Griffiths Chapter 3: Laplace's equation and the mean-value property, the uniqueness theorems, the method of images, separation of variables in Cartesian and spherical coordinates with Fourier's trick, and the multipole expansion.
Subject: Electrodynamics (Griffiths) · 60 slides · diagram-first lesson
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Title
Griffiths, Chapter 3
Laplace's equation, the method of images, separation of variables, and the multipole expansion
Objectives
Deck 3 reduced electrostatics to one scalar equation. This deck is four techniques for actually solving it when you cannot just integrate over the charge.
The theme throughout: you often do not know where the charge is — it is induced, and it depends on the field you are trying to find. These four methods break that circle.
Something has shifted. Every problem so far began 'given the charge distribution', and that is a luxury real problems rarely offer — because much of the charge is induced on conductors, and how it arranges itself depends on the very field you are trying to find.
This deck is four escapes from that circle. Uniqueness turns guessing into a legitimate method. Images replace an awkward boundary with a fictitious charge. Separation of variables solves the differential equation head-on. And the multipole expansion tells you what a complicated blob looks like from far away, which is often all you actually need.
It is the most technique-heavy deck in the course, and the techniques are not interchangeable — each fits a particular shape of problem. The final pattern slide is the map, and it is worth a glance now so the four do not blur together.
Section
Section 1
Concept
So far every problem started 'given the charge distribution'. Real problems rarely do.
Figure (svg): A point charge above a grounded metal plate, with question marks over the induced charge on the plate's surface.
The induced charge on the plate depends on the field, and the field depends on the induced charge. You cannot integrate your way out of a loop like that.
The escape: stop hunting for the charge and solve the differential equation for V directly, using what you do know — the potential on the boundaries.
\[ \nabla^2 V = 0 \quad\text{in the charge-free region, with } V \text{ specified on the boundary} \]
Look carefully at the circular dependency, because without seeing the problem the solutions look like arbitrary cleverness.
Put a charge above a grounded plate. Charge is induced on the plate. How much, and where? That depends on the field. And what is the field? It is the field of the original charge plus the field of the induced charge. So you need the induced charge to get the field, and the field to get the induced charge.
There is no integrating your way out of that. There is no formula to substitute into, because the thing you would need to integrate is itself the unknown.
The escape is to stop chasing the charge and solve the differential equation for V instead, using what you genuinely do know: the potential on the boundaries. A grounded plate is at zero volts, and that is a hard fact independent of any charge distribution.
Concept
\[ \frac{d^2V}{dx^2} = 0 \quad\Rightarrow\quad V(x) = mx + b \]
Figure (svg): A straight line between two fixed endpoint values, with a midpoint marked as the average of its two neighbours.
The straight line has two properties worth naming, because they generalise to two and three dimensions where the picture is much harder:
\[ V(x) = \tfrac{1}{2}\big[V(x+a) + V(x-a)\big] \]
One dimension is where the two crucial properties are visible without any hard geometry, so they are worth establishing here before generalising.
The equation says the second derivative is zero, so the solution is a straight line. That is the whole one-dimensional theory.
Now read off the two properties. The value at any point is the average of its two neighbours, equally spaced either side — obvious for a straight line, and it is the property that survives into three dimensions. And there are no local maxima or minima in the interior; the extreme values sit at the two ends.
Both reappear almost unchanged in two and three dimensions, where they are far less obvious. Establishing them here makes the general case a generalisation rather than a surprise.
Worked example
Two parallel plates a distance d apart, one held at zero and one at V₀, with nothing in between. Find V everywhere in the gap.
Write Laplace's equation in the only variable that matters, in volts
Why: Nothing depends on y or z, because the plates are large and uniform, so the partial differential equation collapses to an ordinary one.
\[ \frac{d^2V}{dx^2} = 0 \quad\Rightarrow\quad V(x) = mx + b \]
Apply the two boundary conditions to fix the two constants
Why: A second-order equation needs exactly two conditions, and here they are the two plate potentials.
\[ V(0) = 0 \Rightarrow b = 0, \qquad V(d) = V_0 \Rightarrow m = \frac{V_0}{d} \]
\[ V(x) = \frac{V_0}{d}x \]
Differentiate to get the field, in volts per metre
Why: Minus the gradient of a straight line is a constant — a uniform field, pointing from the high-potential plate to the low one.
\[ \mathbf{E} = -\frac{dV}{dx}\hat{\mathbf{x}} = -\frac{V_0}{d}\hat{\mathbf{x}} \]
Verify against the capacitor result from Deck 3, in volts per metre
Why: A parallel-plate capacitor has uniform field E = V/d between the plates, which is exactly what came out. And the potential is linear, so its value halfway across is V₀/2 — the average of the two boundary values, as the mean-value property demands.
The simplest boundary-value problem in the book, and worth doing carefully because the structure is identical to much harder ones.
Notice why the partial differential equation collapses to an ordinary one: the plates are large and uniform, so nothing depends on y or z. Recognising when a problem is effectively one-dimensional is a real skill in itself.
A second-order equation needs exactly two boundary conditions, and here they are the two plate potentials. That matching of constants to conditions is worth watching, because in the separation-of-variables problems later the same accounting applies with many more constants floating about.
The verification connects to something already known: a uniform field of magnitude V over d, which is the parallel-plate capacitor result from the last deck. And the potential is linear, so its midpoint value is the average of the two boundary values — the mean-value property in its simplest form.
Intuition
The mean-value property is not just a theorem — it is an algorithm.
Figure (svg): An animated relaxation grid: the fixed boundary points stay lit while each interior point pulses in turn as it is replaced by the average of its neighbours.
Fix the boundary values. Then sweep through the interior replacing each point by the average of its four neighbours, over and over. The grid relaxes to the solution of Laplace's equation.
It converges because the true solution is the one function for which nothing changes when you do this — the mean-value property is precisely a fixed-point condition. This is how field solvers still work today, with refinements.
This turns the mean-value property from a theorem into something operational, which usually makes it stick.
The procedure is what the animation shows. Fix the boundary values, then sweep through the interior replacing each point with the average of its four neighbours, and repeat until nothing changes.
Why does it converge? Because the true solution is precisely the function for which this operation changes nothing. The mean-value property is a fixed-point condition, and the iteration is hunting for that fixed point.
Real field solvers still work essentially this way, with refinements for speed. It is easy to assume numerical methods are something separate from the theory; here the numerical method is the theory.
Concept
Figure (svg): Left: a soap film stretched on a bent wire loop. Right: a sphere drawn around a point in space, with the potential at the centre marked as the average over the sphere.
\[ V(\mathbf{r}) = \frac{1}{4\pi R^2}\oint_{\text{sphere}} V\,da \]
A soap film stretched on a bent wire takes exactly this shape. Poke it in the middle and it springs back — it has no way to hold a bump.
So the solution to Laplace's equation is the smoothest possible function that meets the boundary values. Nothing interesting happens in the interior; all the structure is at the edges.
The same two properties survive, and in higher dimensions they stop being obvious and become genuinely surprising.
In three dimensions, the value at a point equals the average over the surface of any sphere centred there — any radius at all, as long as the sphere stays inside the charge-free region.
The soap-film analogy is the best two-dimensional intuition available. Stretch a film on a bent wire loop and it takes exactly this shape. Poke it in the middle and it springs back, because it has no way to hold a bump.
The conclusion is worth stating plainly: the solution is the smoothest function consistent with the boundary values. Nothing interesting happens in the interior — all the structure lives at the edges. Which is precisely why specifying the boundary is enough to determine everything.
Counterexample
No maxima or minima in a charge-free region has a startling consequence.
Discussion prompt
Could you arrange fixed charges so that a positive test charge sits in a stable equilibrium, held from every direction?
Figure (svg): A saddle-shaped surface with a ball at the centre, stable along one axis and unstable along the perpendicular one.
This is why ion traps use oscillating fields, and why the classical 'planetary' atom cannot be held together electrostatically. Real traps cheat by making the field time-dependent, which Laplace's equation no longer governs.
This is the most striking consequence of the no-maxima property, and the subject is more memorable for it.
A stable equilibrium for a positive charge needs a local minimum of the potential — a bowl it can settle into from every direction. But Laplace's equation forbids local minima in a charge-free region, and the region immediately around the trapped charge is charge-free.
So every equilibrium point is a saddle: stable along some directions, unstable along at least one other. There is always an escape route.
Hence ion traps use oscillating fields, since a time-varying field is not governed by Laplace's equation. And the classical planetary atom cannot be held together electrostatically, which was one of the pressures that led to quantum mechanics.
Prediction
Predict first
A charge-free region has V = 0 everywhere on its boundary. What is V inside?
Correct: Zero everywhere inside
Why: V = 0 everywhere inside satisfies Laplace's equation and matches every boundary value, so by the first uniqueness theorem it is THE solution. You can also see it from the mean-value property: the maximum and minimum of V must both occur on the boundary, and both are zero, so V is squeezed to zero everywhere in between.
This is the workhorse behind the whole chapter. Any two solutions with the same boundary values have a difference that satisfies exactly these conditions — so the difference is zero and the solutions are the same.
\[ V_1, V_2 \text{ both solve it} \;\Rightarrow\; V_3 = V_1 - V_2 \text{ has } \nabla^2V_3=0, \; V_3=0 \text{ on the boundary} \;\Rightarrow\; V_3 \equiv 0 \]
Commit first, because the argument is worth more than the answer and guessing wrong is a good way to engage with it.
The answer is zero everywhere inside, and there are two routes to it. By uniqueness: the constant zero satisfies Laplace's equation and matches every boundary value, so it is the solution. By the mean-value property: the maximum and the minimum both occur on the boundary, both are zero, so everything in between is squeezed to zero.
The equation at the bottom is why this matters so much. It is the proof of the uniqueness theorem in one line — take two supposed solutions, subtract them, and the difference satisfies exactly these conditions, so it is identically zero and the two solutions were the same all along.
This innocuous-looking special case is doing all the work for the entire chapter.
Concept
Here is the theorem that turns guessing into a legitimate method.
first uniqueness theorem — The solution of Laplace's equation in a volume is uniquely determined once V is specified on the whole boundary of that volume. There is exactly one such function — no more, no less.
Figure (svg): A closed region with the potential specified all around its boundary, and a single question mark inside indicating the one possible interior solution.
The practical consequence is enormous: if you can produce, by any means at all, a function that satisfies Laplace's equation and matches the boundary values, it is THE answer. Guessing, luck, and clever tricks are all legitimate methods.
The method of images, next section, is nothing but this theorem exploited aggressively.
This is the theorem that makes everything else in the deck legitimate, so it is worth stating precisely.
Specify V on the entire boundary of a region and the solution inside is uniquely determined. Exactly one function fits — no more and no less.
The consequence is liberating. If you can produce, by any means whatsoever, a function that satisfies Laplace's equation in the region and matches every boundary value, it is the answer. Not an answer — the answer.
So guessing is a legitimate method. So is borrowing a solution from a different problem, or inventing fictitious charges that are not there. The method of images, coming next, is nothing but this theorem exploited aggressively, and without it the whole trick would be indefensible.
Worked example
Griffiths' Example 3.1, proved in three lines with the uniqueness theorem.
State the setup: a cavity inside conducting material, with no charge in the cavity
Why: The cavity wall is part of the conductor, so the whole boundary of the cavity sits at one and the same potential V₀, in volts.
Propose a candidate solution: V equals V-zero everywhere inside
Why: A constant satisfies Laplace's equation trivially, since all its second derivatives vanish, and it matches the boundary value everywhere on the wall.
\[ \nabla^2 V_0 = 0 \quad\text{and}\quad V = V_0 \text{ on the boundary} \;\checkmark \]
Invoke uniqueness
Why: Laplace's equation with these boundary values has exactly ONE solution. We have found a solution, so it is the only one.
Verify what this implies for the field, in newtons per coulomb
Why: V constant inside means the gradient is zero, so E = 0 throughout the cavity — no matter what charges sit outside the conductor, and no matter how strangely the cavity is shaped. This is the shielding result from Deck 3, now proved rather than argued.
This proves the shielding result that was only argued informally in the last deck, and the proof is three lines.
The cavity wall is part of the conductor, so the entire boundary of the cavity sits at one potential. Propose the constant function equal to that value. It satisfies Laplace's equation trivially, since all its second derivatives vanish, and it matches the boundary everywhere.
By uniqueness, that is the solution. There is no other.
Then take the gradient: V constant means E is zero throughout the cavity — whatever charges sit outside the conductor, and however strangely the cavity is shaped. Notice how much work the uniqueness theorem just did. A result about arbitrary geometries, proved without a single integral.
Concept
The first theorem needs V on the boundary. With conductors you often know something different: the total charge on each one.
second uniqueness theorem — In a volume surrounded by conductors and containing a specified charge density, the field is uniquely determined once the total charge on each conductor is given.
So you may specify either the potential of each conductor, or its total charge, and the field is pinned down either way. That is the licence to solve real problems, where a plate is grounded (V known) or isolated with fixed charge (Q known).
| what you know on each conductor | which theorem | typical phrasing in a problem |
|---|---|---|
| its potential | first | 'a grounded plate' or 'held at V₀' |
| its total charge | second | 'an isolated conductor carrying charge Q' |
| neither | neither | the problem is underspecified |
The first theorem needs the potential on every boundary. Real problems often hand you something different, and this covers that case.
With conductors, you may know the total charge on each one rather than its potential — an isolated conductor carrying charge Q, say, rather than one wired to earth.
The theorem says that is enough: specifying the total charge on each conductor determines the field uniquely.
The table is the practical translation, and it is worth drilling because the phrasing in problems is formulaic. 'Grounded' or 'held at V-zero' means the first theorem. 'Isolated, carrying charge Q' means the second. A problem that gives neither is underspecified, and no unique answer exists.
Sorting
Sort into buckets
Grounded always means the first theorem: 'grounded' is a statement that V = 0 there. Isolated with known charge always means the second.
A quick classification, and the rule is simple once said.
Grounded always means the first theorem, because 'grounded' is a statement about potential. Isolated with known charge always means the second.
The real point of the exercise is reading problem statements for those keywords, since which theorem applies determines what you are allowed to assume and what has to come out as an answer.
Pattern
Compare this with the alternative — integrating over an unknown induced charge distribution — and it is clear why Chapter 3 is organised around clever guesses.
One caution that costs people marks: the solution is only unique in the region where you imposed the conditions. A construction valid outside a sphere usually says nothing at all about the inside.
This is the workflow, and the fourth step is the one that takes some getting used to.
Write the boundary conditions precisely. Find a solution by any means available. Check it satisfies Laplace's equation in the region and matches every boundary value. Then declare victory — because there is no other answer, how you found it is irrelevant.
Set that against the alternative of integrating over an unknown induced charge distribution, and the attraction is obvious. This is why Chapter 3 is organised around clever guesses rather than direct computation.
One caution that costs marks. The solution is unique only in the region where the conditions were imposed. A construction valid outside a sphere tells you nothing whatsoever about the inside.
Check
Think it through.
Check your understanding
In a charge-free region, the potential at the centre of an imaginary sphere is 10 V, and the potential is 12 V at one point on that sphere. What must be true?
Answer: A
Why: In a charge-free region the value at the centre equals the AVERAGE over the sphere. If the average is 10 V and one point reads 12 V, some other part of the sphere must read below 10 V to bring the average back down.
This tests the mean-value property directly.
The value at the centre is the average over the sphere. If the average is ten and one point on the sphere reads twelve, something elsewhere on it must read below ten to compensate. That is the whole argument.
Option D is the interesting one, because it looks plausible: the centre is below one boundary point, so is it a minimum? No — Laplace's equation forbids interior minima outright. Being below some neighbours and above others is exactly the saddle behaviour the equation demands.
Section
Section 2
Intuition
Optics has the same structure, and the analogy is not superficial.
Figure (svg): An object in front of a mirror with its reflected image behind the glass, drawn beside a charge above a conducting plane with its image charge below.
In both cases you replace a boundary you cannot easily handle with a fictitious source behind it that reproduces the boundary's effect exactly.
And in both cases the image is genuinely not there. Reach behind the mirror and you find nothing; dig into the conductor and you find no charge at −d, only a spread of induced charge on the surface. The image is a computational device that gives exactly the right answer in the region you care about.
'Image charge' sounds like loose terminology. It is not — the structural parallel with optics is exact, and seeing that makes the method feel principled rather than like a conjuring trick.
In both cases you replace a boundary you cannot handle directly with a fictitious source placed behind it, chosen so that it reproduces the boundary's effect exactly in the region you care about.
And in both cases the source is genuinely not there. Reach behind the mirror and you find nothing. Dig into the conductor and you find no charge at minus d — only a spread of induced charge sitting on the surface.
Fictitious does not mean approximate, though. The field the image produces in the real region is exactly right, and that is what the uniqueness theorem guarantees.
Concept
A point charge q sits a distance d above an infinite grounded conducting plane. Find the potential above the plane.
Figure (svg): An animation in which the grounded plane and its induced charge fade away and an image charge fades in below in their place.
The trick: throw the plane away and put a charge −q at the mirror position, a distance d below where the plane was.
Two point charges of equal magnitude and opposite sign produce V = 0 on the plane midway between them — exactly the boundary condition the grounded plane imposed.
The canonical example, and the logic is worth getting clear before the algebra.
The problem: a point charge a distance d above an infinite grounded plane, with the induced charge on the plane unknown.
The trick: throw the plane away entirely and put a charge of minus q at the mirror position, the same distance below where the plane used to be. Two point charges, no conductor at all.
It works because two equal and opposite charges produce V equal to zero on the plane midway between them — which is precisely the boundary condition the grounded plane imposed. Same boundary condition, same region, so by uniqueness the same field.
In the animation, the region above the plane is identical in both pictures, and that is the entire claim being made.
Worked example
Write the potential of the two-charge configuration, in volts
Why: Ordinary superposition of two point charges, with the plane at z = 0 and the real charge at z = d.
\[ V(x,y,z) = \frac{1}{4\pi\varepsilon_0}\left[\frac{q}{\sqrt{x^2+y^2+(z-d)^2}} - \frac{q}{\sqrt{x^2+y^2+(z+d)^2}}\right] \]
Check it satisfies Laplace's equation in the region above the plane
Why: Both terms are point-charge potentials, which satisfy Laplace's equation everywhere except at their own charge. The real charge is a genuine source; the image charge sits BELOW the plane, outside the region of interest, so it introduces no false source there.
Check the boundary conditions
Why: On the plane z = 0 the two square roots are equal, so V = 0 as required. Far away both terms vanish, which matches the physical expectation.
\[ V(x,y,0) = \frac{1}{4\pi\varepsilon_0}\left[\frac{q}{\sqrt{x^2+y^2+d^2}} - \frac{q}{\sqrt{x^2+y^2+d^2}}\right] = 0 \;\checkmark \]
Verify by uniqueness, in volts
Why: Laplace's equation holds in the upper region, the boundary values match everywhere, and by the first uniqueness theorem this is therefore THE potential above the plane. The image charge is a fiction — it does not exist — but the potential it helps write down is exact where it matters.
The one rule you must never break: image charges go outside the region where you are computing V. Put one inside and you have changed the charge distribution and are solving a different problem.
Three checks, and together they turn this from a plausible story into a proof.
First, write down the two-charge potential. Second, check it satisfies Laplace's equation in the region of interest — it does, because both terms are point-charge potentials, and the only genuine source above the plane is the real charge. The image sits below the plane, outside the region, so it introduces no false source anywhere that matters.
Third, check the boundary. On the plane the two distances are equal, so the two terms cancel and V is zero, as required. Far away, both terms vanish.
All three pass, so by uniqueness this is the potential above the plane. Which gives one unbreakable rule: image charges go outside the region where you are computing. Put one inside and you have changed the charge distribution and are solving a different problem entirely.
Worked example
The image trick also delivers what looked impossible at the start of the deck — the induced charge distribution itself.
Use the conductor boundary condition from Deck 3, in coulombs per square metre
Why: Surface charge equals epsilon-zero times the field just outside, which is minus epsilon-zero times the normal derivative of V.
\[ \sigma = -\varepsilon_0\left.\frac{\partial V}{\partial z}\right|_{z=0} \]
Differentiate the two-charge potential and evaluate on the plane
Why: Both terms contribute equally on the plane, which is why the result is exactly twice one term.
\[ \sigma(x,y) = \frac{-q\,d}{2\pi(x^2+y^2+d^2)^{3/2}} \]
Figure (svg): A graph of induced surface charge density across the plane: a negative dip directly under the charge, tailing off to zero in both directions.
Integrate over the whole plane to find the total induced charge, in coulombs
Why: Use polar coordinates on the plane, with the area element s ds dphi.
\[ Q_{\text{ind}} = \int_0^{2\pi}\!\!\int_0^{\infty}\frac{-qd}{2\pi(s^2+d^2)^{3/2}}s\,ds\,d\phi = -q \]
Verify against the field-line picture, in coulombs
Why: The total induced charge is exactly −q — which is what the image charge was. That is no coincidence: every field line leaving the positive charge must terminate on the plane, and each line ends on the same amount of negative charge it started from.
This is the payoff that looked impossible at the start of the deck — the induced charge distribution itself, computed exactly.
The route is the boundary condition from the last deck. Surface charge equals epsilon-zero times the field just outside a conductor, which is minus epsilon-zero times the normal derivative of V. Differentiate the two-charge potential, evaluate on the plane, and there it is.
The shape matches intuition: a negative peak directly beneath the charge, tailing off with distance, because the induced charge crowds toward whatever is attracting it.
Then integrate over the whole plane. The total comes to exactly minus q, independent of the height d. The field-line reading explains why — every line leaving the positive charge has to terminate somewhere on the plane, and the total charge it lands on equals the charge it left from.
Raising the charge spreads the induced charge over a wider area but never changes the total, which is worth predicting for yourself before checking.
Concept
The force on the real charge is the force the image would exert — because the field above the plane really is the two-charge field.
\[ \mathbf{F} = -\frac{1}{4\pi\varepsilon_0}\frac{q^2}{(2d)^2}\,\hat{\mathbf{z}} \]
Always attractive: the charge is pulled toward the plate it induced charge on. This is the force that makes a charged balloon stick to a wall.
The energy, though, is not the two-charge energy. Only half of it:
\[ W = -\frac{1}{4\pi\varepsilon_0}\frac{q^2}{4d} \]
Two results, and the second contains a trap, so keep them separate in your head.
The force is straightforward. Since the field above the plane really is the two-charge field, the force on the real charge is exactly what the image would exert. It is always attractive, which is why a charged balloon sticks to a neutral wall.
The energy is not the two-charge energy. It is half of it, and the next slide explains why.
The asymmetry is worth expecting: the image gives the field and the force exactly, but energies need care, because anything quadratic in the field depends on where the field actually exists.
Trap
Having replaced the plane by an image charge, a student computes the energy the same way:
Use the two-point-charge energy formula with separation 2d, in joules
Why: The force came out right from the image picture, so the energy should too.
\[ W \overset{?}{=} -\frac{1}{4\pi\varepsilon_0}\frac{q^2}{2d} \]
This is twice the correct answer.
In the image configuration there is field energy below the plane too. In the real problem that region is inside a conductor, where the field is zero.
Figure (svg): Two panels: the image configuration with field energy both above and below the plane, and the real configuration with field only above it.
Halve the two-charge energy, or integrate the field energy over the upper half space only, in joules
Why: Both routes give the same answer, since the field above the plane is identical in the two configurations and it is only the lower half that differs.
\[ W = \frac{\varepsilon_0}{2}\int_{z>0}E^2\,d\tau = -\frac{1}{4\pi\varepsilon_0}\frac{q^2}{4d} \]
Rule of thumb: the image gives the field and the force exactly, but energies need care, because energy is quadratic and lives everywhere the field does.
The error is natural, because the force came out right from the image picture and the energy feels like it should follow.
The resolution is about where the field lives. In the image configuration there is field on both sides of the plane, and field energy is stored in both halves. In the real problem, the region below the plane is inside a conductor, where the field is exactly zero and no energy is stored at all.
So the real configuration has half the field energy of the imagined one, and the answer is half as large.
Two routes to the right number: halve the two-charge energy, or integrate the field energy over the upper half space only. They agree, because the field above the plane is identical in the two configurations and only the lower half differs.
The general lesson: images reproduce the field in your region, so anything computed from the field in your region is safe. Anything computed from the field everywhere is not.
Prediction
Figure (svg): A grounded conducting sphere of radius R with a point charge q at distance a from its centre, and a question mark inside the sphere.
Predict first
For a charge q at distance a from the centre of a grounded sphere of radius R, where must the image charge sit?
Correct: At distance R²/a from the centre, on the same side as q
Why: This is Griffiths' Example 3.2. The image charge is q' = -Rq/a placed at b = R²/a from the centre, along the line to q. Since a is greater than R, b is less than R, so the image lands safely inside the sphere — outside the region where the potential is being computed, exactly as the method requires.
Notice it is not a simple mirror reflection. The sphere curves, so both the position and the size of the image must be adjusted.
Guess before revealing. The naive guess is a simple mirror reflection, and seeing why that fails is the instructive part.
A plane reflects symmetrically. A sphere curves, so both the position and the size of the image have to be adjusted — the image is neither at the mirror point nor of equal magnitude.
The answer is a charge of minus R over a times q, placed at R squared over a from the centre. And note the check that makes it legitimate: since a is greater than R for a charge outside the sphere, R squared over a is less than R, so the image lands safely inside the sphere and therefore outside the region where V is being computed.
That check is not a formality. If the image landed in the region of interest, the whole construction would be invalid.
Worked example
Griffiths' Example 3.2, worked through.
Place a trial image charge of unknown size at an unknown distance inside the sphere, in coulombs and metres
Why: Two unknowns, and the boundary condition V = 0 on the whole sphere is the constraint that fixes both.
\[ q' = -\frac{R}{a}q, \qquad b = \frac{R^2}{a} \]
Write the potential of the two-charge configuration, in volts
Why: The distances are measured from the real charge and from the image charge to the field point.
\[ V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\left(\frac{q}{\mathfrak{r}} + \frac{q'}{\mathfrak{r}'}\right) \]
Confirm this vanishes on the sphere
Why: For any point on the surface, the ratio of the distance to q and the distance to q' comes out to a/R at every angle. That is the geometric fact that makes the whole trick work.
\[ \frac{\mathfrak{r}}{\mathfrak{r}'} = \frac{a}{R} \;\text{ on } r = R \quad\Rightarrow\quad V = 0 \;\checkmark \]
Verify with the force, and with two limits, in newtons
Why: The force is that between q and q' separated by a − b, giving an attraction of magnitude q²Ra/[4πε₀(a² − R²)²]. As a becomes very large the force falls off as one over a cubed — faster than Coulomb, because the sphere looks neutral from far away. As a approaches R the force diverges, as it must when a charge is pressed against a conducting surface.
\[ F = -\frac{1}{4\pi\varepsilon_0}\frac{q^2Ra}{(a^2-R^2)^2} \]
Two unknowns — the image's size and its position — against one boundary condition, V equal to zero over the entire sphere. That condition has to hold at every point of the surface, which is what pins down both.
The geometric fact that makes it work is worth naming. For an image placed at R squared over a, the ratio of the distance to the real charge and the distance to the image comes out the same at every point on the sphere, equal to a over R. That constant ratio is exactly what lets the two terms cancel everywhere at once.
The verification uses two limits, and both mean something. Far away, the force falls off as one over a cubed rather than one over a squared — faster than Coulomb, because from a distance the grounded sphere looks neutral. Very close, as a approaches R, the force diverges, which is what you would expect when a charge is pressed against a conducting surface.
Concept
A charge in the corner between two perpendicular grounded planes needs more than one image — and the signs alternate.
Figure (svg): A charge in the upper right quadrant formed by two grounded planes, with three image charges in the other three quadrants, signs alternating.
Reflect in each plane, then reflect the reflections
Why: One image handles the horizontal plane and one the vertical, but each image spoils the OTHER plane's boundary condition — so a third image is needed to repair it.
\[ -q \text{ at } (-x_0,y_0), \quad -q \text{ at } (x_0,-y_0), \quad +q \text{ at } (-x_0,-y_0) \]
Check both planes: on each one, the four charges pair up into two equal-and-opposite contributions, so V = 0 as required.
This only closes off for angles that divide 180 degrees evenly. At 60 degrees you need five images; at an arbitrary angle the image series never terminates and the method fails.
The corner problem shows what happens when one image is not enough, and the reasoning generalises.
Reflect the real charge in the horizontal plane and you fix that plane's boundary condition — but the new image spoils the vertical plane's condition. Reflect in the vertical plane instead and the same thing happens in reverse. A third image, the reflection of the reflections, repairs both at once.
Check the result: on each plane, the four charges pair up into two equal-and-opposite contributions, so V is zero everywhere on both.
The limitation matters. This only closes off for angles that divide a hundred and eighty degrees evenly. At sixty degrees you need five images. At an arbitrary angle the series never terminates and the method fails — which is exactly where separation of variables takes over.
Error analysis
A student wants the potential inside a grounded spherical shell with a point charge inside it, and writes:
Annotate
On: \( V = \frac{1}{4\pi\varepsilon_0}\left(\frac{q}{\mathfrak{r}} + \frac{q'}{\mathfrak{r}'}\right), \quad q' = -\frac{R}{a}q \;\text{ at }\; b = \frac{R^2}{a} \)
The general lesson, and it is the one rule of this method: after placing any image, verify two things — that it lies outside your region, and that the boundary condition really is satisfied at every point, not just the convenient ones.
This one is subtler than a straightforward mistake, and the subtlety is the lesson.
The exterior result has been copied across to the interior problem without being re-derived. For a charge inside the sphere, a is less than R, which flips the geometry — and the argument that made V vanish on the sphere was built on the exterior configuration.
The image does still land outside the region of interest, so the method is not doomed. But that has to be checked rather than inherited, and so does the boundary condition itself.
The rule applies to every image problem. After placing any image, verify two things explicitly: that it lies outside your region, and that the boundary condition really holds at every point rather than just at the convenient ones.
Pattern
| geometry | image | position |
|---|---|---|
| charge above a grounded plane | -q | mirror point, distance d on the far side |
| charge outside a grounded sphere | -Rq/a | R²/a from the centre, on the same side |
| charge in a grounded corner (two planes) | three images | the three reflections, signs alternating |
The method only works when the boundary is simple enough for a few point charges to reproduce it. For anything else, the next section takes over.
The procedure, plus the three cases you will actually meet.
Identify the boundary and what it demands, usually V equal to zero on a grounded surface. Guess a small number of images placed outside your region, with unknown sizes and positions. Impose the boundary condition and solve for the unknowns. Verify. Then compute whatever you need — field, induced charge from epsilon-zero times the surface field, force from the image, and energy with the halving caution.
Be honest about the range of the method: it works only when the boundary is simple enough for a few point charges to reproduce it. That is a small set of geometries, but they are common ones, and when it works it is much the fastest route.
Check
One question before separation of variables.
Check your understanding
A charge +q sits 3 cm above an infinite grounded plane. What is the total charge induced on the plane?
Answer: A
Why: Integrating the induced surface density over the whole plane gives exactly -q, with the distance d cancelling out. Field lines from the charge all terminate on the plane, and each ends on an equal amount of negative charge. Moving the charge higher spreads the induced charge over a wider area but never changes its total.
The key insight is that the total induced charge does not depend on the height, which surprises most people first time.
Integrating the induced density over the whole plane gives exactly minus q, with d cancelling out. The field-line argument is the reason: every line from the charge terminates on the plane.
Option C reflects a real confusion about grounding worth clearing up. Grounded means held at zero potential, not uncharged. Charge flows in from the earth precisely so the plane can stay at zero volts.
Section
Section 3
Concept
When no image charge will reproduce the boundary, solve the differential equation head-on — but only after making a bold restriction.
\[ V(x,y) = X(x)\,Y(y) \]
Substitute the product into Laplace's equation and divide through by XY
Why: Each term then involves only one variable, which is the whole point of the guess.
\[ \frac{1}{X}\frac{d^2X}{dx^2} + \frac{1}{Y}\frac{d^2Y}{dy^2} = 0 \]
The first term depends only on x, the second only on y, and they sum to zero for all x and y. The only way that can happen is if each is a constant.
\[ \frac{1}{X}\frac{d^2X}{dx^2} = k^2, \qquad \frac{1}{Y}\frac{d^2Y}{dy^2} = -k^2 \]
One partial differential equation has become two ordinary ones — and ordinary differential equations you can solve.
Separation of variables is the workhorse when images fail, and the opening move deserves explaining because it looks arbitrary.
The guess is that the solution factorises into a function of x times a function of y. Nothing guarantees a particular problem has such a solution, and most individual product solutions will not match the boundary conditions anyway. The justification comes later, from superposition.
The algebra is the elegant part. Substitute the product, divide through by it, and you get one term depending only on x plus another depending only on y, summing to zero for all x and y. The only way that can happen is if each term is separately constant.
So one partial differential equation becomes two ordinary ones — and ordinary differential equations are things you can already solve.
Concept
Separation of variables produces one solution per value of n. None of them, on its own, matches the boundary data. The rescue is that Laplace's equation is linear.
\[ \nabla^2 V_1 = 0, \;\; \nabla^2 V_2 = 0 \quad\Rightarrow\quad \nabla^2(aV_1 + bV_2) = 0 \]
Figure (svg): Three simple sine-shaped solutions being added together to produce a more complicated profile that matches a square boundary shape.
So an infinite sum of separated solutions is still a solution — and with infinitely many coefficients to choose, you have enough freedom to match any boundary function.
This is also why superposition is used so freely everywhere else in the course: every equation in electrostatics is linear in the fields.
This rescues the method from the obvious objection to the previous slide.
No single product solution matches the boundary data. But Laplace's equation is linear, so any sum of solutions is also a solution. Sum infinitely many, each with an adjustable coefficient, and there is enough freedom to match essentially any boundary function.
The picture shows the idea: individually simple sine shapes adding up to something that matches a square profile.
Linearity has been quietly doing this work all along — superposition of fields, superposition of potentials, and now superposition of solutions to a differential equation. It is the same property every time.
Concept
The separation constant can be positive or negative, and choosing wrongly wastes an hour. The boundary conditions decide.
| the solutions | their behaviour | use them in the direction that... |
|---|---|---|
| e^{kx}, e^{-kx} | grow and decay, never repeat | runs off to infinity, or has just two boundary values |
| sin(ky), cos(ky) | oscillate, can vanish repeatedly | is bounded between two grounded walls |
The rule: oscillating solutions belong to the direction with the confining boundaries, because only a sine can vanish at two different places without being identically zero.
Exponentials belong to the direction that runs away, where you need decay rather than repetition. Getting this backwards gives sines where you need decay and no way to satisfy the conditions.
Choosing the separation constant's sign wrongly costs an hour, and the rule for choosing right is simple once stated.
Exponentials grow and decay but never repeat. Sines and cosines oscillate and can vanish repeatedly.
So: oscillating solutions belong to the direction with confining boundaries, because only a sine can vanish at two different places without being identically zero. A combination of real exponentials can vanish at most once.
Exponentials belong to the direction that runs off to infinity, where you need decay rather than repetition. Decide which direction is which before choosing, rather than choosing and discovering the problem three steps later.
Worked example
Griffiths' Example 3.3. Two grounded plates at y = 0 and y = a, running off to positive x, closed at x = 0 by a strip held at V₀.
Figure (svg): A semi-infinite slot: two horizontal grounded plates with a vertical strip at the closed end held at V0, and the interior shaded.
List the four boundary conditions before doing anything else
Why: Every constant in the final answer is fixed by one of these, so writing them down first tells you what the solution must look like.
Choose the signs: oscillation in y, decay in x, in volts
Why: The y direction is confined between two grounded plates, so it takes the sine. The x direction runs to infinity and must decay, so it takes the exponential.
\[ V = \big(Ae^{kx} + Be^{-kx}\big)\big(C\sin ky + D\cos ky\big) \]
Apply the first, second and fourth conditions
Why: Condition four kills the growing exponential. Condition one kills the cosine. Condition two forces the sine to vanish at y = a, which quantises k.
\[ A = 0, \quad D = 0, \quad \sin ka = 0 \;\Rightarrow\; k = \frac{n\pi}{a}, \; n = 1,2,3,\ldots \]
\[ V_n(x,y) = C_n\,e^{-n\pi x/a}\sin\!\left(\frac{n\pi y}{a}\right) \]
Superpose all the allowed n, because no single one can match the last condition
Why: Laplace's equation is linear, so any sum of solutions is a solution. This is the step that makes the method powerful.
\[ V(x,y) = \sum_{n=1}^{\infty}C_n\,e^{-n\pi x/a}\sin\!\left(\frac{n\pi y}{a}\right) \]
Verify the three easy conditions before going on, in volts
Why: Every term vanishes at y = 0 and at y = a, and every term decays as x grows. Three of the four conditions are satisfied by construction; only the strip at x = 0 remains, and that is what Fourier's trick is for.
The canonical Cartesian separation problem, and the first step is the one to insist on.
List all four boundary conditions before doing anything else. Every constant in the final answer is fixed by one of them, so writing them down tells you in advance what shape the solution has to take.
Then the sign choice. The y direction is confined between two grounded plates, so it takes sines. The x direction runs to infinity and must decay, so it takes a decaying exponential.
Three of the four conditions are then satisfied by construction. Vanishing at both plates quantises the separation constant to integer multiples of pi over a — and it is worth noticing that quantisation here is pure boundary-condition bookkeeping, with no quantum mechanics involved anywhere.
Only the strip at x equal to zero is left, and no single term can match it. That is what superposition and Fourier's trick are for.
Concept
At x = 0 all the exponentials equal 1, and the last boundary condition becomes a demand that a sum of sines reproduce a given function.
\[ \sum_{n=1}^{\infty}C_n\sin\!\left(\frac{n\pi y}{a}\right) = V_0(y) \]
Multiply both sides by another sine and integrate across the slot
Why: Sines of different n are ORTHOGONAL over this interval: the integral of the product is zero unless the two match. That single fact isolates one coefficient at a time.
\[ \int_0^a \sin\!\left(\frac{n\pi y}{a}\right)\sin\!\left(\frac{m\pi y}{a}\right)dy = \begin{cases}0, & n\neq m\\ a/2, & n = m\end{cases} \]
\[ C_n = \frac{2}{a}\int_0^a V_0(y)\sin\!\left(\frac{n\pi y}{a}\right)dy \]
This works for any V₀(y), not just a constant. Orthogonality is what makes separation of variables a general method rather than a lucky trick.
This is the mechanism that makes the method general rather than a lucky guess, so it deserves proper weight.
At the open end all the exponentials equal one, and the condition becomes a demand that a sum of sines reproduce a given function.
The trick is to multiply both sides by one particular sine and integrate across the slot. Because sines of different n are orthogonal, every term in the sum dies except the one that matches — so a single coefficient gets isolated by a single integral.
And this works for any boundary function, not just the constant used here. That generality is what makes separation of variables a method rather than a trick, and orthogonality is the property doing the work.
Worked example
Put a constant V₀ into the coefficient integral, in volts
Why: The constant comes straight out and leaves an elementary integral of a sine.
\[ C_n = \frac{2V_0}{a}\int_0^a\sin\!\left(\frac{n\pi y}{a}\right)dy = \frac{2V_0}{n\pi}\big(1 - \cos n\pi\big) \]
Notice that the even terms die
Why: For even n the cosine equals 1 and the bracket vanishes. Only odd harmonics survive — a symmetry of the boundary data showing up in the answer.
\[ C_n = \begin{cases}0, & n \text{ even}\\[4pt] \dfrac{4V_0}{n\pi}, & n \text{ odd}\end{cases} \]
\[ V(x,y) = \frac{4V_0}{\pi}\sum_{n=1,3,5,\ldots}^{\infty}\frac{1}{n}\,e^{-n\pi x/a}\sin\!\left(\frac{n\pi y}{a}\right) \]
Verify the answer at both ends, in volts
Why: Deep inside the slot (x large) every term has decayed and V goes to zero, as required — and the n = 1 term decays slowest, so far from the end the field is essentially a single sine. At x = 0 the series is the Fourier series of a constant, which does converge to V₀ across the opening. This particular series even sums in closed form to (2V₀/π)arctan[sin(πy/a)/sinh(πx/a)], which you can check numerically.
The arithmetic is short; the features of the answer are what matter.
The coefficient integral is elementary, and the even terms all vanish because the bracket contains one minus cosine of n pi. Only odd harmonics survive — a symmetry of the boundary data showing up in the answer.
Now read the solution physically. Each term decays as you move into the slot, and higher harmonics decay faster. So deep inside, the n equals one term dominates completely and the field is essentially a single sine.
That is a genuinely useful piece of physics. Sharp features in the boundary data are high-frequency information, and high frequencies die fastest as you move away from the boundary. Nature smooths its own boundary data. The closed form is a curiosity, and one you can check numerically if you like.
Picture it
Figure (svg): An animation in which a square profile is built up from sine harmonics one at a time, each new term sharpening the corners.
Each extra odd harmonic sharpens the corners of the square profile at x = 0. This is why the series needs infinitely many terms there — and why one term is nearly perfect once you are a slot-width inside.
Physically: the sharp corners of the boundary data are high-frequency information, and high frequencies decay fastest as you move into the slot. Nature smooths its own boundary data.
Let the animation do the work here: one term, then three, then nine, each sharpening the corners of the square profile.
Convergence is fastest in the middle and slowest at the corners, which is why infinitely many terms are needed to represent a discontinuous edge exactly.
Connect that to the physics from the previous slide. Those corners are the high-frequency content, and they are precisely what decays fastest as you move into the slot. One term is nearly perfect once you are a slot-width inside, even though nine are not enough at the opening.
Concept
Same method, different coordinates. For problems with azimuthal symmetry — nothing depending on the angle phi — the general solution is:
\[ V(r,\theta) = \sum_{l=0}^{\infty}\left(A_l r^l + \frac{B_l}{r^{l+1}}\right)P_l(\cos\theta) \]
The Legendre polynomials P_l play the role the sines played in the slot: they are the orthogonal set that the boundary data gets expanded in.
| l | P_l(cos θ) | shape |
|---|---|---|
| 0 | 1 | uniform in all directions |
| 1 | cos θ | one lobe up, one down — a dipole pattern |
| 2 | (3cos²θ − 1)/2 | two nodes — a quadrupole pattern |
| 3 | (5cos³θ − 3cos θ)/2 | three nodes |
The r-dependence comes in two flavours per l: a growing one and a decaying one. Which you keep is decided, again, by the boundary conditions.
Same method, different coordinates — this is a translation rather than a new technique.
For problems with azimuthal symmetry, meaning nothing depends on the angle around the axis, the general solution is a sum over Legendre polynomials with two radial behaviours available for each term.
The Legendre polynomials play exactly the role the sines played in the slot: they are the orthogonal set the boundary data gets expanded in. Look at the shapes in the table — the l equals one term is a dipole pattern, l equals two a quadrupole pattern, and so on.
The two radial behaviours per term are the thing to watch. One grows with r, one decays. Which you keep is decided by the boundary conditions, and that is the subject of the trap a couple of slides from now.
Socratic
Fourier's trick looked like a computational convenience. It is more than that.
Discussion prompt
Suppose the sines were NOT orthogonal. What would go wrong when you tried to find the coefficients?
There is a second property doing quiet work here: completeness. The sines can represent any reasonable boundary function, so the method never fails for lack of building blocks. Legendre polynomials have both properties too, which is why the spherical case works identically.
Orthogonality looks like a computational convenience. It is more fundamental than that, and the counterfactual is the clearest way to see it.
If the sines were not orthogonal, multiplying by one and integrating would leave contributions from every other term. Instead of one equation per coefficient you would get an infinite system of coupled equations, every coefficient tangled with all the others.
Orthogonality decouples them completely: each coefficient comes from its own single integral, computed independently of the rest.
There is a second property quietly doing work too — completeness. The sines can represent any reasonable boundary function, so the method never fails for want of building blocks. Legendre polynomials have both properties, which is precisely why the spherical case works identically.
Worked example
Griffiths' Example 3.6, with a simple choice of boundary data so the series terminates.
A hollow sphere of radius R has its surface potential set to V₀(θ) = k cos θ, with no charge inside.
Prune the general solution for the interior region
Why: The origin is inside, so every decaying term would diverge there and must be discarded before anything else is done.
\[ V(r,\theta) = \sum_{l=0}^{\infty}A_l r^l P_l(\cos\theta) \]
Match the surface data against the Legendre polynomials, in volts
Why: The boundary function is k cos θ, and P₁(cos θ) IS cos θ, so only the l = 1 term can contribute — no integration needed.
\[ \sum_l A_l R^l P_l(\cos\theta) = k\cos\theta = kP_1(\cos\theta) \]
\[ A_1 R = k \;\Rightarrow\; A_1 = \frac{k}{R}, \qquad \text{all other } A_l = 0 \]
\[ V(r,\theta) = \frac{k}{R}\,r\cos\theta = \frac{k}{R}z \]
Verify the answer three ways, in volts
Why: At r = R it gives k cos θ, matching the boundary. It satisfies Laplace's equation, since z is a linear function whose second derivatives all vanish. And E = −∇V = −(k/R)ẑ, a perfectly uniform field inside the sphere — which is a genuinely surprising and correct result for this boundary data.
When the boundary data is a single Legendre polynomial, the answer is a single term. When it is not, you integrate against each P_l in turn — the exact analogue of Fourier's trick.
The boundary data here is chosen so the series terminates, which lets the method be seen clearly without heavy integration.
Prune first. The origin is inside the region, so every decaying term would blow up there and has to be discarded before anything else happens.
Then match the surface data. It is k cos theta, and the first Legendre polynomial is cos theta, so only the l equals one term can contribute. No integration needed at all — the coefficient can be read straight off.
The verification is worth all three checks. It matches the boundary at r equals R. It satisfies Laplace's equation, since the answer is linear in z. And its gradient is a constant, so the field inside is perfectly uniform — which is a genuinely surprising result for this boundary data and worth sitting with for a moment.
Trap
Solving for the potential inside a hollow sphere with specified surface potential, a student keeps the full general solution:
Write V with both the growing and the decaying radial terms and fit the surface data
Why: The general solution has both, so keeping both looks safe and complete.
\[ V(r,\theta) = \sum_l\left(A_l r^l + \frac{B_l}{r^{l+1}}\right)P_l(\cos\theta) \]
But the decaying term blows up at r = 0 — and the origin is inside the region being solved. This 'solution' has an infinite potential at the centre with no charge there to produce it.
Kill the terms that misbehave anywhere inside your region, before fitting anything:
| region | keep | kill | because |
|---|---|---|---|
| inside a sphere (origin included) | A_l r^l | B_l/r^{l+1} | the decaying terms diverge at r = 0 |
| outside a sphere (to infinity) | B_l/r^{l+1} | A_l r^l | the growing terms diverge as r → ∞ |
| a shell between two radii | both | neither | neither r = 0 nor infinity is in the region |
Then fit the surviving coefficients to the boundary data with Legendre orthogonality
Why: Half as many unknowns, and each one determined by a single integral against the corresponding Legendre polynomial.
Do this pruning FIRST. Fitting first and pruning later means solving twice as many equations, most of which you are about to throw away.
Keeping the general solution is normally good practice, which is what makes this trap effective.
Keeping both radial behaviours when the origin is inside your region means keeping terms that diverge at r equal to zero. That would be a potential blowing up at a point where there is no charge — physically impossible, and mathematically the solution to a different problem.
So prune before fitting, using the table. Inside a sphere including the origin, keep only the growing terms. Outside, extending to infinity, keep only the decaying ones. In a shell between two radii, neither r equal to zero nor infinity is in the region, so keep both.
Do this first. Fitting first and pruning later means solving twice as many equations, most of which are about to be thrown away.
Worked example
Griffiths' Example 3.8. An uncharged metal sphere of radius R is dropped into a uniform field. Find the potential outside.
Figure (svg): A sphere in an initially uniform field, with the field lines bending to meet the sphere perpendicularly, positive induced charge on top and negative below.
Set the sphere's potential to zero and write the two boundary conditions, in volts
Why: The sphere is an equipotential and you may choose it to be zero. Far away the field is undisturbed, so the potential must approach that of a uniform field.
\[ \text{(i) } V = 0 \text{ at } r = R, \qquad \text{(ii) } V \to -E_0 r\cos\theta \text{ as } r\to\infty \]
Keep only the decaying radial terms — plus whatever condition (ii) demands
Why: The region is outside the sphere, so pure r^l terms would blow up at infinity, except that condition (ii) explicitly requires one term growing like r.
Impose the surface condition to link the two coefficient families
Why: Setting the whole series to zero at r = R fixes each B_l in terms of the corresponding A_l.
\[ B_l = -A_l R^{2l+1} \]
Match the far-field condition, which contains only cos theta
Why: Since P₁(cos θ) is exactly cos θ, only the l = 1 term can survive; every other A_l is zero and A₁ = −E₀.
\[ V(r,\theta) = -E_0\left(r - \frac{R^3}{r^2}\right)\cos\theta \]
Verify at the surface, far away, and via the induced charge, in volts and coulombs per square metre
Why: At r = R the bracket vanishes, so V = 0 on the sphere as required. For large r the R³/r² term is negligible and V returns to −E₀r cos θ. And the induced surface charge, σ = −ε₀ ∂V/∂r at r = R, comes out as 3ε₀E₀cos θ — positive on top, negative underneath, integrating to zero total charge, exactly as an uncharged sphere demands.
The showpiece of the spherical method, and it combines several ideas, so it is worth taking slowly.
The two boundary conditions are unusual because the second one lives at infinity rather than on a surface. Far away the field must be undisturbed, so V has to approach minus E-zero r cos theta.
The sphere is an equipotential and you are free to set it to zero, which then also makes the equatorial plane zero by symmetry.
The far-field condition contains only cos theta, and since the first Legendre polynomial is exactly cos theta, only l equals one survives. Every other coefficient is zero, and the infinite series collapses to a single term.
The verification is worth doing in full. At the surface the bracket vanishes, so V is zero there. Far away the correction term dies and the uniform field returns. And the induced surface charge works out to three epsilon-zero E-zero cos theta — positive on top, negative underneath, integrating to zero net charge, exactly as an uncharged sphere requires.
Pattern
Steps three and four are what turn an intimidating infinite series into two or three surviving terms. Most exam problems collapse to a single value of n or l.
Six steps, and the third and fourth are where the time gets saved.
Choose coordinates matching the boundaries. Write the general separated solution. Prune anything that misbehaves inside your region. Apply the easy conditions — the ones every term satisfies — which fixes the form and quantises the constants. Then superpose and use orthogonality for whatever is left. Finally check the limits.
It looks more complex than it usually is in practice. Most exam problems collapse to a single value of n or l once the pruning and the easy conditions are done. The infinite series is a safety net rather than a workload.
Check
Think before clicking.
Check your understanding
You are solving Laplace's equation inside a rectangular box with three grounded walls and one wall held at a given potential. Which functions belong in the direction perpendicular to the two grounded parallel walls?
Answer: A
Why: A confined direction with a grounded wall at each end needs a function that vanishes at two distinct points. Only the oscillating solutions can do that non-trivially, and the requirement that they vanish at both ends is what quantises the separation constant to nπ/a.
This tests the sign choice, which is the step most likely to derail an otherwise correct solution.
A direction confined between two grounded walls needs a function that vanishes at two distinct points without being identically zero. Only oscillating solutions can manage that.
Option B is worth walking through. A combination of real exponentials can be made to vanish at one point, but forcing it to vanish at a second point as well drives it to zero everywhere. That is the concrete reason exponentials cannot serve a confined direction.
Section
Section 4
Concept
From far enough away, any localised charge distribution looks like a point charge. The multipole expansion says exactly how good that approximation is, and what the corrections are.
Figure (svg): An irregular blob of charge with a far-away field point, and three simplified pictures beside it: a point charge, a dipole, and a quadrupole.
\[ V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\sum_{n=0}^{\infty}\frac{1}{r^{(n+1)}}\int (r')^n P_n(\cos\alpha)\,\rho(\mathbf{r}')\,d\tau' \]
Each term dies faster than the one before it. Far away the first non-zero term dominates completely, and everything after it is a small correction.
The multipole expansion answers a practical question: what does a complicated charge distribution look like from far away?
The answer is a series, each term dying faster than the last. Far enough away, the first non-vanishing term dominates completely and everything after it is a small correction.
The picture is the summary — the real blob, and then the sequence of simpler configurations that approximate it: a point charge, a dipole, a quadrupole.
One important point: the expansion is exact, not approximate. It is an infinite series that converges whenever you are further out than the largest extent of the charge. Truncating it is an approximation; the series itself is not.
Concept
The multipole series is not a new idea — it is one algebraic identity applied to the separation distance.
Figure (svg): A source point at distance r-prime from the origin, a field point at distance r, and the angle alpha between them, with the separation distance marked.
Write the separation distance with the law of cosines
Why: Factor out r, since it is the large quantity, leaving a small parameter r'/r inside.
\[ \mathfrak{r} = r\sqrt{1 + \left(\frac{r'}{r}\right)^2 - 2\frac{r'}{r}\cos\alpha} \]
Expand one over the separation in powers of that small parameter
Why: The coefficients of this expansion are precisely the Legendre polynomials — which is where they come from, and why they keep appearing in spherical problems.
\[ \frac{1}{\mathfrak{r}} = \frac{1}{r}\sum_{n=0}^{\infty}\left(\frac{r'}{r}\right)^n P_n(\cos\alpha) \]
Substitute that into the integral for V and you have the multipole expansion. It is exact — an infinite series, not an approximation — and it converges whenever r is bigger than the largest r' in the distribution.
This demystifies the expansion by showing it is one algebraic identity rather than a new idea.
Write the separation distance with the law of cosines, then factor out the large distance so that what remains inside involves the small ratio of source distance to field distance.
Expanding one over that square root in powers of the small ratio produces coefficients that are exactly the Legendre polynomials. That is where they come from, and it explains why the same polynomials keep turning up in every spherical problem in the book — not a coincidence, just what falls out of this geometry.
Substituting the expansion into the potential integral gives the multipole series directly. Note the convergence condition: r has to exceed the largest source distance, which is exactly the statement that you are outside the charge distribution.
Concept
\[ V_{\text{mon}} = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}, \qquad Q = \int\rho\,d\tau' \]
The n = 0 term. Since P₀ = 1, the integral is just the total charge, and this term is the field of a point charge with all that charge at the origin.
If the total charge is non-zero, this term dominates everything at large r and the distribution is, from far away, simply a point charge.
If the total charge is zero, this term vanishes entirely and the leading behaviour comes from the next term down. That is the interesting case — and the one Chapter 4 lives in, since atoms and molecules are neutral.
The leading term, and the simplest.
Since the zeroth Legendre polynomial is one, the integral is just the total charge, and the term is the potential of a point charge with all of it at the origin.
So if the total charge is non-zero, this term dominates at large distance and the distribution looks like a point charge. Which is why a distant charged sphere, cube or irregular blob all look identical.
The interesting case is when the total charge is zero, because then this term vanishes entirely and the leading behaviour comes from the next one down. That is the situation for atoms and molecules, which is why the next term matters so much in Chapter 4.
Concept
\[ V_{\text{dip}} = \frac{1}{4\pi\varepsilon_0}\frac{\mathbf{p}\cdot\hat{\mathbf{r}}}{r^2}, \qquad \mathbf{p} = \int \mathbf{r}'\,\rho(\mathbf{r}')\,d\tau' \]
dipole moment — A vector measuring how far apart the positive and negative charge sit, weighted by how much of each there is. Units: coulomb metres. For two point charges q and -q separated by a vector d pointing from the negative to the positive charge, p = qd.
Figure (svg): Two point charges separated by a distance, with the dipole moment vector drawn pointing from the negative charge to the positive one.
Watch the direction: p points from minus to plus, while the field between the charges points from plus to minus. They are opposite, and mixing them up flips every sign in Chapter 4.
The definition needs care, because the direction convention causes persistent errors.
The dipole moment is the charge-weighted average position — an integral of position times density. For two point charges it reduces to q times the separation vector, and that vector points from the negative charge to the positive one.
That direction is the opposite of the field between the charges, which runs from positive to negative. Mixing the two flips signs throughout Chapter 4, so it is worth saying explicitly: p points from minus to plus.
Units are coulomb metres, and a slide near the end of this deck gives a feel for typical magnitudes.
Worked example
Griffiths' Example 3.10. Two charges ±q separated by d; find the approximate potential far away.
Write the exact potential as a difference of two point-charge terms, in volts
Why: No approximation yet — this is exact for any distance.
\[ V = \frac{1}{4\pi\varepsilon_0}\left(\frac{q}{\mathfrak{r}_+} - \frac{q}{\mathfrak{r}_-}\right) \]
Use the law of cosines and expand for r much larger than d
Why: The two distances differ from r by about half of d cos theta each, in opposite senses.
\[ \mathfrak{r}_{\pm} \approx r \mp \frac{d}{2}\cos\theta \quad\Rightarrow\quad \frac{1}{\mathfrak{r}_{\pm}} \approx \frac{1}{r}\left(1 \pm \frac{d}{2r}\cos\theta\right) \]
Subtract, keeping only the leading surviving term
Why: The one-over-r pieces cancel exactly, because the total charge is zero. What is left is the dipole term.
\[ V \approx \frac{1}{4\pi\varepsilon_0}\frac{qd\cos\theta}{r^2} \]
Verify against the general dipole formula, in volts
Why: The dipole moment here is p = qd pointing along the axis from minus to plus, so p · r-hat = qd cos θ, which reproduces the result exactly. And the answer vanishes on the equatorial plane where cos θ = 0 — consistent with the equidistant-charges argument that V = 0 there.
This is a controlled approximation, and the control is the part worth attending to.
Write the exact potential first — a difference of two point-charge terms, valid at any distance. Then expand for a field point much further away than the separation.
The two distances differ by about d cos theta, and that small difference is the entire answer, because the leading one-over-r pieces cancel exactly. They cancel because the total charge is zero, which is the monopole term vanishing.
The verification against the general formula closes the loop, and the check that the potential vanishes on the equatorial plane is worth doing independently: there, both charges are equidistant, so the contributions cancel exactly with no approximation involved.
Worked example
Four charges at the corners of a square, alternating in sign. What survives at large distance?
Figure (svg): Four charges at the corners of a square: plus at top-left and bottom-right, minus at top-right and bottom-left.
Compute the monopole moment, in coulombs
Why: Just add the charges with their signs.
\[ Q = q - q - q + q = 0 \]
Compute the dipole moment, in coulomb metres
Why: Sum each charge times its position vector. The two positive charges sit at opposite corners, so their position vectors cancel against each other, and the same happens for the negatives.
\[ \mathbf{p} = \sum_i q_i\mathbf{r}_i = 0 \]
Conclude that the leading behaviour is quadrupole
Why: With the first two moments gone, the n = 2 term is the first survivor.
\[ V \sim \frac{1}{r^3}, \qquad E \sim \frac{1}{r^4} \]
Verify by checking a point far away on the diagonal, in volts
Why: Along a diagonal the two nearer charges have opposite signs and nearly cancel, as do the two further ones — and the residual difference falls off much faster than either a single charge or a dipole would. Physically the arrangement is 'neutral, and also symmetric enough to have no net separation of charge', so it takes a third-order description to say anything about it at all.
This shows the hierarchy in action, and it is a good test of whether the definitions have landed.
Add the charges: zero, so no monopole. Compute the charge-weighted positions: the two positives sit at opposite corners so their position vectors cancel, and the same happens for the negatives, so no dipole either.
With the first two moments gone, the leading behaviour is quadrupole — potential falling as one over r cubed, field as one over r to the fourth.
The physical reading is worth having. This arrangement is neutral, and also symmetric enough to have no net separation of charge. It takes a third-order description to say anything about it at all, which is precisely what 'quadrupole' means.
Concept
Take minus the gradient of the dipole potential in spherical coordinates and you get:
\[ \mathbf{E}_{\text{dip}}(r,\theta) = \frac{p}{4\pi\varepsilon_0 r^3}\big(2\cos\theta\,\hat{\mathbf{r}} + \sin\theta\,\hat{\boldsymbol{\theta}}\big) \]
Figure (svg): The dipole field pattern with the field along the axis drawn twice as long as the field at the same distance in the equatorial plane, in the opposite direction.
Chapter 4 uses this constantly: every polarised atom is a tiny dipole, and this is the field each one contributes.
Three features of this field are worth memorising, because it turns up constantly in Chapter 4.
It falls off as one over r cubed, faster than a point charge. On the axis it points along p and has twice the magnitude of the equatorial field at the same distance. And in the equatorial plane it points against p.
The factor of two between axial and equatorial is the detail most often got wrong, and the picture makes it visible.
Where it is going: every polarised atom in Chapter 4 is a tiny dipole, and this is the field each one contributes. The whole of the next deck is built by superposing enormous numbers of these.
Concept
The dipole moment of a charge distribution is not always a property of the distribution alone.
Figure (svg): A single positive charge shown twice: once at the origin with a zero dipole moment, and once displaced from the origin with a non-zero dipole moment.
Move the origin and the dipole moment changes — unless the total charge is zero, in which case p is genuinely origin-independent.
\[ \mathbf{p}_{\text{new}} = \mathbf{p}_{\text{old}} - Q\,\mathbf{a} \quad\text{(origin shifted by } \mathbf{a}\text{)} \]
So 'the dipole moment' is only a well-defined property of neutral objects. For charged ones it depends on your choice of origin, and the usual convention is to place the origin at the centre of the distribution.
This catches people out, so it is better met here than in a problem.
A single point charge at the origin has zero dipole moment. Move the origin and the same charge now has a non-zero one. So the dipole moment is not always a property of the distribution alone.
The rule is in the formula: shifting the origin changes p by the total charge times the shift. So if the total charge is zero, p is genuinely origin-independent and is a real property of the object.
Which means 'the dipole moment' is a well-defined phrase only for neutral objects — fortunate, since in Chapter 4 everything is neutral. For charged objects the convention is to put the origin at the centre of the distribution.
Comparison
Comparison matrix
| term | V falls off as | E falls off as | vanishes when |
|---|---|---|---|
| monopole | 1/r | 1/r² | total charge is zero |
| dipole | 1/r² | 1/r³ | charge centres coincide |
| quadrupole | 1/r³ | 1/r⁴ | the distribution is spherically symmetric |
The rule: use the first non-vanishing term. Everything after it is a correction that dies faster, and at large r those corrections are negligible.
A neutral atom has no monopole term, so its leading behaviour is dipole. A symmetric neutral atom has no permanent dipole either — until a field pulls its charges apart, which is exactly where Chapter 4 begins.
This grid is the summary, and it encodes the working rule.
The rule: use the first non-vanishing term. Everything after it dies faster and is negligible at any appreciable distance.
Read the last column, which says when each term disappears. Total charge zero kills the monopole. Coincident charge centres kill the dipole. Spherical symmetry kills the quadrupole.
Then look ahead. A neutral atom has no monopole term, so its leading behaviour is dipole — and a symmetric neutral atom has no permanent dipole either, until a field pulls its charges apart. That last sentence is exactly where Chapter 4 begins.
Estimation
A water molecule has a permanent dipole moment because its two hydrogens sit on the same side of the oxygen, at an angle of about 105 degrees.
Predict first
A water molecule's dipole moment is about 6.2 × 10⁻³⁰ C·m. If you modelled it as one elementary charge displaced by a distance d, how big would d be?
Correct: About 0.04 nm — a fraction of the molecule's size
Why: d = p/e = (6.2 × 10⁻³⁰)/(1.60 × 10⁻¹⁹) ≈ 3.9 × 10⁻¹¹ m, that is 0.039 nm. A water molecule is roughly 0.1 nm across, so this is a substantial fraction of its size — which is why water is such a strongly polar solvent and has such a high dielectric constant.
The unit chemists use is the debye, about 3.34 × 10⁻³⁰ C·m, so water comes in at roughly 1.85 D. Anything above about 1.5 D counts as strongly polar.
Chapter 4 is built on exactly this quantity: the dipole moment per unit volume of a material is its polarisation.
Numbers make the abstraction concrete, and this one is reassuring rather than surprising.
Water's dipole moment divided by the elementary charge gives an effective separation of about four hundredths of a nanometre. The molecule itself is around a tenth of a nanometre across, so the separation is a substantial fraction of its size.
Which is why water is so strongly polar, such a good solvent for salts, and has such a high dielectric constant — all of which the next deck explains properly.
The unit chemists actually use is the debye, worth recognising since it turns up everywhere else. Anything above about one and a half debye counts as strongly polar.
Real world
Discussion prompt
Why is the first non-vanishing multipole moment usually the only one anyone quotes?
These four show the expansion is a working tool across several fields rather than a textbook exercise.
The nuclear physics one is worth a moment, because it shows what the higher terms are actually for. A nucleus's monopole moment is just its charge, and its dipole moment vanishes by symmetry, so the quadrupole moment is the first thing carrying shape information — it tells you whether the nucleus is spherical or deformed.
The prompt makes the general principle explicit: the first non-vanishing moment is the one anyone quotes, because each successive term dies by another factor of r and at any real distance the first survivor dominates by orders of magnitude.
Pattern
| the problem | the technique | the clue |
|---|---|---|
| charge near a grounded plane or sphere | method of images | a simple grounded boundary |
| potential specified on a box or slot | separation of variables, Cartesian | flat walls, rectangular region |
| potential specified on a sphere | separation of variables, spherical | spherical boundary, angular data |
| field far from a localised blob | multipole expansion | you only need large-r behaviour |
| known charge with high symmetry | Gauss's law (Deck 2) | no induced charge to worry about |
| known charge, no symmetry | direct integration for V | everything is specified up front |
The single most useful question when you are stuck: do I know where all the charge is? If yes, integrate or use Gauss. If no, because some of it is induced, you are in Chapter 3 territory.
This is the map for the whole deck, and the slide to come back to when a problem will not start.
The most useful diagnostic question is at the bottom: do I know where all the charge is? If yes, integrate or use Gauss's law — that is Chapter 2. If not, because some of it is induced, you are in Chapter 3 territory and one of these four techniques applies.
Then the clue column picks between them. A simple grounded boundary means images. Flat walls with specified potentials mean Cartesian separation. A spherical boundary with angular data means spherical separation. Needing only the large-r behaviour means multipoles.
Check
Last question.
Check your understanding
A neutral molecule has its positive and negative charge centres separated by a small distance. At a distance far from the molecule, how does its electric field fall off?
Answer: A
Why: Neutral means the monopole term is zero, so the leading term is the dipole one. The dipole POTENTIAL falls off as 1/r² and its FIELD, being the gradient, falls off as 1/r³. This is why intermolecular forces are so much shorter-ranged than the force between ions.
This tests whether the hierarchy has been understood rather than memorised.
Neutral means no monopole, so the dipole term leads. The dipole potential falls as one over r squared, and the field, being its gradient, falls as one over r cubed.
That distinction between potential and field falloff is exactly what the wrong options probe, so it is worth being able to state both rather than just recognising the right answer.
The physical consequence is worth adding: it is why intermolecular forces are so short-ranged compared with the force between ions, which is monopole against monopole and reaches right across a solution.
Exit ticket
Discussion prompt
In one sentence each: what does the uniqueness theorem let you get away with, and why does the image charge have to sit outside your region?
Next deck: what happens when the space between your conductors is filled with matter that polarises.
Two questions, and they are a genuine check on whether the chapter's logic has landed.
Uniqueness lets you get away with guessing, because once a function satisfies Laplace's equation and matches every boundary value, it is provably the only one. Method does not matter; verification does.
The image must sit outside your region because inside it would be a real source — it would change the charge density there, and you would be solving Poisson's equation with charge that is not actually present.
Producing both of those without prompting is what lets you apply the method to a geometry you have not seen before, which is the real test.
Recap
| the result | the formula |
|---|---|
| mean value property | V at a point = average of V over any sphere around it |
| image for a grounded plane | −q at the mirror point |
| image for a grounded sphere | q' = −Rq/a at b = R²/a |
| induced charge on the plane | σ = −qd/[2π(s² + d²)^{3/2}], totalling −q |
| slot problem | V = (4V₀/π)Σ_odd (1/n)e^{−nπx/a}sin(nπy/a) |
| sphere in a uniform field | V = −E₀(r − R³/r²)cos θ |
| dipole potential and field | p·r̂/(4πε₀r²) ; falls as 1/r³ |
Next deck: Chapter 4, where matter itself becomes a sea of tiny dipoles.
Take stock of what has changed. You arrived able to compute fields from known charge. You leave able to handle the far more common situation where the charge is induced and unknown.
Go down the results table and check each entry is attached to a method rather than memorised as a formula — the image positions, the slot solution, the sphere in a uniform field, the dipole potential.
Then look ahead to the final deck. Everything so far has assumed empty space between the conductors. Real space is full of matter that polarises, and Chapter 4 is about what that matter does to every result derived so far.
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