Problem 5.169 — Rayleigh Range of Gaussian Beam

Problem Statement

Determine the electric field for the configuration described: Determine the electric field for the configuration described: Find the Rayleigh range $z_R$ for a Gaussian beam with $w_0 = 1.0$ mm at $\lambda = 1064$ nm. $$z_R = \frac{\pi w_0^2}{\lambda} = \frac{\pi\times(1.0\times10^{-3})^2}{1064\times10^{-9}} = \frac{\pi\times10^{-6}}{1.064\times10^{-6}} = \fra

Given Information

• See problem statement for all given quantities.

Physical Concepts & Formulas

Gauss’s law relates the electric flux through any closed surface to the total enclosed charge. It is one of Maxwell’s four equations and is especially powerful when the charge distribution has spherical, cylindrical, or planar symmetry, because the flux integral then simplifies to $E \cdot A = Q_\text{enc}/\varepsilon_0$.

• $\oint \mathbf{E}\cdot d\mathbf{A} = Q_{\text{enc}}/\varepsilon_0$ — Gauss’s law
• $E = Q/(4\pi\varepsilon_0 r^2)$ — field outside a sphere
• $E = \sigma/\varepsilon_0$ — field between infinite parallel plates

Step-by-Step Solution

Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Worked Calculation

$$z_R = \frac{\pi w_0^2}{\lambda} = \frac{\pi\times(1.0\times10^{-3})^2}{1064\times10^{-9}} = \frac{\pi\times10^{-6}}{1.064\times10^{-6}} = \fra

Given Information

• Geometry and charge distribution as given in the problem
• Permittivity of free space $\varepsilon_0 = 8.85\times10^{-12}\,\text{C}^2\text{N}^{-1}\text{m}^{-2}$
• Coulomb constant $k = 1/(4\pi\varepsilon_0) = 9\times10^9\,\text{N m}^2\text{C}^{-2}$

Physical Concepts & Formulas

The electric field $\vec{E}$ at any point is the force per unit positive test charge placed at that point. For symmetric charge distributions, Gauss’s Law — $\oint \vec{E}\cdot d\vec{A} = Q_{\text{enc}}/\varepsilon_0$ — is the most powerful tool: choose a Gaussian surface matching the symmetry, and the flux integral reduces to $E \times A$. For arbitrary distributions, direct integration using Coulomb’s law with superposition gives $\vec{E} = \int k\,dq\,\hat{r}/r^2$. The direction of $\vec{E}$ always points away from positive charges and toward negative charges.

• $\vec{E} = \dfrac{kq}{r^2}\hat{r}$ — Coulomb’s law for a point charge
• $\oint \vec{E}\cdot d\vec{A} = \dfrac{Q_{\text{enc}}}{\varepsilon_0}$ — Gauss’s Law
• For a uniformly charged sphere (outside): $E = \dfrac{kQ}{r^2}$
• For an infinite line charge: $E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$
• For an infinite plane: $E = \dfrac{\sigma}{2\varepsilon_0}$

Step-by-Step Solution

Step 1 — Identify symmetry: Spherical, cylindrical, or planar symmetry determines which Gaussian surface to use.

Step 2 — Choose Gaussian surface: Select a surface where $\vec{E}$ is either parallel or perpendicular to $d\vec{A}$ everywhere, and where $|E|$ is constant on the surface.

Step 3 — Apply Gauss’s Law:

$$

$$

Step 4 — Solve for $E$:

$$

$$

Step 5 — Direction: By symmetry, $\vec{E}$ points radially outward (for positive charge) or inward (for negative).

Worked Calculation

Substituting all values with units:

For a sphere of radius $R$, charge $Q$, at exterior point $r > R$:

$$

Answer

$$\boxed{E = \dfrac{kQ}{r^2}\quad(r > R)}$$

Physical Interpretation

The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.