Maximum Stored Energy in a Coaxial Line at the Dielectric Breakdown Limit: 04-BS-9, December 2017, Q3

Question 3: Maximum Stored Energy in a Coaxial Line at the Dielectric Breakdown Limit (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
Inner radius $a$	2 mm $=2\times10^{-3}$ m
Outer radius $b$	4 mm $=4\times10^{-3}$ m
Relative permittivity $\varepsilon_r$	2.25
Maximum allowed field $E_{\max}$	$10^7$ V/m (occurs at $r=a$, where $E$ peaks)

Find. The maximum energy $U$ that can be stored in a 1 m length without exceeding $E_{\max}$.

Coaxial line filled with dielectric $\varepsilon_r=2.25$; the radial field is largest at the inner conductor, so $E_{\max}$ constrains the design there.

Approach. The peak field occurs at $r=a$: $E_{\max}=\lambda/(2\pi\varepsilon a)$, which fixes the line charge $\lambda$. Integrate the energy density $\tfrac12\varepsilon E(r)^2$ over the annulus (equivalently, use $U=\tfrac12CV^2$) to get the stored energy per unit length.

Line charge density at the field limit. With $\varepsilon=\varepsilon_r\varepsilon_0$: $$\lambda=2\pi\varepsilon aE_{\max}=2\pi(2.25\times8.85\times10^{-12})(2\times10^{-3})(10^7)$$ $$\lambda=\boxed{2.502\times10^{-6}\ \text{C/m}}$$
Voltage across the line and capacitance per unit length. $$V=\frac{\lambda}{2\pi\varepsilon}\ln\!\frac{b}{a}=E_{\max}\,a\,\ln\!\frac{b}{a}=(10^7)(0.002)\ln2=\boxed{1.386\times10^{4}\ \text{V}}$$ $$C'=\frac{2\pi\varepsilon}{\ln(b/a)}=\frac{2\pi(1.9913\times10^{-11})}{0.69315}=1.8047\times10^{-10}\ \text{F/m}$$
Stored energy in 1 m. $$U=\tfrac12C'V^2=\tfrac12(1.8047\times10^{-10})(1.386\times10^{4})^2$$ $$\boxed{U=1.734\times10^{-2}\ \text{J}=17.34\ \text{mJ}}$$

Quantity	Result
Line charge $\lambda$	$2.502\times10^{-6}$ C/m
Voltage $V$	$1.386\times10^4$ V
Max stored energy (1 m length)	$1.734\times10^{-2}$ J (17.34 mJ)

Concept & Theory

Topic. Energy storage in a coaxial capacitor limited by dielectric breakdown field.

Key relations. $E(r)=\lambda/(2\pi\varepsilon r)$ (max at $r=a$); $C'=2\pi\varepsilon/\ln(b/a)$; $U=\tfrac12C'V^2$.

Why this works. Because the field in a coax falls as $1/r$, the inner conductor surface always sees the highest field for a given charge — so the breakdown constraint is applied there, which uniquely fixes $\lambda$ (and hence $V$ and $U$) for a target maximum field, independent of what happens at larger $r$.

Common pitfall. Forgetting the relative permittivity $\varepsilon_r$ (using $\varepsilon_0$ alone understates $\lambda$, $V$, and $U$ all by factors of $\varepsilon_r$ and $\varepsilon_r^2$ respectively where they appear), or applying $E_{\max}$ at $r=b$ instead of the true peak at $r=a$.

Source. Sadiku, Elements of Electromagnetics, Ch. 5 (coaxial capacitance, energy storage, dielectric strength).

Question 3 of 8: Maximum Stored Energy in a Coaxial Line at the Dielectric Breakdown Limit

Question 3: Maximum Stored Energy in a Coaxial Line at the Dielectric Breakdown Limit (20 marks)