Minimum Plate Area for a Capacitor at the Air Breakdown Limit: 04-BS-9, December 2016, Q3

Question 3: Minimum Plate Area for a Capacitor at the Air 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
Plate separation $d$	$0.5$ mm $=5\times10^{-4}$ m
Max allowable field $E_{\max}$	$10^6$ V/m
Stored energy $U$	$1$ J

Find. The minimum plate area $A$ that keeps the field at or below $E_{\max}$ while storing $U$.

Parallel-plate air capacitor; the field between the plates must not exceed the air breakdown limit $E_{\max}$.

Approach. The maximum field fixes the maximum voltage $V_{\max}=E_{\max}d$; express the stored energy in terms of that voltage and the (unknown) capacitance $U=\tfrac12 C V_{\max}^2$, substitute $C=\varepsilon_0 A/d$, and solve for $A$ — the SMALLEST area that can hold $1$ J without exceeding the field limit.

Express stored energy at the field limit in terms of area. At the maximum allowable field, $V_{\max}=E_{\max}d$ and $C=\varepsilon_0 A/d$, so $$U=\frac12CV_{\max}^2=\frac12\left(\frac{\varepsilon_0A}{d}\right)(E_{\max}d)^2=\frac12\varepsilon_0AdE_{\max}^2$$
Solve for the minimum area. Smaller $A$ would force a higher field (for the same $U$) than $E_{\max}$ allows, so this is the minimum: $$A=\frac{2U}{\varepsilon_0dE_{\max}^2}=\frac{2(1)}{(8.85\times10^{-12})(5\times10^{-4})(10^6)^2}$$ $$A=\boxed{452.0\ \text{m}^2}$$

Quantity	Result
Minimum plate area $A$	$452.0$ m$^2$

Check: a $452\ \text{m}^2$ plate (a square of side $\approx21.3$ m) is enormous for a real capacitor — this follows directly from the given data (storing a full joule at a modest breakdown field over a sub-millimetre gap is intrinsically a large-area problem), not from an arithmetic slip; the numbers are taken exactly as printed.

Concept & Theory

Topic. Parallel-plate capacitance, dielectric breakdown limit, and stored electrostatic energy.

Key relations. $C=\varepsilon_0A/d$; $U=\tfrac12CV^2$; breakdown constraint $V\le E_{\max}d$.

Why this works. For a fixed gap $d$ and fixed stored energy $U$, increasing the plate area lets the same energy be stored at a lower voltage (hence lower field) because capacitance grows with area; the MINIMUM area is exactly the one that reaches, but does not exceed, the breakdown field — any smaller area would require a higher voltage than the dielectric can sustain.

Common pitfall. Solving for area using an assumed voltage or charge instead of eliminating them via the breakdown-field constraint; or using $U=\tfrac12QV$ with an unstated $Q$ rather than the two knowns actually given ($U$ and $E_{\max}$).

Source. Sadiku, Elements of Electromagnetics, Ch. 5 (capacitance, dielectric strength, energy in a capacitor).

Question 3 of 8: Minimum Plate Area for a Capacitor at the Air Breakdown Limit

Question 3: Minimum Plate Area for a Capacitor at the Air Breakdown Limit (20 marks)