E Field and Potential Between Two Charged Layers: 04-BS-9, May 2014, Q1

Question 1: E Field and Potential Between Two Charged Layers (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
Layer thickness $t$ (each)	$1\times10^{-6}$ m
Gap between layers	$d=1\times10^{-6}$ m
Charge densities	$\rho_A=+1\ \text{C/m}^3$, $\rho_B=-1\ \text{C/m}^3$
Relative permittivity $\varepsilon_r$ (everywhere)	$10$

Find. (i) $E$ in the neutral gap; (ii) the potential difference between the two layers' outside (far) faces.

Two thin charged layers (+ρ, −ρ), each thickness t, separated by a neutral gap d. E points from the + layer toward the − layer.

Approach. Treat the whole span as one dielectric ($\varepsilon=\varepsilon_r\varepsilon_0$) with a free-charge density that is nonzero only inside the two layers. Integrate $dD/dx=\rho_{\text{free}}$ across the structure (with $D=0$ far outside, since the two layers carry equal and opposite total charge), then get $E=D/\varepsilon$ and $V=\int E\,dx$.

Part (i): D and E in the gap. Starting from $D=0$ outside layer A, integrating $\rho_A=1\ \text{C/m}^3$ across the layer's own thickness $t$ gives the areal density carried into the gap, where $D$ stays constant (no charge there): $$D_{\text{gap}}=\rho_A\,t=(1)(1\times10^{-6})$$ $$D_{\text{gap}}=1\times10^{-6}\ \text{C/m}^2$$ $$E=\frac{D_{\text{gap}}}{\varepsilon_r\varepsilon_0}=\frac{1\times10^{-6}}{(10)(8.85\times10^{-12})}$$ $$E=\boxed{1.130\times10^{4}\ \text{V/m, directed from the +layer toward the −layer}}$$
Part (ii): potential across the whole span. $D(x)$ rises linearly from 0 to $D_{\text{gap}}$ across layer A, stays flat across the gap, then falls linearly back to 0 across layer B (consistent check: it returns to exactly zero, confirming the equal-and-opposite-charge assumption). Integrating $E=D/\varepsilon$ over each region and summing: $$\int D\,dx = \underbrace{\tfrac12\rho_A t^2}_{\text{layer A}}+\underbrace{\rho_A t\,d}_{\text{gap}}+\underbrace{\tfrac12\rho_A t^2}_{\text{layer B}}=2\times10^{-12}\ \text{C/m}$$ $$V=\frac{\int D\,dx}{\varepsilon_r\varepsilon_0}=\frac{2\times10^{-12}}{(10)(8.85\times10^{-12})}$$ $$V=\boxed{2.26\times10^{-2}\ \text{V}=22.6\ \text{mV}}$$

Quantity	Result
$D$ in the gap	$1\times10^{-6}$ C/m²
$E$ in the gap	$1.130\times10^4$ V/m
$V$, outside face to outside face	$22.6$ mV

Concept & Theory

Topic. Gauss's law / $D$-field integration for a pair of finite-thickness charged slabs (a distributed-charge capacitor).

Key relations. $dD/dx=\rho_{\text{free}}$; $E=D/\varepsilon_r\varepsilon_0$; $V=\int E\,dx$.

Why this works. Because the two layers carry equal and opposite total areal charge, $D$ must return to zero once both layers have been traversed — exactly like a parallel-plate capacitor's field vanishing outside its plates. Between the layers, where there is no free charge, $D$ (and hence $E$) stays constant, which is the "neutral region" value asked for.

Common pitfall. Treating the layers as infinitesimally thin sheets and using $E=\sigma/2\varepsilon$ — that formula ignores the internal field variation used to get the total potential drop across the layers themselves, which is needed for part (ii).

Source. Sadiku, Elements of Electromagnetics, Ch. 5 (Gauss's Law and Applications, dielectrics).

Question 1 of 8: E Field and Potential Between Two Charged Layers

Question 1: E Field and Potential Between Two Charged Layers (20 marks)