Question 1 of 8: Concentric Spherical Capacitor — Breakdown-Limited Charge
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2013 — 04-BS-9 Basic Electromagnetics. Three-hour, closed-book exam (approved Casio/Sharp calculator only). Aids given: $\varepsilon_0=8.85\times10^{-12}$ F/m, $\mu_0=4\pi\times10^{-7}$ H/m, $e=1.6\times10^{-19}$ C. Format: eight questions offered; any five constitute a complete paper and only the first five appearing in the answer book are marked. All eight are solved below for completeness.
Reference texts: Sadiku, Elements of Electromagnetics / Hayt & Buck, Engineering Electromagnetics — electrostatics, Gauss's law, capacitance, magnetostatics, Faraday's law, Maxwell's equations and displacement current; Young & Freedman, University Physics with Modern Physics — dipole fields, torque on a current loop, electromagnetic wave propagation in dielectrics.
2 mm $\Rightarrow$ dielectric outer radius $b=12$ mm
Outer shell inner radius $d$
15 mm (air gap from $b$ to $d$)
Dielectric relative permittivity $\varepsilon_r$
2.25
Dielectric breakdown field
$10^7$ V/m
Air breakdown field
$10^6$ V/m
Find. The largest charge $Q$ the capacitor can hold before either the dielectric layer or the air gap electrically breaks down.
Concentric spherical shells: dielectric sleeve ($a$ to $b$) then air gap ($b$ to $d$) before the outer shell.
Approach. For a spherical charge distribution, Gauss's law gives $D(r)=Q/(4\pi r^2)$, continuous across the dielectric/air interface (no free charge there); the local field is $E(r)=D(r)/\varepsilon$, which is largest at the smallest radius of each region. Compute the charge that would just reach breakdown in each region separately, then take the smaller value — that is the true (lowest) upper bound.
Field profile from Gauss's law. A concentric spherical Gaussian surface of radius $r$ encloses charge $Q$ regardless of the dielectric, so
$$D(r)=\frac{Q}{4\pi r^2}, \qquad E(r)=\frac{D(r)}{\varepsilon(r)}=\frac{Q}{4\pi\varepsilon(r) r^2}.$$
In each region $E(r)$ decreases with $r$, so the breakdown risk is highest at the inner radius of that region: $r=a$ in the dielectric, $r=b$ at the start of the air gap.
Charge limit set by the dielectric ($a\le r\le b$). Breakdown first occurs at $r=a=0.010$ m:
$$Q_{\text{diel}}=E_{\text{bd,diel}}\cdot 4\pi\varepsilon_0\varepsilon_r a^2 = (10^7)(4\pi)(8.85\times10^{-12})(2.25)(0.010)^2$$
$$Q_{\text{diel}} \approx \boxed{2.50\times10^{-7}\ \text{C} = 250\ \text{nC}}$$
Charge limit set by the air gap ($b\le r\le d$). Breakdown first occurs at $r=b=0.012$ m:
$$Q_{\text{air}}=E_{\text{bd,air}}\cdot 4\pi\varepsilon_0 b^2 = (10^6)(4\pi)(8.85\times10^{-12})(0.012)^2$$
$$Q_{\text{air}} \approx \boxed{1.60\times10^{-8}\ \text{C} = 16.0\ \text{nC}}$$
Governing limit. Since $Q_{\text{air}}<Q_{\text{diel}}$, the air gap breaks down first as charge is added. The lowest upper bound on stored charge is therefore
$$Q_{\max}=\boxed{16.0\ \text{nC}}.$$