Field Discontinuity Across a Sphere's Thin Surface-Charge Layer: 04-BS-9, May 2018, Q1

Question 1: Field Discontinuity Across a Sphere's Thin Surface-Charge 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
Point charge at centre	$+2e$
Sphere radius $R$	$1\times10^{-10}$ m
Volume charge (uniform density, fills $r\le R$)	total $-e$
Surface charge (infinitely thin layer at $r=R$)	total $-e$

Find. The difference between the electric field magnitude just inside ($r=R^-$) and just outside ($r=R^+$) the thin surface-charge layer.

Point charge +2e at the centre, uniform volume charge −e filling the sphere out to R, and a thin surface layer −e exactly at R. The field jumps from a positive outward value just inside the layer to exactly zero just outside it.

Approach. Apply Gauss's law with two spherical Gaussian surfaces at $r=R^-$ (enclosing the point charge and all of the volume charge, but not yet the thin surface layer) and at $r=R^+$ (enclosing everything, including the surface layer); the difference is exactly the field jump a thin charged layer produces.

Enclosed charge and field just inside the surface layer. At $r=R^-$ the Gaussian surface encloses the point charge and the ENTIRE volume charge (it fills the sphere out to the same radius $R$), but not yet the surface layer itself: $$Q_{\text{enc}}(R^-)=+2e+(-e)=+e$$ $$E(R^-)=\frac{Q_{\text{enc}}(R^-)}{4\pi\varepsilon_0R^2}=\frac{e}{4\pi\varepsilon_0R^2}$$ directed radially OUTWARD (net enclosed charge is positive).
Enclosed charge and field just outside the surface layer. At $r=R^+$ the Gaussian surface now also encloses the surface layer: $$Q_{\text{enc}}(R^+)=+2e+(-e)+(-e)=0$$ $$E(R^+)=0$$ — the system is exactly neutral once the surface layer is included, so the field vanishes immediately outside it.
Field difference. Substituting $e=1.6\times10^{-19}$ C and $R=1\times10^{-10}$ m into Step 1 (using $1/4\pi\varepsilon_0=8.99\times10^{9}$ N·m$^2$/C$^2$ from the given $\varepsilon_0$): $$E(R^-)=\frac{(8.99\times10^{9})(1.6\times10^{-19})}{(1\times10^{-10})^2}$$ $$\boxed{\Delta E=E(R^-)-E(R^+)=1.438\times10^{11}\ \text{V/m}}$$

Quantity	Result
$E(R^-)$, just inside the surface layer	$1.438\times10^{11}$ V/m, radially outward
$E(R^+)$, just outside the surface layer	$0$ V/m
Field difference $\Delta E$	$1.438\times10^{11}$ V/m

Concept & Theory

Topic. Gauss's law and the boundary condition a thin surface-charge layer imposes on the normal electric field.

Key relations. Gauss's law $Q_{\text{enc}}=\varepsilon_0\oint\vec E\cdot d\vec A$; boundary condition $E_{\text{above}}-E_{\text{below}}=\sigma/\varepsilon_0$ across any thin charged layer.

Why this works. The field just inside a thin charged layer "sees" everything enclosed except that layer's own charge, while the field just outside it also sees the layer — the difference between the two is therefore exactly the layer's own contribution, completely independent of whatever other charge sits deeper inside. Here that layer's contribution happens to null the field entirely because the whole four-part system (point + volume + surface) is net neutral.

Common pitfall. Forgetting to include the volume charge when computing the "just inside" field (using only the point charge), or assuming a net-neutral assembly must have zero field everywhere — Gauss's law only guarantees zero OUTSIDE all the charge, not between its layers.

Source. Sadiku, Elements of Electromagnetics, Ch. 3 (Gauss's law, boundary conditions, thin surface charge).

Question 1 of 8: Field Discontinuity Across a Sphere's Thin Surface-Charge Layer

Question 1: Field Discontinuity Across a Sphere's Thin Surface-Charge Layer (20 marks)