Minimum Capacitor Plate Radius for a Target Stored Energy: 04-BS-9, December 2015, Q3

Question 3: Minimum Capacitor Plate Radius for a Target Stored 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 separation $d$	$0.5$ mm $=5\times10^{-4}$ m
Relative permittivity $\varepsilon_r$	$2.5$
Maximum field in dielectric $E_{\max}$	$5\times10^{6}$ V/m
Target stored energy $U$	$1$ J

Find. The smallest plate radius $R$ that can store $1$ J without exceeding $E_{\max}$.

Circular parallel-plate capacitor with dielectric filling the gap; the field between the plates is uniform when fringing is neglected.

Approach. The uniform field between large parallel plates gives a uniform energy density $u=\tfrac12\varepsilon E^2$; set $E=E_{\max}$ (the largest field the dielectric tolerates) and solve for the smallest volume — hence radius — that stores $1$ J at that density.

Permittivity of the dielectric. $$\varepsilon=\varepsilon_r\varepsilon_0=(2.5)(8.85\times10^{-12})$$ $$\varepsilon=\boxed{2.2125\times10^{-11}\ \text{F/m}}$$
Energy density at the breakdown field. $$u=\frac12\varepsilon E_{\max}^2=\frac12(2.2125\times10^{-11})(5\times10^{6})^2$$ $$u=\boxed{276.6\ \text{J/m}^3}$$
Volume needed to store 1 J at that density. $$\text{Vol}=\frac{U}{u}=\frac{1}{276.6}$$ $$\text{Vol}=\boxed{3.616\times10^{-3}\ \text{m}^3}$$
Solve for the plate radius, $\text{Vol}=\pi R^2 d$. $$R=\sqrt{\frac{\text{Vol}}{\pi d}}=\sqrt{\frac{3.616\times10^{-3}}{\pi(5\times10^{-4})}}$$ $$R=\boxed{1.517\ \text{m}}$$

Quantity	Result
$\varepsilon$	$2.2125\times10^{-11}$ F/m
Energy density $u$ at $E_{\max}$	$276.6$ J/m$^3$
Required volume	$3.616\times10^{-3}$ m$^3$
Smallest plate radius $R$	$1.517$ m

Check: a $\sim$1.5 m plate radius for a $0.5$ mm gap is physically large, but it follows directly from the low energy density ($\approx277$ J/m$^3$) permitted by this dielectric/breakdown-field combination — the arithmetic is not an artifact.

Concept & Theory

Topic. Electrostatic energy density and stored energy in a dielectric-filled parallel-plate capacitor.

Key relations. Energy density $u=\tfrac12\varepsilon E^2$; total stored energy $U=u\cdot\text{Vol}=\tfrac12\varepsilon E^2(\pi R^2 d)$.

Why this works. Neglecting fringing, the field between the plates is uniform, so the total energy is simply the (uniform) energy density times the volume it fills. Fixing $E$ at its maximum allowed value gives the highest energy density the dielectric can sustain, so the volume — and hence radius, at fixed gap $d$ — found this way is the smallest one that still reaches the target energy without breaching $E_{\max}$; any smaller plate would either exceed the breakdown field or store less than 1 J.

Common pitfall. Solving via capacitance and voltage ($U=\tfrac12 CV^2$) without recognizing $V=E_{\max}d$ is fixed by the breakdown condition, which is algebraically equivalent but easy to mis-set-up; or multiplying by $\varepsilon_0/\varepsilon_r$ instead of $\varepsilon_r\varepsilon_0$.

Source. Sadiku, Elements of Electromagnetics, Ch. 5 (capacitance, energy in an electrostatic field).

Question 3 of 8: Minimum Capacitor Plate Radius for a Target Stored Energy

Question 3: Minimum Capacitor Plate Radius for a Target Stored Energy (20 marks)