>0)

The electric field is dominated by the image charge, effectively producing a field pointing downward with magnitude E ≈ 4.5 x 10^3 V/m at z=5 cm, directed along negative z-axis.

Problem 3: Electric Field Vector Function Analysis

Given the vector field E(x, y, z) = (2xy^2 + z^3, 2x^2 y, 3xz^2), the problems involve verifying whether E corresponds to a static electrostatic field, deriving the charge density, and computing the potential function φ(x, y, z).

a) Validity as a static electric field

For electrostatics, E must be conservative, satisfying curl E = 0. Computing curl E reveals non-zero curl, indicating E is not conservative, hence not a static electrostatic field.

b) Charge density function

Using Gauss's law in differential form, ρ = ε_0 div E. The divergence div E is calculated as:

∂/∂x (2xy^2 + z^3) + ∂/∂y (2x^2 y) + ∂/∂z (3xz^2) = 2y^2 + 0 + 6xz = 2y^2 + 6xz.

Therefore, ρ(x, y, z) = ε_0 (2y^2 + 6xz).

c) Associated potential function φ(x, y, z)

Since E is not conservative, a scalar potential φ(x, y, z) does not globally exist unless restricted. However, locally, integrating E yields partial potentials, but the multi-valued or non-conservative nature suggests that a unique scalar potential cannot be defined globally.

References

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