Electric Field at the Surface of a Uniformly Charged Electron Beam: 04-BS-9, December 2017, Q2

Question 2: Electric Field at the Surface of a Uniformly Charged Electron Beam (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Beam diameter	1 mm $\Rightarrow$ radius $R=5\times10^{-4}$ m
Electron number density $n$ (uniform)	$1.33\times10^{12}$ /m$^3$

Find. The magnitude and direction of $\vec E$ at $r=R$ (the beam's surface), treating the beam as an infinitely long uniformly charged cylinder.

Electron beam modeled as an infinite uniformly charged cylinder; by symmetry, $E$ is radial and its magnitude depends only on $r$. Because the volume charge is negative, $\vec E$ points radially INWARD.

Approach. Apply Ampere-style cylindrical Gauss's law: for a uniform volume charge density inside radius $R$, $E(r)\cdot2\pi rL=Q_{\text{enc}}/\varepsilon_0$; evaluate at $r=R$ where all of the charge in a length $L$ is enclosed.

Volume charge density of the electron beam. Each electron carries $-e$; with number density $n$: $$\rho_v=-ne=-(1.33\times10^{12})(1.6\times10^{-19})=\boxed{-2.128\times10^{-7}\ \text{C/m}^3}$$
Field at the surface, from cylindrical Gauss's law. Enclosed charge per unit length at $r=R$ is $\rho_v\pi R^2$, so $$E(R)\cdot2\pi RL=\frac{\rho_v\pi R^2L}{\varepsilon_0}\ \Rightarrow\ E(R)=\frac{\rho_vR}{2\varepsilon_0}$$ $$|E(R)|=\frac{(2.128\times10^{-7})(5\times10^{-4})}{2(8.85\times10^{-12})}$$ $$\boxed{E=6.011\ \text{V/m}}$$

Quantity	Result
Volume charge density $\rho_v$	$-2.128\times10^{-7}$ C/m$^3$
Field magnitude at the surface	$6.011$ V/m
Direction	radially INWARD (toward the beam axis, since the enclosed charge is negative)

Concept & Theory

Topic. Gauss's law in cylindrical symmetry for a uniformly charged (negative) beam.

Key relations. $E(r)=\rho_vr/(2\varepsilon_0)$ for $r\le R$ inside a uniformly charged infinite cylinder; the sign of $\rho_v$ sets the field direction.

Why this works. Cylindrical symmetry guarantees $\vec E$ is purely radial and depends only on the enclosed charge per unit length, exactly as spherical symmetry does for Gauss's law with a sphere — the only difference is the $1/r$ (not $1/r^2$) falloff structure and the cylindrical Gaussian surface.

Common pitfall. Treating the given quantity as a charge density directly instead of first converting the electron NUMBER density $n$ to a charge density via $\rho_v=-ne$; or reporting the field as outward, forgetting the beam's net charge is negative.

Source. Sadiku, Elements of Electromagnetics, Ch. 3 (Gauss's law, cylindrical symmetry).

Question 2 of 8: Electric Field at the Surface of a Uniformly Charged Electron Beam

Question 2: Electric Field at the Surface of a Uniformly Charged Electron Beam (20 marks)