Circular Capacitor — Capacitance and Breakdown-Limited Energy: 04-BS-9, December 2013, Q5

Question 5: Circular Capacitor — Capacitance and Breakdown-Limited Energy (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
Plate radius $r$	$5$ cm $=0.05$ m
Plate separation $d$	$1$ mm $=1\times10^{-3}$ m
Relative permittivity $\varepsilon_r$	$2.5$
Breakdown field $E_{\text{bd}}$	$10^7$ V/m

Find. (i) capacitance $C$; (ii) the maximum energy that can be stored before dielectric breakdown.

Circular parallel-plate capacitor, radius r, gap d filled with dielectric εr; breakdown field 10⁷ V/m.

Approach. Compute $C$ from the parallel-plate formula, find the maximum voltage from the breakdown field ($V_{\max}=E_{\text{bd}}d$), then evaluate the stored energy $U=\tfrac12 CV^2$ at that maximum voltage.

Part (i): Capacitance. Plate area $A=\pi r^2=\pi(0.05)^2=7.854\times10^{-3}\ \text{m}^2$: $$C=\frac{\varepsilon_0\varepsilon_r A}{d}=\frac{(8.85\times10^{-12})(2.5)(7.854\times10^{-3})}{1\times10^{-3}}$$ $$C=\boxed{173.8\ \text{pF}}$$
Maximum safe voltage. Breakdown occurs when the uniform field $E=V/d$ reaches $E_{\text{bd}}$: $$V_{\max}=E_{\text{bd}}\,d=(10^7)(1\times10^{-3})$$ $$V_{\max}=\boxed{10\,000\ \text{V}}$$
Part (ii): Maximum stored energy. $$U_{\max}=\tfrac12 C V_{\max}^2=\tfrac12(173.8\times10^{-12})(10\,000)^2$$ $$U_{\max}=\boxed{8.69\ \text{mJ}}$$

Quantity	Result
$C$	$173.8$ pF
$V_{\max}$	$10\,000$ V
$U_{\max}$	$8.69$ mJ

Concept & Theory

Topic. Parallel-plate capacitance and dielectric-breakdown-limited stored energy.

Key relations. $C=\varepsilon_0\varepsilon_r A/d$; uniform field between plates $E=V/d$; stored energy $U=\tfrac12 CV^2=\tfrac12 QV$.

Why this works. For plate dimensions much larger than the gap (here $r\gg d$, so fringing is negligible), the field between the plates is uniform, so breakdown occurs everywhere at once exactly when $V/d$ reaches the dielectric's rated breakdown field — that voltage is therefore the largest voltage (and hence energy) the capacitor can safely hold.

Common pitfall. Using the breakdown field directly in the energy formula without first converting it to a voltage via $V_{\max}=E_{\text{bd}}d$; forgetting the factor of $\varepsilon_r$ in $C$.

Source. Sadiku, Elements of Electromagnetics, Ch. 5 (Dielectrics, Boundary Conditions, Capacitance).

Question 5 of 8: Circular Capacitor — Capacitance and Breakdown-Limited Energy

Question 5: Circular Capacitor — Capacitance and Breakdown-Limited Energy (20 marks)