Boundary Conditions on E and V

Electrostatics keeps promising smooth fields — and then almost every problem you meet plants an idealized surface in the middle of the picture. A capacitor plate, the skin of a conductor, a cell membrane decorated with charge, the interface between two dielectrics: all are modeled as mathematical sheets of zero thickness carrying charge per unit area $\sigma$ . Right at such a sheet the field cannot behave smoothly, and pretending otherwise is how sign errors and phantom factors of two are born. What you need are the exact crossing rules — which pieces of $\mathbf{E}$ and $V$ glide through unchanged, which pieces jump, and by precisely how much. The rules come from the two structural facts this chapter has established, each interrogated on a tiny geometric witness. Gauss's law, applied to a wafer-thin pillbox straddling the sheet, dictates the fate of the perpendicular component: $E_\perp$ jumps by $\sigma/\varepsilon_0$ . The vanishing curl of the electrostatic field, applied to a thin rectangular loop, protects the parallel component: $\mathbf{E}_\parallel$ crosses untouched. The potential behaves more gently still — crossing an infinitesimally thin sheet costs no work, so $V$ is continuous, though its slope inherits the field's kink. These four statements are the matching conditions that stitch every boundary-value problem together. When E2 solves Laplace's equation region by region — separation of variables, image charges, multipole matching — the constants in each region are fixed by exactly these conditions. Learn them as a reflex now, and the entire boundary-value apparatus ahead becomes bookkeeping.

Key takeaways

Across any surface carrying charge density $\sigma$ , the normal component of the field jumps: $E_\perp^{\text{above}}-E_\perp^{\text{below}}=\sigma/\varepsilon_0$ (pillbox + Gauss).
The tangential component is continuous: $\mathbf{E}_\parallel^{\text{above}}=\mathbf{E}_\parallel^{\text{below}}$ (thin loop + $\nabla\times\mathbf{E}=0$ ). Both facts combine into one vector statement: $\mathbf{E}_{\text{above}}-\mathbf{E}_{\text{below}}=(\sigma/\varepsilon_0)\,\hat{\mathbf{n}}$ .
The potential is continuous across the sheet, $V_{\text{above}}=V_{\text{below}}$ , because $\mathbf{E}$ stays finite while the crossing path shrinks to zero — but its normal derivative jumps by $-\sigma/\varepsilon_0$ .
The field acting on the sheet itself is the average of the two sides, $\tfrac12(\mathbf{E}_{\text{above}}+\mathbf{E}_{\text{below}})$ — a layer of charge cannot push on itself.
These are the matching conditions of every E2 boundary-value solution: solve the PDE in each region, then glue with $V$ continuous and the prescribed kink in $\partial V/\partial n$ .
Key idea: the normal component of E jumps by σ/ε₀ while the tangential component is continuous; equivalently V is continuous but ∂V/∂n jumps by −σ/ε₀.

The core insight

The normal component of $\mathbf{E}$ jumps by $\sigma/\varepsilon_0$ across a charged surface while the tangential component is continuous; equivalently, $V$ is continuous but its normal derivative jumps by $-\sigma/\varepsilon_0$ — the complete crossing rules for electrostatics.

Where this lives today

Every finite-element field solver on Earth executes this article at scale. Packages like COMSOL, Ansys Maxwell, and the open-source FEniCS discretize a device into millions of cells and then impose exactly equations (4) and (5) at every material interface — $V$ continuous, normal derivative jumping with the surface charge (or its dielectric generalization). Semiconductor TCAD tools that design the transistors in the chip you are reading this on spend most of their runtime enforcing crossing rules at oxide-silicon interfaces, where sheets of trapped charge shift threshold voltages by creasing the potential precisely as the second figure shows. When a simulation "converges", it means the creases finally match. Biology runs the same mathematics wet. A cell membrane is a charged double layer a few nanometres thick; electrophysiology models (Poisson-Boltzmann solvers, molecular-dynamics codes with implicit solvent) treat it as a boundary surface with prescribed $\sigma$ , matching potentials across it to compute the transmembrane fields that gate ion channels. And at the exotic end, the "metallic" boundary conditions imposed on the plates in Casimir-force calculations are this article's conductor limit, applied to the quantum vacuum — the crossing rules of 1873, still the first thing any field theory must respect at a surface.

The physical picture

Picture field lines threading a charged sheet. Some lines are born on the sheet's charge (or die there, if $\sigma<0$ ) — and those births are visible only in the perpendicular traffic: extra flux leaves the neighborhood of the sheet, so $E_\perp$ must be different on the two sides, by an amount that counts the charge, $\sigma/\varepsilon_0$ . Lines skimming parallel to the sheet neither begin nor end on it, so the tangential field glides through unchanged. The visual consequence: field lines kink as they cross a charged surface, the way a light ray kinks entering glass — the parallel part preserved, the perpendicular part rescaled.

Figure

Field lines kink at a charged sheet — parallel component preserved, normal component jumps

The potential sees the same event more calmly. Walk straight through the sheet: the work per unit charge is $\mathbf{E}\cdot\mathrm{d}\boldsymbol{\ell}$ , and $\mathbf{E}$ , however discontinuous, stays finite — so as the walk shrinks to zero length, the work vanishes and $V$ arrives on the far side with the same value it left with. What $V$ cannot hide is its slope: $-\partial V/\partial n$ is $E_\perp$ , so the graph of $V$ has a corner at the sheet — continuous in value, with two different slopes meeting at a crease of size $\sigma/\varepsilon_0$ . That corner picture is worth internalizing: charged sheets show up in potential problems not as cliffs but as creases.

Figure

The potential has a corner, not a cliff — V continuous, slope jumps by σ/ε₀

One more physical note before the derivations. The sheet's own charge feels a field too — but not the full field of either side, because a patch of charge cannot push on itself. Remove the patch's own (symmetric) contribution and what remains is the average of the two sides, $\tfrac12(\mathbf{E}_{\text{above}}+\mathbf{E}_{\text{below}})$ . That average is what enters the force per area on any charged surface, and it is the origin of the famous factor of $\tfrac12$ in the electrostatic pressure on a conductor.

The four crossing rules

Set up the geometry once: a surface carries charge density $\sigma$ (coulombs per square metre), with unit normal $\hat{\mathbf{n}}$ pointing from the side we call "below" to the side we call "above". The labels are pure convention; the physics only cares that $\hat{\mathbf{n}}$ and the jump are quoted consistently. The perpendicular story is Gauss's law shrunk onto the surface. Take the flux rule from Gauss's Law in Integral Form and feed it a pillbox of face area $A$ and vanishing height:

E_\perp^{\text{above}}-E_\perp^{\text{below}}=\frac{\sigma}{\varepsilon_0} \qquad (1)

where: $E_\perp=\mathbf{E}\cdot\hat{\mathbf{n}}$ is the normal component on each side and $\varepsilon_0=8.854\times10^{-12}\ \mathrm{F/m}$ . Read it physically: the sheet is a source of normal flux, and the strength of the source is exactly the charge it carries. With $\sigma=0$ the normal field is continuous; with $\sigma>0$ the field on the upper side is more "outgoing" than on the lower by $\sigma/\varepsilon_0$ — whether that means growing, shrinking, or reversing depends on what external fields are also present. The parallel story is the vanishing curl shrunk onto the surface. The Curl of the Electrostatic Field established $\oint\mathbf{E}\cdot\mathrm{d}\boldsymbol{\ell}=0$ for every loop; feed that a thin rectangle straddling the sheet and the two long legs must cancel:

\mathbf{E}_\parallel^{\text{above}}=\mathbf{E}_\parallel^{\text{below}} \qquad (2)

where: $\mathbf{E}_\parallel$ is the (two-component) piece of $\mathbf{E}$ lying in the surface. No amount of surface charge can break this — equation (2) has no $\sigma$ in it. A charged sheet bends field lines only by feeding their perpendicular part. The two rules combine into a single vector equation, the compact form worth memorizing:

\mathbf{E}_{\text{above}}-\mathbf{E}_{\text{below}}=\frac{\sigma}{\varepsilon_0}\,\hat{\mathbf{n}} \qquad (3)

where: the entire discontinuity of the electric field, all three components, is carried by the normal direction, with magnitude $\sigma/\varepsilon_0$ . This is the form that generalizes: in dielectrics $\varepsilon_0\mathbf{E}\to\mathbf{D}$ with free charge on the right (an E3 story), and in full electrodynamics the same equation survives untouched. The potential adds two more rows to the table. Since $V_{\text{above}}-V_{\text{below}}=-\int\mathbf{E}\cdot\mathrm{d}\boldsymbol{\ell}$ across a path of vanishing length with finite integrand,

V_{\text{above}}=V_{\text{below}} \qquad (4)

where: continuity holds across any single layer of charge, however strong. And translating equation (1) into potential language via $E_\perp=-\partial V/\partial n$ ,

\frac{\partial V_{\text{above}}}{\partial n}-\frac{\partial V_{\text{below}}}{\partial n}=-\frac{\sigma}{\varepsilon_0} \qquad (5)

where: $\partial V/\partial n=\hat{\mathbf{n}}\cdot\nabla V$ is the slope of $V$ along the normal. Equations (4) and (5) are the crease picture in formulas: value continuous, slope kinked. They are precisely the data a boundary-value problem needs — and precisely what the E2 machinery will impose, region by region.

Which condition answers which question

you are crossing	use	what it tells you
any charged surface, field language	equation (3)	full vector jump, $(\sigma/\varepsilon_0)\hat{\mathbf{n}}$
any charged surface, potential language	equations (4) + (5)	$V$ glides, slope kinks by $-\sigma/\varepsilon_0$
uncharged interface ( $\sigma=0$ )	all of (1)-(5)	everything continuous; field is smooth
a conductor's surface	(3) with $\mathbf{E}_{\text{below}}=0$	$\mathbf{E}=(\sigma/\varepsilon_0)\hat{\mathbf{n}}$ just outside — owned by Conductors in Electrostatic Equilibrium
dielectric interface	$\hat{\mathbf{n}}\cdot(\mathbf{D}_{\text{ab}}-\mathbf{D}_{\text{be}})=\sigma_f$	normal $\mathbf{D}$ tracks free charge only (E3)
force on the sheet itself	$\mathbf{f}=\sigma\cdot\tfrac12(\mathbf{E}_{\text{ab}}+\mathbf{E}_{\text{be}})$	self-force excluded; the average acts

Derivation AThe pillbox and the loop — both jump conditions from the integral laws

The goal is the complete field crossing rule, equation (3), derived from the two integral statements of electrostatics: Gauss's law for the normal component and the vanishing circulation for the tangential component. The strategy is to build two tiny geometric probes that straddle the surface — a pillbox and a rectangle — and shrink their thickness to zero, watching which contributions survive the limit and which die. The one subtlety to watch: the pillbox's side wall and the rectangle's short legs both vanish because $\mathbf{E}$ is bounded, so their areas and lengths kill them; nothing is being swept under the rug.

Step 1. The pillbox. Center a squat cylinder on the surface: face area $A$ (small enough that $\sigma$ and $\mathbf{E}$ are effectively uniform across it), height $\epsilon\to0$ , one face in the "above" region, one in the "below". Gauss's law says the total outward flux equals $Q_{\text{enc}}/\varepsilon_0=\sigma A/\varepsilon_0$ — the enclosed charge is pure surface charge, since any volume density contributes $\rho A\epsilon\to0$ .

Step 2. Evaluate the flux face by face. Top face (outward normal $+\hat{\mathbf{n}}$ ): $+E_\perp^{\text{above}}A$ . Bottom face (outward normal $-\hat{\mathbf{n}}$ ): $-E_\perp^{\text{below}}A$ . Side wall: area $\propto\epsilon\to0$ with $|\mathbf{E}|$ bounded, so its flux vanishes. Why boundedness matters: the field of a surface layer is finite (it is a volume layer viewed from afar); only line charges and point charges make $\mathbf{E}$ blow up, and none are present in the pillbox. Therefore $\big(E_\perp^{\text{above}}-E_\perp^{\text{below}}\big)A=\sigma A/\varepsilon_0$ , which is equation (1).

Step 3. The loop. Lay a thin rectangle perpendicular to the surface: long legs of length $\ell$ running parallel to the surface (one just above, one just below, along an arbitrary tangential direction $\hat{\mathbf{t}}$ ), short legs of height $\epsilon\to0$ . The electrostatic circulation theorem gives $\oint\mathbf{E}\cdot\mathrm{d}\boldsymbol{\ell}=0$ around it.

Step 4. Walk the rectangle. Top leg: $+\mathbf{E}_{\text{above}}\cdot\hat{\mathbf{t}}\,\ell$ . Bottom leg (walked backwards): $-\mathbf{E}_{\text{below}}\cdot\hat{\mathbf{t}}\,\ell$ . Short legs: length $\epsilon\to0$ , bounded integrand, zero contribution. So $\big(\mathbf{E}_{\text{above}}-\mathbf{E}_{\text{below}}\big)\cdot\hat{\mathbf{t}}=0$ for every tangential direction $\hat{\mathbf{t}}$ — which is equation (2), all tangential components at once. Why no charge appears: circulation counts no sources; surface charge can feed flux but never circulation, so the parallel field is protected no matter how large $\sigma$ grows.

Step 5. Assemble. The difference vector $\mathbf{E}_{\text{above}}-\mathbf{E}_{\text{below}}$ has zero projection on every tangential direction (Step 4) and normal projection $\sigma/\varepsilon_0$ (Step 2); hence it is $(\sigma/\varepsilon_0)\hat{\mathbf{n}}$ — equation (3). Verified: the isolated infinite sheet from Applications of Gauss's Law and Symmetry gives $\pm\sigma/2\varepsilon_0$ on the two sides — jump $\sigma/\varepsilon_0$ with tangential parts zero on both sides, matching (3) exactly; Griffiths eqs. 2.31-2.33.

Go deeper · Derivation B · 1 minWhy the potential glides through while its slope kinks

Derivation BWhy the potential glides through while its slope kinks

The goal is the potential pair, equations (4) and (5): continuity of $V$ across a charged surface, and the jump of its normal derivative. The strategy for continuity is a shrinking path integral — the same "bounded integrand times vanishing interval" logic that killed the pillbox wall; for the slope, we simply translate the already-proven field jump into gradient language. Watch where the argument would fail: it needs $\mathbf{E}$ finite in the gap, which a single charged layer guarantees but an idealized dipole layer does not.

Step 1. Continuity. Take points $a$ (below) and $b$ (above), separated by distance $\epsilon$ straight through the surface. By the definition of the potential, $V(b)-V(a)=-\int_a^b\mathbf{E}\cdot\mathrm{d}\boldsymbol{\ell}$ . The integrand is bounded — say $|\mathbf{E}|\le E_{\max}$ near the surface — so $|V(b)-V(a)|\le E_{\max}\,\epsilon\to0$ . Hence $V_{\text{above}}=V_{\text{below}}$ , equation (4). Why a dipole layer is the exception: two sheets $\pm\sigma$ separated by $d\to0$ with $\sigma d$ held fixed drive $E_{\max}\to\infty$ in the gap, the product $E_{\max}\epsilon$ stays finite, and $V$ genuinely jumps — the one structure that puts a cliff, not a crease, in the potential.

Step 2. The slope. On each side of the surface the field is the gradient of the potential, $E_\perp=-\hat{\mathbf{n}}\cdot\nabla V=-\partial V/\partial n$ , taking the derivative on that side.

Step 3. Substitute into equation (1): $-\partial V_{\text{above}}/\partial n+\partial V_{\text{below}}/\partial n=\sigma/\varepsilon_0$ , i.e.

\frac{\partial V_{\text{above}}}{\partial n}-\frac{\partial V_{\text{below}}}{\partial n}=-\frac{\sigma}{\varepsilon_0}

which is equation (5). The minus sign is the downhill convention at work: positive $\sigma$ makes the potential fall away faster on the upper side than the slope it arrived with.

Step 4. Consistency of the pair. Equations (4) and (5) together say the graph of $V$ along the normal is continuous with a concave kink of size $\sigma/\varepsilon_0$ — and they are exactly equivalent to (1) and (2): tangential continuity of $\mathbf{E}$ is automatic once $V$ is continuous along the surface (differentiate (4) tangentially). Nothing new was smuggled in; the potential language is a repackaging, optimized for boundary-value work. Verified: for the charged spherical shell (next derivation) the explicit solution has $V$ continuous at $r=R$ while $\mathrm{d}V/\mathrm{d}r$ jumps from $0$ to $-\sigma/\varepsilon_0$ ; Griffiths eq. 2.36.

Go deeper · Derivation C · 1 minMatching in action — the charged shell rebuilt from boundary conditions alone

Derivation CMatching in action — the charged shell rebuilt from boundary conditions alone

The goal is a dress rehearsal for E2: recover the potential of a uniformly charged spherical shell (radius $R$ , charge density $\sigma$ ) without integrating over the charge at all — solve Laplace's equation in each charge-free region and let the boundary conditions fix every constant. The strategy showcases the division of labor: the PDE proposes a family of solutions; regularity and the reference at infinity discard half; continuity of $V$ and the slope-kink rule (5) nail what remains. This is the template every separation-of-variables solution follows.

Step 1. Solve where there is no charge. Inside ( $r<R$ ) and outside ( $r>R$ ) the shell, $\nabla^2V=0$ with spherical symmetry, whose general solution (from Laplace's Equation) is $V(r)=a+b/r$ in each region — four constants total: $a_{\text{in}},b_{\text{in}},a_{\text{out}},b_{\text{out}}$ .

Step 2. Spend the global conditions. Regularity at the origin forces $b_{\text{in}}=0$ (a $1/r$ inside would announce a point charge that is not there). The reference $V(\infty)=0$ forces $a_{\text{out}}=0$ . Two constants left: $V_{\text{in}}=a_{\text{in}}$ , $V_{\text{out}}=b_{\text{out}}/r$ .

Step 3. Impose continuity at the shell, equation (4): $a_{\text{in}}=b_{\text{out}}/R$ .

Step 4. Impose the slope kink, equation (5), with $\hat{\mathbf{n}}=\hat{\mathbf{r}}$ : $\partial V_{\text{out}}/\partial r-\partial V_{\text{in}}/\partial r=-b_{\text{out}}/R^2-0=-\sigma/\varepsilon_0$ , so $b_{\text{out}}=\sigma R^2/\varepsilon_0$ and then $a_{\text{in}}=\sigma R/\varepsilon_0$ . Why the derivative condition carries the charge: the PDE never saw $\sigma$ — all the source information enters through the crease, which is how boundary conditions encode physics the interior equation cannot know.

Step 5. Assemble and audit:

V(r)=\begin{cases}\dfrac{\sigma R}{\varepsilon_0}, & r\le R\\[2mm] \dfrac{\sigma R^2}{\varepsilon_0\,r}, & r\ge R\end{cases}

With total charge $Q=4\pi R^2\sigma$ the outside form is $Q/4\pi\varepsilon_0 r$ — the point-charge potential, as Applications of Gauss's Law and Symmetry found from the field side; the inside is the constant plateau that Potential of a Localized Charge Distribution computed by direct integration. The field jump checks too: $E_{\text{out}}(R)=\sigma/\varepsilon_0$ , $E_{\text{in}}=0$ , jump $\sigma/\varepsilon_0$ ✓. Verified: same result as the direct superposition integral (Griffiths Ex. 2.7) with zero integration performed here — the boundary conditions carried all of it; matching against the point-charge limit at $r\gg R$ is exact.

Limiting cases

Case	Result	Meaning
isolated charged sheet	$E_\perp=\pm\sigma/2\varepsilon_0$ on the two sides	symmetric split; jump is still $\sigma/\varepsilon_0$
conductor surface	$\mathbf{E}_{\text{out}}=(\sigma/\varepsilon_0)\hat{\mathbf{n}}$ , $\mathbf{E}_{\text{in}}=0$	whole jump on one side — see Conductors in Electrostatic Equilibrium
uncharged interface ( $\sigma=0$ )	all components of $\mathbf{E}$ and $V$ continuous	the field is smooth; nothing to match
dielectric interface, no free charge	$\varepsilon_1E_{1\perp}=\varepsilon_2E_{2\perp}$ , $\mathbf{E}_\parallel$ continuous	normal $\mathbf{D}$ is the continuous object (E3)
dipole (double) layer	$V$ itself jumps by $D/\varepsilon_0$	the lone exception to continuity of $V$ ; infinite $E$ in the gap
force on the layer	$\mathbf{f}=\sigma\,\tfrac12(\mathbf{E}_{\text{ab}}+\mathbf{E}_{\text{be}})$	self-force excluded; average field acts

Common trapThe potential jumps across a charged sheet — no, it creases

The single most common boundary-condition error is granting $V$ a discontinuity it does not have. Across any single layer of surface charge — however large $\sigma$ — the potential is continuous: crossing zero thickness against finite field costs zero work. What jumps is the field's normal component, i.e. the slope $\partial V/\partial n$ : the graph of $V$ has a crease, not a cliff. (The only structure that cliffs the potential is an idealized dipole layer, where the field in the vanishing gap is infinite.) A second, quieter trap: when computing the force per unit area on the charged surface itself, do not use the full field of either side — the layer cannot push on itself, and the field that acts on it is the average $\tfrac12(\mathbf{E}_{\text{above}}+\mathbf{E}_{\text{below}})$ . Forgetting the average is exactly a factor-of-two error, and it is why the electrostatic pressure on a conductor is $\sigma^2/2\varepsilon_0$ and not $\sigma^2/\varepsilon_0$ — the accounting Conductors in Electrostatic Equilibrium completes.

Worked examples

1 · Reading σ off a field jump (confidence builder)

Measurements just above and just below a large charged plane give $\mathbf{E}_{\text{above}}=(120\,\hat{\mathbf{x}}+80\,\hat{\mathbf{z}})$ V/m and $\mathbf{E}_{\text{below}}=(120\,\hat{\mathbf{x}}-45\,\hat{\mathbf{z}})$ V/m, with $\hat{\mathbf{z}}$ the upward normal. (a) Check the data are consistent with electrostatics. (b) Find $\sigma$ . (c) Find the force per unit area on the sheet.

Step 1. Consistency is the tangential rule: $\mathbf{E}_\parallel$ is $120\,\hat{\mathbf{x}}$ V/m on both sides — continuous ✓. If the parallel components had differed, no static surface charge could explain the data.

Step 2. The normal jump reads off the charge: $E_\perp^{\text{above}}-E_\perp^{\text{below}}=80-(-45)=125$ V/m, so $\sigma=\varepsilon_0\times125=8.854\times10^{-12}\times125=\boxed{1.11\times10^{-9}\ \mathrm{C/m^2}\approx1.1\ \mathrm{nC/m^2}}$ . Positive, as expected for a sheet whose normal field points away on top and toward it below… almost: note the below-side normal field is $-45$ V/m, i.e. toward the sheet — an external downward field of $17.5$ V/m is superposed on the sheet's own $\pm62.5$ V/m. The boundary condition never needed to untangle that; it counts only the jump.

Step 3. Force per area uses the average field: $\mathbf{E}_{\text{avg}}=\tfrac12\big[(120+120)\hat{\mathbf{x}}+(80-45)\hat{\mathbf{z}}\big]=(120\,\hat{\mathbf{x}}+17.5\,\hat{\mathbf{z}})$ V/m. Then $\mathbf{f}=\sigma\mathbf{E}_{\text{avg}}=1.107\times10^{-9}\times(120\,\hat{\mathbf{x}}+17.5\,\hat{\mathbf{z}})=\boxed{(1.33\,\hat{\mathbf{x}}+0.19\,\hat{\mathbf{z}})\times10^{-7}\ \mathrm{N/m^2}}$ — the external field's $17.5$ V/m does the pushing along $\hat{\mathbf{z}}$ ; the sheet's own $\pm62.5$ V/m cancelled out of the average, exactly as self-force exclusion demands.

2 · The charged shell, audited in volts (exam level)

A spherical shell of radius $R=10.0$ cm carries uniform $\sigma=35.4\ \mathrm{nC/m^2}$ . Using only the boundary-condition solution of the third derivation: (a) find $V$ at the surface and the total charge $Q$ ; (b) find $\mathbf{E}$ just inside and just outside; (c) verify both crossing rules numerically.

Step 1. The surface (and interior plateau) potential: $V(R)=\sigma R/\varepsilon_0=35.4\times10^{-9}\times0.100/8.854\times10^{-12}=\boxed{400\ \mathrm{V}}$ (399.8 V unrounded). Total charge $Q=4\pi R^2\sigma=4\pi\times(0.100)^2\times35.4\times10^{-9}=\boxed{4.45\ \mathrm{nC}}$ . Cross-check by the point-charge formula: $kQ/R=8.99\times10^9\times4.449\times10^{-9}/0.100=400\ \mathrm{V}$ ✓.

Step 2. Fields from the two branches: inside, $V$ is constant so $\mathbf{E}_{\text{in}}=0$ . Outside, $E_{\text{out}}(R)=\sigma/\varepsilon_0=35.4\times10^{-9}/8.854\times10^{-12}=\boxed{4.00\ \mathrm{kV/m}}$ , radially outward. (Equivalently $kQ/R^2=8.99\times10^9\times4.449\times10^{-9}/0.0100=4.00$ kV/m ✓.)

Step 3. Crossing rule for the field, equation (3): jump $=4.00\ \mathrm{kV/m}-0=\sigma/\varepsilon_0$ ✓, entirely along $\hat{\mathbf{r}}=\hat{\mathbf{n}}$ , with tangential components zero on both sides ✓.

Step 4. Crossing rules for the potential: $V$ is $400$ V on both sides of $r=R$ — continuous ✓ (equation 4). Slopes: $\partial V_{\text{in}}/\partial r=0$ ; $\partial V_{\text{out}}/\partial r=-\sigma R^2/\varepsilon_0 r^2\big|_R=-4.00$ kV/m. Jump $=-4.00\ \mathrm{kV/m}-0=-\sigma/\varepsilon_0$ ✓ (equation 5). The crease is concave toward the charge: the potential leaves the shell flat and falls away outside — a 400-volt plateau creased onto a $1/r$ tail.

Origins · 1 minPeter Gustav Lejeune Dirichlet and how this was discovered

Peter Gustav Lejeune Dirichlet

1805–1859 · Berlin and Göttingen

Dirichlet is the mathematician who made "given the boundary, find the interior" into a well-posed question. Born in Düren to a family of Belgian descent, he carried Fourier's heat-theory manuscripts back from Paris and spent a career in Berlin turning analysis rigorous — the modern definition of a function is essentially his. In potential theory his name attaches to the problem this article's conditions feed: find the harmonic function taking prescribed values on a boundary (the Dirichlet problem), and to the variational route he taught for solving it (the Dirichlet principle, later patched by Hilbert after Weierstrass found the gap). His Berlin lectures on the inverse-square forces of nature, published as the Vorlesungen über die der umgekehrten Verhältnisse des Quadrats der Entfernung wirkenden Kräfte, canonized the boundary-value viewpoint electrostatics still uses: interior equation plus surface data equals unique answer. He succeeded Gauss at Göttingen in 1855; Riemann and Dedekind were shaped in his seminar. Every time E2 pins separation-of-variables constants with " $V$ continuous, slope jumps by $-\sigma/\varepsilon_0$ ", it is solving a Dirichlet-style matching problem.

History

1785

— Coulomb establishes the inverse-square force and observes that the field of a charged conductor stands perpendicular to its surface — boundary behavior read straight off experiment.

1813

— Poisson extends Laplace's equation into charged regions; surface layers enter potential theory as prescribed sources.

1828

— Green's Essay recasts electrostatics as an interior equation steered by boundary data, introducing the surface potential and its normal derivative as the problem's inputs.

1850

— Dirichlet's lectures make the boundary-value problem well-posed: prescribe V on a surface and the harmonic interior is determined.

1873

— Maxwell's Treatise assembles the pillbox and loop arguments in essentially their modern textbook form.

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §2.3.5 · Jackson, Classical Electrodynamics, 3rd ed., §1.6 · Purcell & Morin, Electricity and Magnetism, 3rd ed., §3.2 · Zangwill, Modern Electrodynamics, §3.4

Provenance. All equations are standard textbook results. Verifier should recompute: the pillbox derivation of $E_\perp$ jump (Gauss's law, Griffiths §2.3.5, eq. 2.31), the loop derivation of tangential continuity (Griffiths eq. 2.32), continuity of $V$ and the normal-derivative jump (Griffiths eq. 2.36), the charged-shell matching against the point-charge limit, and the conductor pressure $P=\sigma^2/2\varepsilon_0$ (Griffiths §2.5.3, Jackson §1.6, Purcell §3.2). Constant $\varepsilon_0=8.854\times10^{-12}$ F/m is CODATA-standard. Worked numbers here additionally check: $\sigma=\varepsilon_0\times125\ \mathrm{V/m}=1.11\ \mathrm{nC/m^2}$ ; shell with $R=0.100$ m, $\sigma=35.4\ \mathrm{nC/m^2}$ : $Q=4.45$ nC, $V(R)=400$ V, $E_{\text{out}}(R)=4.00$ kV/m by both $\sigma/\varepsilon_0$ and $kQ/R^2$ .

Builds on: Gauss's Law in Integral Form · The Curl of the Electrostatic Field · The Electrostatic Potential · Continuous Charge Distributions Leads to: Applications of Gauss's Law and Symmetry · Potential of a Localized Charge Distribution

Why is V continuous across a charged surface when E is not?

Because the two objects integrate differently. The potential difference across the surface is $-\int\mathbf{E}\cdot\mathrm{d}\boldsymbol{\ell}$ along a path whose length shrinks to zero, and the integrand — however discontinuous — stays finite: a bounded number times a vanishing interval is zero. The field, by contrast, is a local quantity, free to take different limits from the two sides. The practical picture: $V$ creases (continuous value, kinked slope of size $\sigma/\varepsilon_0$ ) while $E_\perp$ steps. If you ever produce a solution where $V$ itself steps across a single charged layer, a constant got lost in one region — recheck the matching.

Is the potential ever genuinely discontinuous across a surface?

Yes — across an idealized dipole layer (double layer): two sheets $\pm\sigma$ separated by $d\to0$ with the dipole moment per area $D=\sigma d$ held fixed. In the vanishing gap the field grows as $\sigma/\varepsilon_0\to\infty$ , the product of field and path length stays finite, and $V$ jumps by $D/\varepsilon_0$ (Jackson §1.6). This is the lone exception, and it is physical: cell membranes and electrochemical interfaces are dipole layers to excellent approximation, which is why "membrane potential" is a jump. A single layer of charge, no matter how strong, can only crease $V$ ; it takes a double layer to cliff it.

Why exactly does the pillbox's side wall contribute nothing?

Two ingredients: the wall's area scales with the pillbox height $\epsilon$ , and the field on the wall is bounded. Flux is field times area, so the wall's contribution scales as $E_{\max}\times(\text{perimeter})\times\epsilon\to0$ . Nothing subtle is hidden — but both ingredients are needed. If the surface carried an idealized line charge along it, the field nearby would diverge as $1/s$ and boundedness would fail; the pillbox argument then correctly refuses to apply at the line itself. The same "bounded times vanishing" logic kills the short legs of the loop in the tangential argument and drives the continuity of $V$ — it is one lemma wearing three costumes.

Do these boundary conditions survive when fields depend on time?

Better than you might expect. The normal condition (3) comes from Gauss's law, which is exact in full electrodynamics — unchanged. The tangential condition (2) comes from $\oint\mathbf{E}\cdot\mathrm{d}\boldsymbol{\ell}=0$ , which Faraday's law amends to $-\mathrm{d}\Phi_B/\mathrm{d}t$ ; but the loop's area shrinks to zero with its height, and any finite $\partial\mathbf{B}/\partial t$ threads vanishing flux — so tangential $\mathbf{E}$ stays continuous even dynamically. The magnetic field acquires its own mirror-image pair (normal $B$ continuous, tangential $B$ jumping with surface current), and the full quartet governs reflection and transmission of electromagnetic waves at interfaces — the E8 machinery is these arguments verbatim.

What changes at the boundary between two dielectrics?

The bookkeeping variable changes. Inside matter the useful flux field is $\mathbf{D}=\varepsilon_0\mathbf{E}+\mathbf{P}$ , and the pillbox argument applied to $\mathbf{D}$ counts only free charge: $\hat{\mathbf{n}}\cdot(\mathbf{D}_{\text{above}}-\mathbf{D}_{\text{below}})=\sigma_f$ . With no free charge at the interface, normal $\mathbf{D}$ is continuous — which for linear media means $\varepsilon_1E_{1\perp}=\varepsilon_2E_{2\perp}$ : the normal $\mathbf{E}$ does jump, fed by the bound surface charge that polarization piles up at the interface. Tangential $\mathbf{E}$ remains continuous (the loop argument never cared about the medium). Field lines therefore refract at dielectric boundaries with a tangent law, $\tan\theta_1/\tan\theta_2=\varepsilon_1/\varepsilon_2$ — the E3 chapter builds this out.

How are these conditions actually used to solve problems in E2?

As the glue between regions. The strategy of every boundary-value method — separation of variables, images, multipole matching — is: solve Laplace's or Poisson's equation separately in each charge-free region, obtaining families of solutions with free constants; then impose (i) $V$ continuous across each interface, (ii) $\partial V/\partial n$ jumping by $-\sigma/\varepsilon_0$ where surface charge lives (or the Dirichlet data "V given on this conductor"), plus regularity at the origin and the reference at infinity. Counting is exact: the uniqueness theorems descending from Laplace's Equation guarantee the constants are fully determined — as the shell derivation showed, where two matching conditions nailed two surviving constants with no integration at all.

Why does the average field act on the sheet, and what happens to the other half?

Split the field near the sheet into two pieces: the sheet's own contribution ( $\pm\sigma/2\varepsilon_0$ , symmetric, pointing away on both sides for positive $\sigma$ ) and everything else (external, smooth across the sheet). A patch of charge exerts no net force on itself — Newton's third law within the patch — so only the external piece pushes it. But the external piece is precisely the average of the total fields on the two sides: adding $\mathbf{E}_{\text{above}}$ and $\mathbf{E}_{\text{below}}$ cancels the sheet's own equal-and-opposite contribution and doubles the external one. Hence $\mathbf{f}=\sigma\cdot\tfrac12(\mathbf{E}_{\text{above}}+\mathbf{E}_{\text{below}})$ per unit area. Applied to a conductor ( $\mathbf{E}_{\text{in}}=0$ ) this yields the outward electrostatic pressure $P=\sigma^2/2\varepsilon_0$ — worked out where it belongs, in Conductors in Electrostatic Equilibrium.