Charge Density Distribution That Produces a Given Quadratic Field Inside a Slab: 04-BS-9, May 2017, Q7

Question 7: Charge Density Distribution That Produces a Given Quadratic Field Inside a Slab (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.

Check: the field-expression numerator is read as the standard textbook notation $\rho_0d/\varepsilon_0$, using the reference density $\rho_0=1$ C/m$^3$ given separately in the same sentence. This reading is confirmed self-consistent below: the derived $\rho_v(x)$ evaluates to exactly $\pm\rho_0$ at the slab edges $x=\pm d/2$, matching the given reference value.

Given.

Quantity	Value
Field inside the slab	$E(x)=\dfrac{\rho_0d}{\varepsilon_0}\left[\left(\dfrac{x}{d}\right)^2-\dfrac14\right]$, $\|x\|\le d/2$
Field outside the slab	$E=0$, $\|x\|>d/2$
Slab half-thickness parameter $d$	$10^{-6}$m
Reference density $\rho_0$	$1$ C/m$^3$

Find. The volume charge density $\rho_v(x)$ inside the slab.

Given field $E(x)$ (downward parabola, vanishing at both slab faces) and the derived linear charge density $\rho_v(x)=2\rho_0x/d$ that produces it.

Approach. In one dimension, Gauss's law reduces to $dE/dx=\rho_v(x)/\varepsilon_0$. Differentiate the given $E(x)$ directly.

Differentiate $E(x)$. $$\frac{dE}{dx}=\frac{\rho_0d}{\varepsilon_0}\cdot\frac{2x}{d^2}=\frac{2\rho_0x}{\varepsilon_0d}$$
Apply the 1-D Gauss's-law relation $\rho_v=\varepsilon_0\,dE/dx$. $$\rho_v(x)=\varepsilon_0\cdot\frac{2\rho_0x}{\varepsilon_0d}=\boxed{\frac{2\rho_0}{d}\,x,\qquad|x|\le d/2}$$ a charge density that varies LINEARLY across the slab, zero at the centre.
Consistency check at the slab edges. At $x=\pm d/2$: $$\rho_v\!\left(\pm\frac{d}{2}\right)=\frac{2\rho_0}{d}\left(\pm\frac{d}{2}\right)=\boxed{\pm\rho_0=\pm1\ \text{C/m}^3}$$ matching the given reference density exactly — confirming the reading of the field expression — and the odd symmetry ($\rho_v(-x)=-\rho_v(x)$) integrates to zero net charge, consistent with $E=0$ outside the (electrically neutral) slab.

Quantity	Result
Charge density $\rho_v(x)$	$\dfrac{2\rho_0}{d}x=2\times10^6\,x\ \text{C/m}^3$ (with $x$ in m), for $\|x\|\le d/2$
$\rho_v$ at $x=+d/2$	$+1$ C/m$^3$
$\rho_v$ at $x=-d/2$	$-1$ C/m$^3$
Net charge in slab (per unit area)	$0$ (odd/antisymmetric distribution)

Concept & Theory

Topic. Recovering a volume charge density from a given 1-D electric field profile, using the differential (point) form of Gauss's law.

Key relations. $\nabla\cdot\vec E=\rho_v/\varepsilon_0$, which for a field depending only on $x$ reduces to $dE/dx=\rho_v/\varepsilon_0$.

Why this works. Gauss's law in differential form is a LOCAL statement — it relates the charge density at a point directly to how fast the field is changing there, with no need to know the field's origin or shape elsewhere; differentiating the given (quadratic) $E(x)$ is therefore a direct, exact route to $\rho_v(x)$ without ever having to guess or assume a source distribution.

Common pitfall. Trying to integrate $E(x)$ (or reasoning from the potential) instead of differentiating — the density is the DERIVATIVE of the field, not its integral; also forgetting to check the boundary self-consistency ($\rho_v(\pm d/2)=\pm\rho_0$), which is what confirms the field expression was read correctly.

Source. Sadiku, Elements of Electromagnetics, Ch. 3 (Gauss's law, differential form; Poisson's/continuity relations for 1-D charge distributions).

Question 7 of 8: Charge Density Distribution That Produces a Given Quadratic Field Inside a Slab

Question 7: Charge Density Distribution That Produces a Given Quadratic Field Inside a Slab (20 marks)