Coaxial Capacitor — Capacitance and Breakdown Voltage: 04-BS-9, May 2014, Q2

Question 2: Coaxial Capacitor — Capacitance and Breakdown Voltage (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 cylinder radius $a$	$0.5\times10^{-3}$ m (1 mm dia.)
Outer cylinder radius $b$	$3\times10^{-3}$ m (6 mm dia.)
Relative permittivity $\varepsilon_r$	$2.5$
Breakdown field $E_{\text{bd}}$	$10^7$ V/m
Section length $L$	$1$ m

Find. (i) capacitance $C$ of the 1 m section; (ii) maximum applied voltage before breakdown.

Coaxial cylinders, inner radius a, outer radius b, dielectric-filled gap.

Approach. Use the standard coaxial-capacitor formula for $C$, then find $V_{\max}$ from the fact that the field is largest at the inner conductor's surface, $E(a)=V/(a\ln(b/a))$.

Part (i): capacitance. $$C=\frac{2\pi\varepsilon_r\varepsilon_0 L}{\ln(b/a)}=\frac{2\pi(2.5)(8.85\times10^{-12})(1)}{\ln(6)}$$ $$C=\boxed{77.6\ \text{pF}}$$
Part (ii): breakdown-limited voltage. The radial field $E(r)=V/(r\ln(b/a))$ is largest at $r=a$ (the inner surface), so breakdown first occurs there: $$V_{\max}=E_{\text{bd}}\,a\,\ln(b/a)=(10^7)(0.5\times10^{-3})(1.7918)$$ $$V_{\max}=\boxed{8.96\ \text{kV}}$$

Quantity	Result
$C$ (1 m section)	$77.6$ pF
$V_{\max}$	$8.96$ kV

Question 2 of 8: Coaxial Capacitor — Capacitance and Breakdown Voltage

Question 2: Coaxial Capacitor — Capacitance and Breakdown Voltage (20 marks)