Maximum Stored Energy in a Parallel-Plate Capacitor with a Partial Dielectric Layer: 04-BS-9, May 2018, Q7

Question 7: Maximum Stored Energy in a Parallel-Plate Capacitor with a Partial Dielectric Layer (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 area $A$	50 cm$^2=5\times10^{-3}$ m$^2$
Total plate separation $d$	1 mm $=1\times10^{-3}$ m
Dielectric thickness $t_d$ / $\varepsilon_r$	$0.75\times10^{-3}$ m / 2.5 (glued to one plate)
Remaining air-gap thickness $t_{\text{air}}$	$d-t_d=0.25\times10^{-3}$ m
Max allowed field, air	$1\times10^6$ V/m
Max allowed field, dielectric	$1\times10^7$ V/m

Find. The maximum electric energy that can be stored without exceeding either material's breakdown field.

Series stack: air gap (upper) plus a 0.75 mm dielectric slab (εr=2.5) glued to the lower plate. The same D field threads both layers, so whichever material's own breakdown field is reached first sets the overall energy limit.

Approach. The two layers are in series, so the SAME displacement field $D$ threads both (no free charge sits at the dielectric–air interface): $E_{\text{air}}=D/\varepsilon_0$ and $E_{\text{diel}}=D/(\varepsilon_r\varepsilon_0)=E_{\text{air}}/\varepsilon_r$. Since $\varepsilon_r>1$, $E_{\text{air}}$ is always the larger of the two for a given $D$ — check which material's OWN limit is reached first as $D$ rises from zero.

Identify the binding constraint. If the AIR gap were driven to its own limit $E_{\text{air}}=1\times10^6$ V/m, the dielectric would only be at $E_{\text{air}}/\varepsilon_r=4\times10^5$ V/m — well under its own $1\times10^7$ V/m limit. Conversely, driving the DIELECTRIC to its own limit would require $E_{\text{air}}=\varepsilon_r\times10^7=2.5\times10^7$ V/m, far above air's $1\times10^6$ V/m limit. So the air gap governs: $$\boxed{E_{\text{air}}=1.000\times10^6\ \text{V/m (binding constraint)}}$$
Displacement field and the dielectric's field at this condition. $$D=\varepsilon_0E_{\text{air}}=(8.85\times10^{-12})(1\times10^6)=8.850\times10^{-6}\ \text{C/m}^2$$ $$E_{\text{diel}}=\frac{D}{\varepsilon_r\varepsilon_0}=\frac{E_{\text{air}}}{\varepsilon_r}=\frac{1\times10^6}{2.5}=4.000\times10^5\ \text{V/m}$$
Voltage across the whole capacitor. $$V=E_{\text{air}}t_{\text{air}}+E_{\text{diel}}t_d=(1\times10^6)(2.5\times10^{-4})+(4\times10^5)(7.5\times10^{-4})=250+300$$ $$V=550.0\ \text{V}$$
Maximum stored energy. Summing the energy density $u=\tfrac12DE$ over each layer's volume: $$U=\tfrac12D\,E_{\text{air}}(A\,t_{\text{air}})+\tfrac12D\,E_{\text{diel}}(A\,t_d)=5.531\times10^{-6}+6.638\times10^{-6}$$ $$\boxed{U=1.217\times10^{-5}\ \text{J}=12.17\ \mu\text{J}}$$

Quantity	Result
Displacement field $D$	$8.850\times10^{-6}$ C/m$^2$
$E_{\text{air}}$ / $E_{\text{diel}}$	$1.000\times10^6$ V/m / $4.000\times10^5$ V/m
Voltage across capacitor $V$	550.0 V
Maximum stored energy $U$	$1.217\times10^{-5}$ J (12.17 μJ)

Concept & Theory

Topic. Energy storage in a series-layered (partially filled) parallel-plate capacitor, limited by dielectric breakdown.

Key relations. $D$ continuous across a charge-free interface; $E=D/\varepsilon$ in each layer; energy density $u=\tfrac12DE=\tfrac12\varepsilon E^2$.

Why this works. Because the same displacement field $D$ threads both layers in series, the LOWER-permittivity material always carries the higher $E$ for a given $D$ — which is exactly why the air gap, not the dielectric, is the practical breakdown-limiting layer here even though the dielectric's own quoted breakdown strength is ten times higher.

Common pitfall. Assuming the dielectric governs simply because its quoted field limit is the bigger number, without checking the $D$-continuity ratio; or applying each material's OWN maximum field independently (double-counting headroom the series physics doesn't allow).

Source. Sadiku, Elements of Electromagnetics, Ch. 5 (dielectrics, boundary conditions, capacitance, energy storage).

Question 7 of 8: Maximum Stored Energy in a Parallel-Plate Capacitor with a Partial Dielectric Layer

Question 7: Maximum Stored Energy in a Parallel-Plate Capacitor with a Partial Dielectric Layer (20 marks)