Question 2 of 8: Electric Field at the Surface of a Charged Electron Beam
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2015 — 04-BS-9 Basic Electromagnetics. Three-hour, closed-book exam (approved Casio/Sharp calculator only). Aids given: $\varepsilon_0=8.85\times10^{-12}$ F/m, $\mu_0=4\pi\times10^{-7}$ H/m, $e=1.6\times10^{-19}$ C. Format: eight questions offered; any five constitute a complete paper and only the first five appearing in the answer book are marked. All eight are solved below for completeness.
Reference texts: Sadiku, Elements of Electromagnetics / Hayt & Buck, Engineering Electromagnetics — Coulomb's law and superposition, Gauss's law, Biot–Savart law, magnetic field of current sheets, Faraday's law, capacitance and field energy, mutual inductance; Young & Freedman, University Physics with Modern Physics — light propagation and reflection geometry.
Question 2: Electric Field at the Surface of a Charged Electron Beam (20 marks)
Find. The magnitude and direction of $\vec E$ at the beam's surface ($r=a$).
Cross-section of the electron beam (radius a); the negative space charge draws the field radially inward toward the axis.
Approach. Model the beam as a uniform line of moving charge with linear charge density $\lambda=I/v$ (current equals charge per unit length times drift speed). Apply Gauss's law for an infinite line/cylindrical charge at $r=a$, the surface enclosing the beam's entire charge per unit length.
Linear charge density magnitude.
$$\lambda=\frac{I}{v}=\frac{10^{-6}}{6\times10^7}$$
$$\lambda=\boxed{1.667\times10^{-14}\ \text{C/m}}$$
Electric field at $r=a$ (Gauss's law, cylindrical symmetry).
$$E=\frac{\lambda}{2\pi\varepsilon_0 a}=\frac{1.667\times10^{-14}}{2\pi(8.85\times10^{-12})(10^{-6})}$$
$$E=\boxed{299.7\ \text{V/m}}$$, directed radially inward toward the beam axis — since the carriers are electrons, the beam's net charge is negative, so field lines terminate on it rather than emanate from it.
