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04-BS-9 · December 2014

Question 3 of 8: Capacitance of a Coaxial Transmission Line

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — December 2014 — 04-BS-9 Basic Electromagnetics. Three-hour, closed-book exam (approved Casio/Sharp calculator only). Aids given: $\epsilon_0=8.85\times10^{-12}$ 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 — electrostatics, Gauss's law, capacitance, magnetostatics, Faraday's law, plane waves; Young & Freedman, University Physics with Modern Physics — induced EMF, AC quantities, torque on a current loop.

Question 3: Capacitance of a Coaxial Transmission Line (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.

QuantityValue
Inner cylinder radius $a$$1\times10^{-3}$ m
Outer cylinder radius $b$$2\times10^{-3}$ m
Relative permittivity $\varepsilon_r$$2.25$
Section length $L$$2$ m

Find. The capacitance $C$ of the 2 m section.

a b εr = 2.25, L = 2 m
Coaxial cylinders, inner radius a = 1 mm, outer radius b = 2 mm, dielectric-filled gap.

Approach. Apply the standard coaxial-capacitor formula, derived from Gauss's law on a cylindrical Gaussian surface and integrating $E$ from $a$ to $b$. The geometry enters only through the radius ratio $b/a=2$, so this is a single-step application of the formula once $\varepsilon_r$ and $L$ are substituted.

  1. Capacitance. $$C=\frac{2\pi\varepsilon_r\varepsilon_0 L}{\ln(b/a)}=\frac{2\pi(2.25)(8.85\times10^{-12})(2)}{\ln(2)}$$ $$C=\boxed{361.0\ \text{pF}}$$ Per unit length this is $C/L=180.5$ pF/m, a purely geometric/material property of the cross-section, so $C$ scales linearly with the section length.
QuantityResult
$C$ (2 m section)$361.0$ pF