Griffiths 2.3-2.5: the potential as a line integral of E, E = -grad V, Poisson and Laplace, boundary conditions at a surface charge, electrostatic energy and where it is stored, the five properties of conductors, and capacitance.
Subject: Electrodynamics (Griffiths) · 60 slides · diagram-first lesson
Open the interactive version of this deck · Homework for this lesson
Title
Griffiths, Chapter 2.3-2.5
One scalar function instead of three — plus energy, conductors, and capacitors
Objectives
Zero curl was the licence. This deck spends it: three vector components collapse into one scalar function you can actually solve for.
This is the deck the rest of the course leans on. Chapter 3 is entirely about solving for V; Chapter 4 is about what V does inside matter.
The last deck proved two things about the electrostatic field: its divergence is the charge density over epsilon-zero, and its curl is exactly zero everywhere. The second of those has not been spent yet, and this deck is where it gets cashed in.
Zero curl means the field is conservative, and a conservative field is the gradient of some scalar. So instead of chasing three component functions of position, you can chase one number at each point and differentiate at the end. That is not a small convenience — it is the difference between problems you can do and problems you cannot.
Concretely: by the end of this deck you will compute the field of a charged sphere without a single vector integral, and you will know why the inside of a car is a safe place to be during a lightning storm.
Section
Section 1
Intuition
A vector field is three functions of three variables. A scalar field is one. If you can trade the first for the second, you should.
Figure (svg): On the left, three separate grids of numbers labelled Ex, Ey and Ez. On the right, a single grid labelled V, with an arrow showing that the gradient recovers all three.
The trade is only legal because ∇ × E = 0. A general vector field cannot be written this way — it takes three functions to say three things.
And the payoff compounds: scalars add as numbers, so superposition of potentials needs no vector diagrams at all.
The honest motivation is bookkeeping, and it is worth being blunt about that rather than dressing it up.
A vector field is three separate functions of x, y and z. To specify the electric field everywhere, you must supply three numbers at every point in space. A scalar field supplies one. The picture is the whole argument: three grids of numbers collapsing into one, with the gradient as the arrow that expands it back out again. Nothing is lost in the compression, because you can always recover all three components by differentiating.
The compression is legal only because the curl vanishes. A general vector field genuinely needs three functions; it is a special property of electrostatic fields that one suffices. If you are wondering why the same trick is not used for the magnetic field, that is the reason — its curl is not zero, which is why Chapter 5 needs a vector potential instead of a scalar one.
The second payoff is the one you will feel in every problem set. Potentials are numbers, so superposing them is ordinary addition with signs. No components, no diagrams, no resolving anything into x-hat and y-hat.
Concept
\[ V(\mathbf{r}) \equiv -\int_{\mathcal{O}}^{\mathbf{r}} \mathbf{E}\cdot d\mathbf{l} \]
Figure (svg): A reference point at infinity on the left, a field point on the right, and two different wandering paths between them marked as giving the same answer.
The path-independence is not an assumption; it is the zero-curl result from Deck 2. Without it this definition would not even be well posed.
The reference point O is your choice. Change it and every potential shifts by the same constant — which changes nothing physical, because only differences in V are measurable.
\[ V(\mathbf{b}) - V(\mathbf{a}) = -\int_{\mathbf{a}}^{\mathbf{b}} \mathbf{E}\cdot d\mathbf{l} \]
Read this definition slowly, because every symbol in it is doing work. V at a point is minus the line integral of E, taken from some reference point out to the point you care about. In words: it is the work per unit charge you would have to do to carry a test charge from the reference point to here, against the field.
The picture shows two wildly different wandering paths between the same two points, and the claim is that both give the same number. That is neither obvious nor free — it is exactly the zero-curl result from the last deck. If the field had circulation, the two paths would disagree, and 'the' potential at a point would not exist as a well-defined thing.
The reference point is your choice, like deciding where sea level is on a map. Move it and every potential in the problem shifts by the same constant, and nothing physical changes, because only differences in V ever show up in a measurement. That is why the second equation — the potential difference between two points — is the one that is genuinely observable.
Notation
Annotate
On: \( V(\mathbf{r}) = -\int_{\mathcal{O}}^{\mathbf{r}} \mathbf{E}\cdot d\mathbf{l} \)
Potential is not potential energy. Multiply by a charge and you get energy: U = QV, in joules. V itself is energy per unit charge, in joules per coulomb — that is, volts.
The minus sign deserves real attention, because it is the single most common source of sign errors in this chapter and it is pure convention rather than physics.
It is chosen so that the potential behaves like height in a gravitational field. With the minus sign in place, a positive charge released from rest moves from high V toward low V — it rolls downhill, exactly like a ball. Drop the minus sign and every one of those intuitions inverts.
The second note is the practical one. The integral runs from the reference point to the field point. Writing the limits the other way round picks up a spurious sign, and it happens constantly. Say it out loud each time: from the reference, to the point I care about.
Finally, keep potential and potential energy apart, because the words are nearly identical and the quantities are not. V is joules per coulomb — energy per unit charge, measured in volts. Multiply by an actual charge and you get an energy in joules. Confusing the two produces answers that are off by a factor of the charge, and it is a hard error to spot after the fact.
Concept
The fundamental theorem for gradients, run backwards, inverts the definition:
\[ \mathbf{E} = -\nabla V \]
Figure (svg): An animation showing a test charge circling along one equipotential doing no work, while the field streams outward across the equipotentials.
So a single scalar carries all the information in the vector field. Solve for V, differentiate, and you are done.
The relationship also explains the geometry: gradients are perpendicular to level surfaces, so E is always perpendicular to equipotentials.
This is the inverse of the definition, and it is what makes the whole strategy pay. Having compressed three functions into one, you get all three back by differentiating.
Work through the picture rather than the formula. The circles are equipotentials — surfaces on which V is constant — and the arrows are the field. Two features are worth noticing. The arrows cross the circles at right angles, always and everywhere, which is a general property of gradients: a gradient is perpendicular to the level surfaces of its function. And the arrows point from the crowded region toward the sparse one, that is, from high V to low V, which is the minus sign showing up visually.
There is a practical reading too. Where the equipotentials bunch together, V changes rapidly over a short distance, so the gradient is large and the field is strong. Where they spread apart, the field is weak. A contour map of a mountain works in exactly the same way, and you almost certainly already have that intuition — it just needs connecting.
Picture it
Figure (svg): An animation of a potential landscape with a positive charge sliding down the slope while a negative charge climbs it.
The last point is the one that trips people. Potential energy is QV, so for a negative Q the landscape is effectively flipped upside down.
This is the mental model worth making permanent, because nearly every qualitative question about potential can be answered from it with no algebra at all.
The animation shows both cases. Release a positive charge and it slides downhill, trading potential energy for kinetic energy exactly as a ball would. Release a negative charge on the same slope and it climbs — which looks wrong until you remember that its energy is Q times V with Q negative, so the landscape it actually experiences is the mirror image of the one drawn.
The four bullets each answer a question that turns up on exams. High V is a hilltop. A steep slope is a strong field — note carefully that strength is the steepness, not the height, so a large V does not imply a large E. Flat means zero field, since the gradient of a constant is zero. And the negative-charge case is the one that catches people out.
If any of that feels shaky, try this: point at where on the landscape the field is strongest, then point at where V is largest. They are different places, and noticing that is usually the moment the distinction lands.
Worked example
The result you will reuse more than any other in this course.
Start from the field of a point charge, in newtons per coulomb
Why: Purely radial, so a radial path makes the dot product trivial — and path independence says a radial path is as good as any other.
\[ \mathbf{E} = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\hat{\mathbf{r}} \]
Integrate in from infinity to radius r, in volts
Why: Taking the reference point at infinity is the standard choice for any charge distribution of finite extent.
\[ V(r) = -\int_{\infty}^{r}\frac{1}{4\pi\varepsilon_0}\frac{q}{r'^2}\,dr' = -\frac{q}{4\pi\varepsilon_0}\left[-\frac{1}{r'}\right]_{\infty}^{r} \]
\[ V(r) = \frac{1}{4\pi\varepsilon_0}\frac{q}{r} \]
Verify by differentiating back, in newtons per coulomb
Why: Minus the gradient of q/(4πε₀r) is plus q/(4πε₀r²) r-hat, because the derivative of 1/r is minus 1/r². The two minus signs cancel and the original field returns — which is the check worth doing every time.
Notice the exponent: the field falls off as one over r squared, but the potential falls off only as one over r.
This is the single most reused result in the course, so it is worth deriving rather than taking on trust. Everything later — the multipole expansion, the potential of a shell, the energy of an assembly — is built by superposing copies of this one formula.
The mechanics are easy because the field is radial: a radial path makes the dot product trivial, since E and the step are parallel the whole way, and path independence guarantees any other route would give the same answer. Putting the reference at infinity is the standard choice for any charge distribution of finite size, because the potential naturally falls to zero out there.
The verification step is the habit to install. Differentiate the answer back and check you recover the field you started from. The derivative of one over r is minus one over r squared, and that minus cancels against the minus in E equals minus grad V, leaving the correct positive radial field. Two minus signs cancelling is exactly where errors hide, so this check is cheap insurance every time.
Look at the exponents before moving on, because this is the fact most often mangled: the field falls off as one over r squared, but the potential falls off only as one over r. The potential reaches further.
Concept
Figure (svg): Four charges around a field point, each labelled with a number, and the potential shown as a plain arithmetic sum of four numbers.
\[ V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\sum_i \frac{q_i}{\mathfrak{r}_i}, \qquad V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\int\frac{\rho(\mathbf{r}')}{\mathfrak{r}}\,d\tau' \]
Compare this with the same calculation for E: separation vectors, components, cancellations. Here it is a sum of numbers with signs.
This is why the standard strategy for a hard field problem is: compute V by integration, then differentiate. One scalar integral beats three vector ones.
This is where the strategy justifies itself, so it is worth not rushing.
Imagine computing the field of the four charges in the picture directly: four separation vectors, four sets of components, careful attention to directions, and a vector sum at the end. Now the potential version: four numbers, each with a sign, added together. That is the entire calculation.
Which gives the standard strategy for any hard field problem where symmetry fails. Compute V by integration — one scalar integral — and then take the gradient at the end. One easy integral plus a differentiation beats three hard integrals every time.
One honest caveat. The differentiation at the end can be fiddly, and you need V as a function of position to differentiate it. Computing V only at the single point you care about and then trying to extract E from it will leave you stuck, so keep the position variable alive through the integral.
Trap
Finding the potential midway between a positive and an equal negative charge, a student reasons by analogy with the field:
Note that the two fields add up there, so the potentials must add up too
Why: Between the charges the two field contributions point the same way and reinforce, so it looks as though everything reinforces.
\[ V \overset{?}{=} \frac{1}{4\pi\varepsilon_0}\left(\frac{q}{d/2} + \frac{q}{d/2}\right) = \frac{q}{\pi\varepsilon_0 d} \]
But V carries the sign of the charge, and the two charges have opposite signs.
Add potentials with their signs, not with their directions:
\[ V = \frac{1}{4\pi\varepsilon_0}\left(\frac{+q}{d/2} + \frac{-q}{d/2}\right) = 0 \]
Compare against the field at the same point, in newtons per coulomb
Why: The field there is NOT zero — both contributions point from the positive charge toward the negative one and add. So V = 0 and E is large, at the same point.
The lesson: V = 0 does not mean E = 0, and E = 0 does not mean V = 0. The field is the slope of V, and a function can pass through zero while sloping steeply, or be large and perfectly flat.
The error is reasoning by analogy, and it is worth naming as such. It correctly notices that between a positive and a negative charge the two field contributions reinforce, then assumes the potentials must reinforce too. They do not, because V carries the sign of the charge rather than a direction.
The punchline is the pair of statements at the bottom, and it feels paradoxical the first time. At the midpoint of a dipole, V is exactly zero while E is large. Zero potential does not mean zero field.
The landscape picture stops it being paradoxical. V is the height and E is the slope. A hillside crossing sea level is at zero height while being extremely steep. Height and slope are independent, and neither one constrains the other.
The reverse case is worth holding alongside it: inside a charged spherical shell, E is zero everywhere while V is a large non-zero constant. Flat ground, high up a mountain. Between them, those two examples close off both directions of the confusion.
Prediction
Predict first
Moving from far away toward an isolated negative charge, with the reference at infinity, what does the potential do?
Correct: Decreases toward minus infinity
Why: V = q/(4πε₀r) carries the sign of q. With q negative, V is negative everywhere and becomes more negative as r shrinks. Physically: a positive test charge is attracted inward, so it rolls DOWNHILL as it approaches — meaning the potential must be falling.
Figure (svg): A graph of potential against distance showing a curve that plunges to minus infinity near the origin and approaches zero far away.
A quick sanity rule: near a positive charge V is large and positive; near a negative charge V is large and negative. If your answer breaks that rule, a sign has been lost.
Commit before revealing — the wrong answer here is instructive rather than careless.
The formula settles it immediately: V equals q over four-pi-epsilon-zero r, and with q negative the whole thing is negative, becoming more negative as r shrinks. But the physical argument is the one worth having, because it still works when the formula is not to hand. A positive test charge is attracted toward a negative charge, so it moves inward on its own — and things move spontaneously downhill. So the potential must be falling as you approach.
A quick sanity rule to carry forward: near a positive charge V is large and positive, near a negative charge V is large and negative. Any answer that breaks that rule has lost a sign somewhere.
The graph rewards a look too. V dives toward minus infinity at the charge, not toward zero, and it approaches zero only far away — which is the reference-point choice showing up on the axis.
Socratic
Nothing forces it. Griffiths could have defined V as plus the line integral and the physics would be identical. So why the minus?
Discussion prompt
Suppose the minus sign were dropped. What would then happen to a positive charge released from rest at a point of high V?
So the minus sign is a convention chosen to make the gravitational analogy work. Keep it, and every energy statement in the book reads the way you expect.
This gets asked a lot and rarely answered properly. The honest answer is that nothing in the physics forces it. Griffiths could have defined V with a plus sign and every prediction would come out identical, provided the sign were carried consistently everywhere after.
What the convention buys is that the gravitational analogy survives intact. With the minus sign, positive charges roll downhill in V, high potential means stored energy for a positive charge, and potential energy decreases as things speed up. All of which matches the intuitions already built in mechanics.
It is worth sitting with the counterfactual for a moment. Without the minus sign, a positive charge released from rest would accelerate toward higher V, and the phrase 'potential energy' would be actively misleading. A convention that preserves intuition is not arbitrary in any practical sense, even if it is arbitrary in principle.
Worked example
A result Chapter 3 uses repeatedly, and a clean drill on splitting an integral at a boundary.
Recall the field from Deck 2, in newtons per coulomb
Why: Gauss's law gave a field growing linearly inside and falling as one over r squared outside.
\[ E(r) = \begin{cases} \dfrac{1}{4\pi\varepsilon_0}\dfrac{qr}{R^3}, & r<R \\[8pt] \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}, & r>R\end{cases} \]
Outside is immediate, in volts
Why: The exterior field is identical to a point charge's, so the exterior potential is too.
\[ V(r) = \frac{1}{4\pi\varepsilon_0}\frac{q}{r} \qquad (r>R) \]
Inside, integrate in two pieces: infinity to R, then R to r
Why: The field changes form at the surface, so the integral must be split there. This is the same manoeuvre as the shell, but now the inside field is not zero.
\[ V(r) = -\int_{\infty}^{R}\frac{q\,dr'}{4\pi\varepsilon_0 r'^2} - \int_{R}^{r}\frac{q r'\,dr'}{4\pi\varepsilon_0 R^3} \]
\[ V(r) = \frac{q}{4\pi\varepsilon_0}\left[\frac{1}{R} - \frac{r^2-R^2}{2R^3}\right] = \frac{q}{8\pi\varepsilon_0 R}\left(3 - \frac{r^2}{R^2}\right) \]
Verify at both ends, in volts
Why: At r = R the bracket gives 3 − 1 = 2, so V = q/(4πε₀R) — matching the exterior formula exactly at the surface. At the centre, V = 3q/(8πε₀R), which is one and a half times the surface value: the potential peaks at the centre, as it must, since the field points outward everywhere inside.
This example drills one specific skill: splitting an integral at a boundary where the integrand changes form. That skill comes back constantly, so the reason is worth making explicit rather than treating as a formality.
The field here has two different expressions — growing linearly inside, falling as one over r squared outside. Integrating in from infinity to a point inside means passing through the surface at r equals R, where the integrand changes. Using one formula across the whole range is simply wrong, and it is the standard error on this problem.
The outside case is immediate: outside a spherically symmetric ball the field is identical to a point charge's, so the potential is too. Inside needs the two-piece integral.
Check the result carefully. At r equals R the bracket gives three minus one, so the inside and outside formulas agree exactly at the surface — V is continuous, as it must be. And the centre value is three-halves of the surface value, so the potential peaks at the centre. Does that make sense? The field points outward everywhere inside, so you are always walking downhill as you move out from the centre, which means the centre has to be the high point.
Picture it
Figure (svg): Left: concentric circular equipotentials around a point charge. Right: the figure-of-eight equipotentials of a dipole, with the flat zero-potential plane between the charges.
The dipole's zero-potential plane is worth noting: V = 0 all the way along it, yet the field there is large and points from the positive charge toward the negative one.
Equipotentials are the contour lines of the potential landscape, and this is about learning to read them.
Take the four properties one at a time. They never cross, because a point cannot be at two potentials at once — the same reason contour lines on a map never cross. They are perpendicular to E everywhere, because moving along one costs no work. Their spacing encodes field strength, so crowded means strong. And the surface of any conductor is one of them, which is the property that makes the whole of Chapter 3 possible.
The dipole picture on the right rewards a closer look. The flat plane down the middle sits at V equals zero everywhere, because every point on it is equidistant from the two charges and their contributions cancel exactly. But the field on that plane is large, pointing from the positive charge toward the negative one. That is the same zero-V-with-non-zero-E lesson as the earlier trap, now visible geometrically.
Worth asking yourself: if you carried a test charge along one of these surfaces from one side to the other, how much work would you do? None — and that is what 'equipotential' means in practice.
Concept
Setting V to zero somewhere is a choice, like choosing where sea level is. Three conventions cover almost everything you will meet:
| situation | usual reference | why |
|---|---|---|
| any finite charge distribution | infinity | V naturally goes to zero there |
| an infinite wire or plane | some finite distance | the integral to infinity diverges |
| circuits and conductors | the earth (ground) | it is a real, physically fixed conductor |
Only differences in V are physical. Add 100 volts to every potential in a problem and the field, the forces and every measurement are unchanged.
\[ \mathbf{E} = -\nabla(V + \text{constant}) = -\nabla V \]
The reference is a choice, not a fact, and choosing it badly makes problems harder than they need to be.
Three conventions cover almost everything. Infinity is the default for any charge distribution of finite extent, because V naturally tends to zero there and the choice costs nothing. For infinite objects — a wire, a plane — the integral to infinity diverges and a finite reference is needed instead. And in circuits the earth is used, because it is a physical conductor whose potential really is held fixed.
Then the invariance, made concrete: add a hundred volts to every potential in a problem and nothing measurable changes, because the gradient of a constant is zero and the field is untouched. That is exactly why a bird on a high-voltage wire is unharmed. It sits at a very high potential relative to the ground, but every part of it is at nearly the same potential, so there is no field driving current through it.
Edge cases
Try to set the reference at infinity for an infinite charged wire and something breaks.
\[ V(s) = -\int_{\infty}^{s}\frac{\lambda}{2\pi\varepsilon_0 s'}\,ds' = -\frac{\lambda}{2\pi\varepsilon_0}\big[\ln s'\big]_{\infty}^{s} = \infty \]
The logarithm diverges. The trouble is that an infinite wire has infinite total charge, so 'far from all the charge' is a place that does not exist.
Discussion prompt
What should you do instead, and does the physics actually depend on the choice?
The same caution applies to the infinite plane. Both are idealisations of finite objects that are perfectly well behaved at large distances.
Worth knowing about so that the first divergent integral does not cause a panic. The divergence is real, but it is a symptom of an idealisation rather than a physical catastrophe.
For an infinite wire the field falls off as one over s, and integrating one over s gives a logarithm, which diverges as s goes to infinity. The physical reason is that an infinite wire carries infinite total charge, so 'far away from all the charge' names a place that does not exist.
The repair is to choose a finite reference distance and quote potentials relative to it. The potential difference between any two points is then finite, well defined, and completely independent of the reference chosen — which the log-of-a-ratio form makes obvious.
The general lesson: when an idealised problem produces an infinity, ask which idealisation is responsible before doubting the physics. Here it is 'infinitely long', and any real wire is perfectly well behaved.
Worked example
Griffiths' Example 2.6. A shell of radius R carrying total charge q, with the reference at infinity.
Figure (svg): A spherical shell with the field zero inside and radial outside, and beneath it a graph of V that is flat inside and falls as 1/r outside.
Outside: integrate the point-charge field in from infinity, in volts
Why: Gauss's law already told you that outside a shell the field is identical to a point charge's, so the potential must be too.
\[ V(r) = -\int_{\infty}^{r}\frac{q}{4\pi\varepsilon_0 r'^2}\,dr' = \frac{1}{4\pi\varepsilon_0}\frac{q}{r} \qquad (r > R) \]
Inside: split the integral at the shell, using the correct field in each region
Why: This is the step that must not be skipped. From infinity to R the field is the point-charge field; from R inward it is exactly zero.
\[ V(r) = -\int_{\infty}^{R}\frac{q}{4\pi\varepsilon_0 r'^2}\,dr' - \int_{R}^{r}(0)\,dr' = \frac{1}{4\pi\varepsilon_0}\frac{q}{R} \qquad (r < R) \]
Verify at the boundary and inside, in volts
Why: At r = R both expressions give q/(4πε₀R), so V is continuous across the shell. Inside, V is a non-zero CONSTANT — and since the gradient of a constant is zero, E = 0 inside, exactly as Gauss's law said. Zero field does not mean zero potential; it means unchanging potential.
This one delivers the most counter-intuitive result in the section, so it is worth building to deliberately.
Outside is quick: Gauss's law already established that the exterior field is identical to a point charge's, so the exterior potential is too. Inside is where the interesting part happens.
The inside calculation must be split at the shell. From infinity in to R, the field is the point-charge field and contributes. From R inward, the field is exactly zero and contributes nothing. So the potential inside equals whatever you accumulated on the way to the surface — and then stops changing.
Now the punchline. Inside the shell the field is zero but the potential is not; it is a non-zero constant. That is genuinely confusing at first, and the landscape picture is the resolution: zero field means flat ground, and flat ground can sit at any altitude you like. Zero field means the potential is unchanging, not that it is zero.
The gradient makes it consistent from the other direction. V is constant inside, the gradient of a constant is zero, so E is zero inside. Each statement implies the other.
Tweak it
Move the shell radius and watch both curves. The field jumps at the surface; the potential does not.
Parameter explorer
Slide R and compare: where does E jump, and where does V merely bend? What physical thing produces the jump?
\[ V(r) = \begin{cases} 1/{R}, & r < {R} \\ 1/r, & r > {R}\end{cases} \]
A kink in V means a discontinuity in E, and a discontinuity in E means surface charge. That is the subject of the next section, in one sentence.
Drive this one yourself — the point is felt rather than derived.
As the shell radius moves, watch two different things. The field jumps discontinuously at the surface: zero on the inside, a finite value just outside. The potential does not jump. It merely changes slope, bending from flat to falling.
That contrast is the whole content of the next section, previewed. A discontinuity in E requires surface charge to produce it, and the shell has surface charge. Meanwhile V is a line integral of a finite field over a vanishing distance, so it has no way to jump — it can only kink.
The diagnostic worth carrying away: a kink in the potential graph means surface charge lives at that point. Exactly that reasoning is what finds induced charge on conductors in the next deck.
Concept
Two equal and opposite charges separated by d. Add their potentials as numbers and watch the leading term cancel.
\[ V = \frac{1}{4\pi\varepsilon_0}\left(\frac{q}{\mathfrak{r}_+} - \frac{q}{\mathfrak{r}_-}\right) \]
Figure (svg): A dipole with a distant field point, showing the two nearly parallel lines to the charges and the small path difference between them.
Expand for a field point much further away than the separation, in volts
Why: The two distances differ by about d cos theta, and that small difference is all that survives the subtraction.
\[ V \approx \frac{1}{4\pi\varepsilon_0}\frac{qd\cos\theta}{r^2} = \frac{1}{4\pi\varepsilon_0}\frac{p\cos\theta}{r^2} \]
So a dipole's potential falls as one over r squared, not one over r, and its field as one over r cubed. The leading one-over-r term cancelled because the total charge is zero.
That cancellation is the first term of the multipole expansion, which the next deck develops in full — and it is why neutral atoms still attract each other in Chapter 4.
This previews the multipole expansion and explains a fact that turns up everywhere in chemistry and molecular physics.
Two equal and opposite charges. Superpose their potentials — as signed numbers — and far away you have a difference of two nearly equal terms. The leading one-over-r pieces cancel exactly, because the total charge is zero, and what survives is the small difference between the two distances.
Geometrically that difference is about d cos theta, which is why the answer carries a cosine and why it vanishes on the equatorial plane. In the figure, far away the two lines to the charges are nearly parallel, and the path difference is the projection of the separation onto that direction.
The consequence worth keeping: a dipole's potential falls off as one over r squared and its field as one over r cubed. That is why neutral molecules attract each other only at short range, while ions interact right across a solution.
Concept
Put the two halves together. Take the divergence of E = −∇V and use Gauss's law:
\[ \nabla\cdot\mathbf{E} = \nabla\cdot(-\nabla V) = -\nabla^2 V = \frac{\rho}{\varepsilon_0} \]
\[ \nabla^2 V = -\frac{\rho}{\varepsilon_0} \qquad \text{(Poisson's equation)} \]
Where there is no charge, the right side vanishes and you get Laplace's equation — the single most important differential equation in this book.
\[ \nabla^2 V = 0 \qquad \text{(Laplace's equation)} \]
All of Chapter 3 is a toolbox for solving that one equation subject to boundary conditions. Everything in this deck is what makes it worth solving.
This is where the two halves of the theory join up.
The derivation is two substitutions. Take the divergence of E equals minus grad V, which gives minus the Laplacian of V. Set that equal to Gauss's law's right-hand side. Rearrange, and you have Poisson's equation.
Then look at what happens in empty space. With no charge, the right-hand side vanishes and you are left with Laplace's equation. That innocuous-looking equation is the most important one in the book, and the whole of the next deck is a toolbox for solving it.
Why it matters so much: in most real problems you do not know where the charge is, because it is induced on conductors and depends on the very field you are solving for. Laplace's equation sidesteps that entirely — solve the differential equation subject to boundary conditions instead of integrating over an unknown charge distribution.
Pattern
Figure (svg): A triangle with charge density at the top, potential at the bottom left and field at the bottom right, with labelled arrows connecting each pair in both directions.
| you know | you want | the cheapest route |
|---|---|---|
| ρ, high symmetry | E | Gauss's law directly |
| ρ, no symmetry | E | integrate for V first, then take −∇V |
| E | V | line integral from a reference point |
| V | E | take the negative gradient |
| V | ρ | Poisson: ρ = −ε₀∇²V |
The row worth memorising is the second: when symmetry fails, going through the potential turns three hard integrals into one easy one.
This triangle is the navigation chart for the whole of electrostatics, and knowing it is what stops you getting stuck at the setup stage.
Walk each arrow. Charge to field directly: Gauss's law if there is symmetry, Coulomb integration if not. Charge to potential: a scalar integral, always available and usually easier. Potential to field: differentiate. Field to potential: integrate along a path. Potential back to charge: Poisson's equation.
The row that matters most in practice is the second in the table. When symmetry fails and Gauss's law is useless, going the long way round through the potential turns three hard vector integrals into one manageable scalar one. That is the standard rescue.
Before starting any problem, ask three things: what do I know, what do I want, and which arrow connects them? Most of the difficulty in these problems is choosing the route, not executing it.
Check
Think it through first.
Check your understanding
At a certain point in space the electric field is zero. What can you say about the potential there?
Answer: A
Why: E is minus the gradient of V, so E = 0 means the gradient vanishes: V is momentarily flat. Its actual value depends entirely on the reference point and could be anything. Inside a charged spherical shell, for instance, E is zero everywhere while V is a large non-zero constant.
The trap here is the reflex that zero field implies zero potential. It does not, and the charged shell earlier in this deck is the standing counterexample.
The correct reasoning: E is minus the gradient of V, so E equals zero says the gradient is zero, which says V is locally flat. Flat says nothing at all about the value — ground can be level at any altitude.
Option C is worth a thought even if you did not pick it, because it previews the next deck. A vanishing gradient could in principle be a maximum, a minimum or a saddle — but in a charge-free region Laplace's equation forbids maxima and minima outright, so it has to be a saddle.
Section
Section 2
Concept
Figure (svg): A charged sheet with a strong field above it and a weaker field below, and a small pillbox straddling it showing the difference.
\[ E_{\text{above}}^{\perp} - E_{\text{below}}^{\perp} = \frac{\sigma}{\varepsilon_0} \]
Only the perpendicular component jumps. The parallel component is perfectly continuous across any surface charge.
\[ \mathbf{E}_{\text{above}}^{\parallel} = \mathbf{E}_{\text{below}}^{\parallel} \]
Short section, used constantly, so the two statements need to stay clearly distinct.
Only the perpendicular component of E jumps, and it jumps by exactly sigma over epsilon-zero. The parallel component is perfectly continuous — it does not care that there is charge on the surface at all.
The picture is the argument in miniature: a strong field above, a weaker one below, and the difference between them set by how much charge sits on the sheet. More charge, bigger jump.
A useful anchor is the isolated charged plane from the last deck: sigma over two-epsilon-zero pointing up above it, the same magnitude pointing down below. The difference in the perpendicular components is sigma over epsilon-zero, exactly as this boundary condition requires. Checking a general rule against a case you already know is a habit worth keeping.
Worked example
Two tiny surfaces, two of Deck 1's theorems, and both conditions fall out.
For the perpendicular jump, apply Gauss's law to a thin pillbox straddling the surface
Why: Shrink the pillbox's height to zero so its sides contribute nothing. Only the two lids remain, each of area A in square metres.
\[ (E^{\perp}_{\text{above}} - E^{\perp}_{\text{below}})A = \frac{\sigma A}{\varepsilon_0} \]
Cancel the arbitrary area
Why: The area A was never physical — it must cancel, and it does.
\[ E^{\perp}_{\text{above}} - E^{\perp}_{\text{below}} = \frac{\sigma}{\varepsilon_0} \]
For the parallel component, run a thin rectangular loop straddling the surface
Why: The electrostatic field has zero curl, so the line integral around any closed loop vanishes. Shrink the loop's short ends to nothing and only the two long sides survive.
\[ \oint\mathbf{E}\cdot d\mathbf{l} = (E^{\parallel}_{\text{above}} - E^{\parallel}_{\text{below}})\,l = 0 \]
Figure (svg): A charged surface with a flat pillbox drawn straddling it, and beside it a thin rectangular loop also straddling it.
Verify against a case you already know, in newtons per coulomb
Why: For an isolated charged plane the field is σ/(2ε₀) upward above and σ/(2ε₀) downward below. The difference in the perpendicular components is σ/(2ε₀) − (−σ/(2ε₀)) = σ/ε₀, exactly as the boundary condition requires.
Both conditions come from tools already in hand — this is not new machinery, it is the same two theorems applied to very small surfaces.
For the perpendicular condition, use a pillbox straddling the surface and apply Gauss's law. Shrink its height to zero so the sides contribute nothing, leaving the two lids. The enclosed charge is sigma times the lid area, and the arbitrary area cancels, which is always a good sign.
For the parallel condition, use a thin rectangular loop straddling the surface and exploit zero curl, so the loop integral vanishes. Shrink the short ends to nothing and only the two long sides survive, forcing the parallel components to be equal.
The pattern recurs throughout electromagnetism, so it is worth noticing: a pillbox plus a divergence law gives the perpendicular condition, and a loop plus a curl law gives the parallel one. The same two constructions produce the magnetic boundary conditions in Chapter 5 and the dielectric ones in Chapter 4.
Concept
\[ V_{\text{above}} = V_{\text{below}} \]
V is a line integral of a finite field over a vanishing distance, so it cannot jump. Its normal derivative can, and does:
\[ \frac{\partial V_{\text{above}}}{\partial n} - \frac{\partial V_{\text{below}}}{\partial n} = -\frac{\sigma}{\varepsilon_0} \]
Figure (svg): A graph of potential across a charged surface: the curve is unbroken but has a sharp kink where the surface sits.
Read this as: a kink in V is surface charge. It is the diagnostic you will use throughout Chapter 3 to find induced charge on conductors.
The reasoning behind continuity is short. V is the line integral of E across the surface, the field is finite, and the distance is shrinking to zero. A finite thing integrated over nothing gives nothing, so V cannot jump.
What can and does jump is the normal derivative, and the graph makes it visible: an unbroken line with a sharp corner in it. Keep the difference between a jump and a kink clear — a jump is a break in the curve, a kink is a break in its slope.
The diagnostic is the takeaway, and it is genuinely useful: a kink in V means surface charge is present at that point, and the size of the kink tells you how much.
This is also the practical route to induced charge on a conductor. Measure or compute the field just outside the surface and you have read off the local surface charge density directly.
Comparison
Everything from Deck 2 and this section, in one grid. Any electrostatics problem is an exercise in choosing which cell to start from.
Comparison matrix
| differential form | integral form | |
|---|---|---|
| Gauss | ∇ · E = ρ/ε₀ | flux = Q_enc/ε₀ |
| no circulation | ∇ × E = 0 | loop integral of E = 0 |
| potential | E = −∇V | V = −∫E · dl |
| Poisson | ∇²V = −ρ/ε₀ | V = (1/4πε₀)∫ρ/𝔯 dτ |
The bottom row is the crucial pair: the differential form is what Chapter 3 solves with boundary conditions, while the integral form is what you use when you already know where all the charge is.
Use this grid as a checkpoint. Everything from the last two decks is in it, and filling the blanks unaided is a good sign the structure has landed.
The top two rows are the physics: charge makes the field diverge, and the field has no circulation. The two below are the machinery built on top — the potential, and Poisson's equation relating it back to the charge.
The left and right columns are the same statements in different clothing. The differential forms are local, saying what happens at a point, and they are what you solve when you have boundary conditions. The integral forms are global, and they are what you use when symmetry lets you evaluate a flux or a loop integral by inspection.
Every remaining problem in Chapters 2 and 3 amounts to picking a cell in this grid and starting there.
Trap
A student computes V on both sides of a charged plate and, finding the same value, concludes the plate is uncharged:
Note that V is the same just above and just below, so nothing is there
Why: Charge produces fields, fields produce potential differences — so equal potentials look like an absence of charge.
\[ V_{\text{above}} = V_{\text{below}} \;\overset{?}{\Rightarrow}\; \sigma = 0 \]
But V is continuous across every surface charge, however large. This test can never detect anything.
Look at the derivative, not the value:
Compare the normal derivatives of V on the two sides, in volts per metre
Why: The value of V is continuous; it is the slope that carries the information about surface charge.
\[ \sigma = -\varepsilon_0\left(\frac{\partial V_{\text{above}}}{\partial n} - \frac{\partial V_{\text{below}}}{\partial n}\right) \]
Equivalently, since E = −∇V: measure the field just outside a conductor and you have read its local surface charge directly. Chapter 3 uses this constantly to find induced charge distributions.
The test in the wrong column is not merely mistaken, it is useless — V is continuous across every surface charge no matter how large, so a test based on it can never detect anything.
The fix is to look at the derivative rather than the value. The value of V is continuous; its slope is not, and the discontinuity in the slope is proportional to the surface charge.
Tie it back to the graph from two slides ago: you are looking for a corner, not a break. Sketching V across a charged plate and marking the corner is a good way to make that stick.
The corrected rule in its practical form is the one you will use in the next deck constantly: sigma equals epsilon-zero times the field just outside a conductor. Measuring the field at the surface is measuring the charge there.
Check
One question, then on to energy.
Check your understanding
Just outside the surface of a charged conductor, the field is measured at 5000 N/C pointing straight out. What is the local surface charge density?
Answer: A
Why: Inside a conductor the field is zero, so the boundary condition E_above - E_below = σ/ε₀ becomes simply E_outside = σ/ε₀. Rearranging gives σ = ε₀E = (8.85 × 10⁻¹²)(5000) ≈ 4.4 × 10⁻⁸ C/m². The relation is purely local — the shape never enters.
The key realisation is that inside a conductor the field is zero, so the general boundary condition collapses to something much simpler: the field just outside a conductor equals sigma over epsilon-zero.
Note that this differs by a factor of two from the isolated charged sheet, and the factor has a physical reason rather than being bookkeeping. An isolated sheet sends field out both ways, so each side gets half. A conductor has zero field inside, so all the flux emerges on one side.
Option C is a units check in disguise. Epsilon-zero has units of coulombs squared per newton metre squared, so multiplying by a field in newtons per coulomb gives coulombs per square metre correctly, while dividing does not. Running that check whenever a constant could plausibly go either way is a good reflex.
Section
Section 3
Concept
Push a charge Q from a to b against the field. The work you do is the charge times the potential difference — which is exactly what makes the potential worth defining.
\[ W = Q\big[V(\mathbf{b}) - V(\mathbf{a})\big] \]
Figure (svg): A charge being pushed from a low-potential point to a high-potential point across equipotential lines, with the work marked.
Because the field is conservative, that work does not depend on the route — and if you bring the charge back, you get all of it returned. That stored work is the charge's potential energy, U = QV.
Units check: coulombs times volts equals joules. An electronvolt is exactly this product for one electron across one volt.
This is where the potential stops being an abstraction and becomes something you can put a number on.
The statement is simple: the work you do moving a charge between two points is the charge times the potential difference. Because the field is conservative it does not matter what route you take, and bringing the charge back returns everything you spent.
Do the units out loud, because it makes the definition concrete. Coulombs times volts equals joules. That is also exactly what an electronvolt is — the energy one electron gains crossing one volt — and connecting the unit to the formula usually demystifies it.
In the picture, each vertical line is an equipotential, and the work depends only on which line you start on and which you finish on, not on any of the wiggles in between.
Worked example
Build a configuration one charge at a time, counting the work as you go.
Bring in the first charge, in joules
Why: There is no field yet, so it costs nothing at all.
\[ W_1 = 0 \]
Bring in the second charge against the first one's potential
Why: It arrives at a place where the first charge has established a potential, so the cost is q₂ times that potential.
\[ W_2 = \frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{\mathfrak{r}_{12}} \]
Bring in the third against the potential of BOTH earlier charges
Why: Potentials add as numbers, so the third charge simply pays for two interactions.
\[ W_3 = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1q_3}{\mathfrak{r}_{13}} + \frac{q_2q_3}{\mathfrak{r}_{23}}\right) \]
Generalise, and notice the double count
Why: Summing over all ordered pairs counts every interaction twice — once from each end — so a factor of one half fixes it.
\[ W = \frac{1}{2}\sum_{i=1}^{n}q_i V(\mathbf{r}_i) = \frac{1}{8\pi\varepsilon_0}\sum_{i=1}^{n}\sum_{j\neq i}\frac{q_iq_j}{\mathfrak{r}_{ij}} \]
Verify the one-half by checking two charges, in joules
Why: For n = 2 the double sum has two terms, both equal to q₁q₂/𝔯₁₂, and the one half turns them into one. That matches the direct calculation above, so the factor is right.
The framing that makes this click is construction: the energy of a configuration is the work required to build it, bringing each charge in from infinity one at a time.
The first charge is free, because there is no field yet to fight against. The second costs its charge times the potential the first has established. The third pays for two interactions, because potentials superpose. Continue and the pattern is clear.
The factor of one half needs justifying rather than asserting. Summing over all ordered pairs counts each interaction twice, once from each end, so halving corrects it. Verify with two charges: the double sum has two identical terms and the half turns them into the one term you know is right.
Notice what the sum deliberately leaves out — the terms where a charge would interact with itself. That exclusion is quiet here and becomes the whole story two slides later.
Estimation
Treat a cloud base and the ground as a giant parallel-plate capacitor: area about 10 km², separation about 2 km, and a potential difference of roughly 100 million volts before it breaks down.
Predict first
Roughly how much energy is stored in that cloud-ground capacitor, in joules?
Correct: About 200 million J
Why: C = ε₀A/d = (8.85 × 10⁻¹²)(10⁷ m²)/(2 × 10³ m) ≈ 4.4 × 10⁻⁸ F. Then W = ½CV² = ½(4.4 × 10⁻⁸)(10⁸)² ≈ 2 × 10⁸ J. That is a few hundred megajoules — comparable to a few tonnes of TNT, and consistent with the fact that only a fraction of it discharges in any one stroke.
Notice the shape of the estimate: the capacitance is tiny, but the energy goes as V squared, and 100 million volts squared is a very large number. In electrostatics the voltage almost always dominates.
\[ W = \tfrac{1}{2}CV^2, \qquad C = \frac{\varepsilon_0 A}{d} \]
Estimates like this build a feel for the magnitudes, which is what tells you whether an answer is plausible.
The model is crude and worth flagging as such: treat the cloud base and the ground as two plates of a giant capacitor. Area about ten square kilometres, separation about two kilometres, and a potential difference of order a hundred million volts before the air breaks down.
The capacitance comes out tiny — tens of nanofarads — which might suggest a small energy. But the energy goes as V squared, and a hundred million volts squared is an enormous number. The two effects fight and the voltage wins decisively.
The general lesson: in electrostatics the voltage almost always dominates an estimate, because the capacitances of real objects are small while potentials can be huge. A few hundred megajoules is comparable to a few tonnes of TNT, which is why only a fraction of it discharges in any single stroke.
Concept
Replace the sum with an integral and the charges with a density:
\[ W = \frac{1}{2}\int \rho V\,d\tau \]
Now substitute Poisson's equation for rho, integrate by parts, and something unexpected happens — the charge disappears entirely:
\[ W = \frac{\varepsilon_0}{2}\int_{\text{all space}} E^2\,d\tau \]
Two formulas, always the same number. But they tell completely different stories about where the energy is.
| formula | the story it tells | integrate over |
|---|---|---|
| ½∫ρV dτ | energy lives in the charges | wherever charge is |
| (ε₀/2)∫E² dτ | energy lives in the field | all of space, charge or not |
Going from a sum to an integral is routine. What happens next is not, and it is genuinely surprising.
Substituting Poisson's equation for the charge density and integrating by parts produces a formula in which the charge has vanished completely, replaced by an integral of the field squared over all of space — including regions with no charge in them at all.
Both formulas always give the same number. But they tell incompatible stories about where the energy is: one says it lives in the charges, the other says it lives in the field, spread through empty space.
The difference in the integration region is the sharpest way to see the disagreement. One integral runs only over the charge; the other runs over everywhere. The next two slides resolve which picture to prefer.
Worked example
Griffiths' Example 2.8. Total charge q on a shell of radius R. Both formulas must give the same joules.
Method 1: use the surface-charge version of the energy integral, in joules
Why: The potential is the same constant q/(4πε₀R) everywhere on the shell, so it comes straight out of the integral.
\[ W = \frac{1}{2}\int \sigma V\,da = \frac{1}{2}\cdot\frac{q}{4\pi\varepsilon_0 R}\int\sigma\,da = \frac{q^2}{8\pi\varepsilon_0 R} \]
Method 2: integrate the field energy over all space, in joules
Why: Inside the shell E is zero, so only the outside region contributes anything.
\[ E = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2} \;(r>R), \qquad E^2 = \frac{q^2}{(4\pi\varepsilon_0)^2r^4} \]
\[ W = \frac{\varepsilon_0}{2}\int_R^{\infty}\frac{q^2}{(4\pi\varepsilon_0)^2r^4}\,(4\pi r^2\,dr) = \frac{q^2}{8\pi\varepsilon_0}\int_R^{\infty}\frac{dr}{r^2} \]
\[ W = \frac{q^2}{8\pi\varepsilon_0 R} \]
Verify: the two methods agree, in joules
Why: Identical answers from utterly different integrals — one over a two-dimensional surface where the charge is, one over the infinite three-dimensional region where the field is. That agreement is the strongest evidence that the field really does carry the energy.
The value here is not the answer but the agreement. Two utterly different calculations landing on the same number is the strongest evidence that the field really does carry the energy.
Method one is almost trivial once you notice that the potential is the same constant everywhere on the shell, so it comes straight out of the integral and only the total charge remains.
Method two integrates the field energy over all space. The only subtlety is that the field is zero inside the shell, so only the exterior contributes — and the integral runs from R out to infinity, with the spherical volume element supplying the r squared that makes it converge.
Set the two answers side by side and notice what just happened. One integral was over a two-dimensional surface where the charge sits. The other was over an infinite three-dimensional region that is mostly empty. They agree exactly. That is not a coincidence, and it is the reason physicists talk about energy being stored in fields.
Intuition
The two formulas disagree about the location and agree about the total. Which is right?
Figure (svg): The same shell drawn twice: once with the energy shaded onto the shell itself, once with it shaded into the space outside the shell.
In electrostatics the question is unanswerable — both bookkeepings give every measurable result correctly. In electrodynamics it is settled: energy is in the field, because the field can carry energy away from the charges entirely, as light.
So prefer the field picture. The quantity below is a genuine energy density, in joules per cubic metre, at every point of space:
\[ u = \frac{\varepsilon_0}{2}E^2 \]
Within electrostatics alone the question is not decidable, and it is worth being honest about that. Both bookkeeping schemes reproduce every measurable result, so no experiment confined to statics can tell them apart.
What settles it is electrodynamics. When charges accelerate, the field detaches and carries energy away at the speed of light — as light. That energy is measurably in transit through empty space long after the source has stopped moving, or even been destroyed. Energy that can travel independently of its source was energy in the field.
So prefer the field picture, and take the energy density formula seriously as a local statement: at every point of space, whether or not there is charge there, the field carries this much energy per unit volume.
The two shaded pictures make the contrast visible — the same total, attributed either to the thin shell or to the whole surrounding volume.
Counterexample
Apply the field-energy formula to a single point charge and watch it break.
\[ W = \frac{\varepsilon_0}{2}\int_0^{\infty}\frac{q^2}{(4\pi\varepsilon_0)^2r^4}4\pi r^2\,dr = \frac{q^2}{8\pi\varepsilon_0}\int_0^{\infty}\frac{dr}{r^2} = \infty \]
The integral diverges at the lower limit. A genuine point charge would take infinite energy to assemble.
The discrete formula does not diverge, because its double sum deliberately skips the i = j terms. The continuous formula cannot skip anything, so it keeps the self-energy the discrete one throws away.
Discussion prompt
Does this infinity ever affect a measurable prediction? Why or why not?
Nobody has fully resolved this. In practice: for point charges use the discrete formula and read the answer as interaction energy; the infinite self-energy is a constant that never changes and so never affects a force.
This is a genuine crack in classical electrodynamics, and the subject is more interesting for having it shown rather than hidden.
The calculation is short. Apply the field-energy formula to a single point charge and the integral runs from zero, where the field diverges — so it diverges too. Building a true point charge would take infinite energy.
The discrete formula did not blow up because it deliberately excluded the terms where a charge interacts with itself. The continuous formula has no way to exclude anything, so it keeps the self-energy that the discrete version threw away.
The practical resolution: for point charges use the discrete formula and read the result as interaction energy. The self-energy is an additive constant that never changes as charges move, so it never affects a force and never appears in an answer. The deeper problem is real and unresolved, and it is where quantum field theory takes over.
Trap
Given two charge configurations, a student computes the total energy by adding:
Add the energy of configuration 1 to the energy of configuration 2
Why: Fields superpose and potentials superpose, so energies ought to as well.
\[ W_{\text{tot}} \overset{?}{=} W_1 + W_2 \]
But energy goes as E squared, and squaring destroys superposition.
Square the sum of the fields, not the sum of the squares:
\[ E_{\text{tot}}^2 = (\mathbf{E}_1+\mathbf{E}_2)\cdot(\mathbf{E}_1+\mathbf{E}_2) = E_1^2 + E_2^2 + 2\mathbf{E}_1\cdot\mathbf{E}_2 \]
Keep the cross term — it is the interaction energy, in joules
Why: That extra term is precisely the energy of interaction BETWEEN the two configurations, and it is usually the only part that changes when things move.
\[ W_{\text{tot}} = W_1 + W_2 + \varepsilon_0\int \mathbf{E}_1\cdot\mathbf{E}_2\,d\tau \]
The general lesson: anything quadratic in the field — energy, energy density, pressure — does not superpose. Anything linear — E, V, force on a fixed charge — does.
The root cause is worth naming precisely: superposition holds for anything linear in the field, and energy is quadratic. Squaring destroys superposition.
Write out the expansion so the extra piece is visible. The square of a sum is the sum of the squares plus twice the cross term, and that cross term is exactly the interaction energy between the two configurations.
The cross term is usually the only part that matters. The self-energies of the individual configurations do not change when things move around; the interaction term does. So it is the piece that produces forces.
The rule to carry away: fields, potentials and forces on a fixed charge all superpose. Energy, energy density and electrostatic pressure do not, because all three are quadratic.
Section
Section 4
Sorting
Four expressions, all correct, all for different jobs. Choosing wrongly is the most common way an energy calculation goes astray.
Sort into buckets
Note that the last three are the same physics in different clothing — for a capacitor, all three give identical joules.
Four formulas, all correct, all for different jobs — and choosing the wrong one is the commonest way an energy calculation goes astray.
The distinction that matters most is whether you are building the configuration or moving one charge within a field somebody else made. Building it costs the one-half; moving a single charge through an existing field does not, because you are not paying for the sources.
The last three are the same physics in different clothing. For a capacitor, the continuous formula, the field-energy formula and one-half C V squared all give identical joules — the geometry has simply been packaged into a single number.
If you are unsure which to reach for, ask: am I assembling something, or moving something through something already assembled?
Concept
In a conductor, some electrons are free to roam through the whole material. That single fact forces everything else in this section.
Figure (svg): Left: an insulator with electrons bound to individual atoms. Right: a conductor with a sea of mobile electrons drifting among fixed positive cores.
If a field ever exists inside a conductor, charges move. They keep moving until they have rearranged themselves to cancel it. Electrostatic means that rearrangement has finished.
One fact drives this entire section, so it is worth stating plainly before anything else: in a conductor, some electrons are free to move through the whole material.
The picture contrasts the two cases. In an insulator every electron is tied to its own atom and can only shift slightly within it. In a conductor there is a sea of mobile electrons drifting among fixed positive cores.
The consequence is a definition worth spelling out. If a field ever exists inside a conductor, charges move — and they keep moving as long as it is there. So 'electrostatic equilibrium', the situation this whole section describes, means precisely that the rearranging has already finished and nothing is moving any more.
Every one of the five properties on the next slide follows logically from that single sentence, and deriving them from it beats memorising them as a list.
Concept
Figure (svg): A metal blob with no field inside, charge shown only on its outer surface, and field arrows outside meeting the surface at right angles.
Every one of the five is a consequence of the same fact: free charges move until they have no reason to.
Rather than reading the five cards as a list, derive them from the one fact — they follow in a chain.
The field inside must be zero, because if it were not, charges would still be moving and this would not be equilibrium. Given that, the divergence of the field is zero, so by Gauss's law there is no charge density in the bulk. Any excess charge therefore has nowhere to go but the surface, and mutual repulsion drives it as far apart as it can get.
Zero field also means zero gradient of the potential, so V is constant throughout: the entire object, surface and interior alike, is a single equipotential. That is the property the next deck leans on hardest, because it is exactly the kind of boundary condition Laplace's equation needs.
Finally, the field just outside must be perpendicular to the surface. Any component along the surface would push the surface charge sideways, and it would keep flowing until that component was gone.
Which of the five would still hold if the conductor were a weird lumpy shape? All of them — none of those arguments used symmetry anywhere.
Prediction
Figure (svg): A thick metal shell with an irregular hollow cavity inside it, with nothing in the cavity.
Predict first
An uncharged conductor has an empty cavity of some weird shape carved out of it. There are charges elsewhere in the room. What is the field inside the cavity?
Correct: Zero, whatever the outside charges are doing
Why: This is Griffiths' Example 2.9. The cavity wall is part of the conductor, so it is all at one potential. If a field existed inside the cavity, you could run a path from one wall point through the cavity along the field (gaining potential) and back through the metal (where the potential is constant) — a closed loop with a non-zero line integral, which zero curl forbids. The shape is irrelevant, and so is everything outside.
This is electrostatic shielding, and it is the reason a car is a safe place in a thunderstorm and a coaxial cable does not leak signal.
Commit before revealing, because the correct answer feels too strong to be true and the surprise is what makes it stick.
The field inside the cavity is exactly zero, regardless of the cavity's shape and regardless of what charges are outside the conductor. Nothing outside can reach in.
The argument is elegant and worth following slowly. Suppose a field existed in the cavity. Then you could run a path from one point on the cavity wall, through the cavity along the field direction, gaining potential the whole way, and return through the conducting material where the potential is constant. That closed loop would have a non-zero line integral of E — which zero curl forbids.
This is electrostatic shielding, and it is why a car is a safe place in a thunderstorm, why the mesh in a microwave door keeps the radiation in, and why sensitive instruments live inside metal boxes.
Worked example
Now put a charge q inside the cavity. The answer changes completely.
Figure (svg): A neutral conductor with a cavity containing a positive charge, minus charges lining the cavity wall and plus charges spread over the outer surface.
Draw a Gaussian surface inside the metal, surrounding the cavity
Why: Everywhere on this surface the field is zero, because it lies within the conducting material.
\[ \oint\mathbf{E}\cdot d\mathbf{a} = 0 \quad\Rightarrow\quad Q_{\text{enc}} = 0 \]
Deduce the charge induced on the cavity wall, in coulombs
Why: The enclosed charge is q plus whatever sits on the cavity wall, and it must total zero.
\[ q + q_{\text{wall}} = 0 \quad\Rightarrow\quad q_{\text{wall}} = -q \]
Use overall neutrality to find the outer surface charge, in coulombs
Why: The conductor started neutral, so if minus q has migrated to the cavity wall, plus q must be left on the outer surface.
Verify what an outside observer sees, in newtons per coulomb
Why: Outside, the field is that of a charge +q spread over the outer surface — and if the conductor is a sphere, it is exactly the field of a point charge q at its centre, regardless of where in the cavity the charge actually sits. The conductor has erased all information about the cavity's shape and the charge's position inside it.
The previous slide had an empty cavity. Putting a charge inside changes everything, and the contrast is the lesson.
The method is a Gaussian surface drawn entirely within the conducting material. Everywhere on it the field is zero, so the flux is zero, so the total enclosed charge must be zero. Since the cavity contains plus q, the cavity wall has to carry minus q.
Overall neutrality then forces plus q onto the outer surface — the conductor started neutral, so charge that migrated inward must leave an equal amount behind.
The part worth dwelling on is what an outside observer sees. The exterior field is that of plus q spread over the outer surface, and if the conductor is a sphere it is exactly the field of a point charge at the centre. All information about where the charge sits inside the cavity, and about the cavity's shape, has been erased. The conductor is an information barrier in both directions.
Intuition
On a conductor of uneven shape, the surface charge is not uniform. It piles up where the surface curves most sharply.
Figure (svg): A conductor shaped like a teardrop, with charge sparse on the broad rounded end and densely packed at the pointed end, with long field arrows leaving the point.
The whole conductor is one equipotential. Near a sharp point the potential must fall to its distant value over a much shorter distance — and a steeper potential drop is a stronger field, and hence a larger sigma.
This is why lightning rods are pointed, why high-voltage hardware is deliberately rounded, and why corona discharge starts at edges. It is also why the earlier estimate σ = ε₀E is so useful: it works point by point.
This explains a set of everyday phenomena, so the reasoning is worth having rather than just the fact.
Start from the whole conductor being one equipotential. Near a sharp point, the potential has to fall from the conductor's value to its distant value over a much shorter distance than it does near a broad flat region. A steeper drop over a shorter distance is a larger gradient, which is a stronger field — and since sigma equals epsilon-zero times the field at the surface, a stronger field means more charge piled up there.
The picture shows it directly: sparse charge on the broad rounded end, densely packed charge at the tip, and a much longer field arrow leaving the point.
Hence lightning rods are pointed on purpose, so the field at the tip ionises the air and gives the strike a controlled path. High-voltage hardware is deliberately rounded for the opposite reason. And corona discharge always starts at edges and points.
Concept
Surface charge sits in the field of all the other surface charge. That produces an outward pull on the surface itself.
\[ f = \frac{\sigma^2}{2\varepsilon_0} \qquad \text{(force per unit area, N/m}^2\text{)} \]
Figure (svg): A patch of a charged conductor surface with outward arrows labelled as electrostatic pressure pulling the metal outward.
The factor of one half is not decoration: a patch of charge cannot push on itself, so it feels the average of the fields on its two sides — zero inside, σ/ε₀ outside.
The pressure is always outward, whatever the sign of the charge, because sigma appears squared. This is why a charged soap bubble expands.
The surface charge sits in the field produced by all the other surface charge, so it feels a force — and the direction is always outward.
The factor of one half is not obvious and deserves explanation. A patch of charge cannot exert a force on itself, so what it feels is the average of the fields on its two sides: zero on the inside of the conductor, and sigma over epsilon-zero on the outside. The average of those is half the outside value.
The pressure is outward whatever the sign of the charge, because sigma enters squared. Positive and negative distributions both push their surface outward.
The tangible example is a charged soap bubble, which expands when charged. It is also why very highly charged conductors can physically deform or tear.
Concept
To 'ground' a conductor is to wire it to the earth, which is large enough to supply or absorb any amount of charge without changing its own potential.
Figure (svg): A metal sphere connected by a wire to a ground symbol, with electrons flowing along the wire to neutralise the sphere.
Chapter 3's method of images is built entirely on this: a grounded plane or sphere fixes V = 0 on a surface, and the whole trick is finding a fictitious charge that reproduces that condition.
'Grounded' gets used loosely, so it is worth pinning down what it means operationally: a wire to the earth, which is large enough to absorb or supply any amount of charge without its own potential shifting.
Two consequences matter for solving problems. The conductor is held at V equals zero by definition of the reference. And its charge is not fixed — whatever charge is needed to keep it at zero simply flows in or out through the wire.
That combination is exactly what makes a grounded conductor the ideal boundary condition. You know V precisely, and you let the charge come out as part of the answer rather than having to specify it in advance.
It is also what the entire method of images in the next deck is built on. A grounded plane or sphere pins V to zero on a surface, and the trick is finding a fictitious charge that reproduces that condition.
Concept
Take two conductors, put +Q on one and −Q on the other. The potential difference between them turns out to be proportional to Q.
\[ C \equiv \frac{Q}{V} \]
Figure (svg): Two conductors of arbitrary shape, one positive and one negative, with field lines running from one to the other and the potential difference marked.
The proportionality is not obvious — it follows from the linearity of the whole theory. Double every charge and you double every field and hence every potential difference.
So the ratio C depends only on the geometry: sizes, shapes, separation. Its unit is the farad, one coulomb per volt, and a farad is enormous — real capacitors run in microfarads and picofarads.
The definition looks like it is defining a ratio that could be anything, so the first job is explaining why the ratio is constant.
Put plus Q on one conductor and minus Q on the other. Double every charge and, because every equation in electrostatics is linear, every field doubles and therefore every potential difference doubles. The ratio is untouched. That linearity is why capacitance is a well-defined property at all.
So C depends only on the geometry — sizes, shapes, separations — and not at all on how much charge you happened to put there.
A farad is enormous, which is worth knowing since the unit gives no intuition on its own. A one-farad capacitor holding one volt stores one coulomb, and a coulomb is a huge amount of charge. Real components are measured in microfarads and picofarads.
Worked example
Griffiths' Example 2.10. Two plates of area A held a distance d apart, with d much smaller than the plates' width.
Figure (svg): Two horizontal plates with charge plus Q on top and minus Q below, a uniform field between them, and the separation d marked.
Write the field between the plates, in newtons per coulomb
Why: This is the two-plane superposition result from Deck 2, with surface density sigma = Q/A.
\[ E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A} \]
Integrate the field across the gap to get the potential difference, in volts
Why: The field is uniform, so the line integral is just field times distance.
\[ V = Ed = \frac{Qd}{\varepsilon_0 A} \]
Form the ratio
Why: The charge Q cancels, which it had to — capacitance is a property of the geometry alone.
\[ C = \frac{Q}{V} = \frac{\varepsilon_0 A}{d} \]
Verify the behaviour against intuition, in farads
Why: Bigger plates mean more capacitance; a wider gap means less. Both match the physical picture, since a bigger area holds more charge at the same potential and a wider gap means more work to move charge across. Numerically: plates of 1 m² held 1 mm apart give only 8.85 nF, which is why real capacitors use thin dielectric layers and rolled-up foil.
The workhorse example, and every step reuses something already established.
The field between the plates is the two-plane superposition result from the last deck, with the surface density written as Q over A. The potential difference is that uniform field times the separation, since integrating a constant is just multiplication. Then form the ratio.
The moment worth pausing on is that Q cancels. It had to — capacitance cannot depend on how much charge you imagined putting on the plates — so if it survives, there is an algebra error somewhere. That is a useful self-check.
Finish with the physical reading and a number. Bigger plates hold more charge at the same voltage, so C rises with area; a wider gap means more work per charge, so C falls with separation. Numerically, a full square metre of plate held a millimetre apart gives only about nine nanofarads, which is why real capacitors use very thin dielectric layers and rolled-up foil.
Worked example
Griffiths' Example 2.11. Two concentric metal shells of radii a and b, with a smaller than b.
Figure (svg): Two concentric circles, the inner one positive and the outer negative, with radial field lines only in the gap between them.
Write the field in the gap, in newtons per coulomb
Why: A Gaussian sphere in the gap encloses only the inner shell's charge, so the field there is the point-charge field.
\[ E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2} \qquad (a < r < b) \]
Integrate from the outer shell in to the inner one, in volts
Why: The potential difference is the line integral of E across the gap.
\[ V = -\int_b^a \frac{Q}{4\pi\varepsilon_0 r^2}\,dr = \frac{Q}{4\pi\varepsilon_0}\left(\frac{1}{a}-\frac{1}{b}\right) \]
\[ C = \frac{Q}{V} = 4\pi\varepsilon_0\,\frac{ab}{b-a} \]
Verify by letting the gap become thin, in farads
Why: Put b = a + d with d very small. Then ab/(b − a) becomes a²/d, and 4πε₀a²/d is exactly ε₀A/d with A = 4πa² — the parallel-plate result. A thin spherical gap is a parallel-plate capacitor bent round, which is a good sign the algebra is right.
Structurally identical to the parallel-plate case — find the field, integrate to get the voltage, take the ratio — but with spherical geometry, so it is good practice at recognising the pattern rather than the formula.
A Gaussian sphere drawn in the gap encloses only the inner shell's charge, so the field there is just the point-charge field. Integrating it across the gap produces the difference of two reciprocals, which is where the ab over b minus a comes from.
The verification connects two results that look unrelated, so it is worth doing carefully. Let the gap become thin by writing b as a plus a small d. Then ab over b minus a becomes a squared over d, and four-pi-epsilon-zero times a squared over d is precisely epsilon-zero times the sphere's area over d — the parallel-plate formula.
Which is exactly what you would hope for: a thin spherical gap is a parallel-plate capacitor bent round. If the algebra had not reproduced that, something would be wrong.
Concept
Charge it up a little at a time. Moving dq across a potential difference q/C costs work:
\[ dW = \frac{q}{C}\,dq \quad\Rightarrow\quad W = \int_0^Q \frac{q}{C}\,dq = \frac{Q^2}{2C} \]
\[ W = \frac{1}{2}\frac{Q^2}{C} = \frac{1}{2}CV^2 = \frac{1}{2}QV \]
The factor of one half appears because the first charge crosses an empty gap for free while the last crosses the full potential difference. The average is half the final value.
Cross-check with the field picture: for a parallel plate capacitor, (ε₀/2)E² times the volume Ad gives exactly the same joules. Two routes, one answer — as in Example 2.8.
Derive the one-half rather than quoting it, because whether it belongs there is routinely misremembered.
Charge the capacitor gradually. Moving a small amount of charge across the potential difference that currently exists costs that amount times the current voltage. Integrate from zero up to the final charge and the one-half appears from the integral, not from any physical subtlety.
The intuitive version alongside: the first charge crosses an empty gap for free, the last crosses the full potential difference, and the average over the whole process is half the final value.
Cross-check with the field picture. For a parallel-plate capacitor, the energy density times the volume between the plates gives exactly the same answer — two independent routes agreeing, in the same spirit as the charged-shell example earlier.
Trap
Asked what happens to a capacitor's capacitance when the voltage across it is doubled, a student answers:
Since C = Q/V, doubling V must halve C
Why: The formula has V in the denominator, so it looks as though raising V lowers C.
\[ C = \frac{Q}{V} \;\overset{?}{\Rightarrow}\; V \to 2V \;\Rightarrow\; C \to \tfrac{1}{2}C \]
But doubling the voltage also doubles the charge. The ratio is untouched.
C is a property of the geometry, and nothing else:
\[ C_{\text{plates}} = \frac{\varepsilon_0 A}{d}, \qquad C_{\text{spheres}} = 4\pi\varepsilon_0\frac{ab}{b-a} \]
Notice what appears in those formulas — and what does not
Why: Areas, radii, separations, and epsilon-zero. No Q, no V. Change the voltage and Q follows so that the ratio stays fixed.
The same logic runs backwards: to change a capacitance you must change the plates — move them, resize them, or (Chapter 4) fill the gap with a dielectric.
The mistake comes from reading C equals Q over V as a formula that computes C from measured values, rather than as the statement that a particular ratio happens to be constant.
The refutation is one line: doubling the voltage also doubles the charge, so the ratio is unchanged. The two quantities are locked together.
Look at what appears in the actual formulas — areas, radii, separations, epsilon-zero — and at what does not: no Q and no V anywhere. That absence is the whole point.
Run the logic backwards, because it is what problems actually ask: to change a capacitance you have to change the plates. Move them, resize them, or fill the gap with a dielectric — which is exactly where Chapter 4 picks up.
Real world
Discussion prompt
Your phone screen detects a finger by a change in capacitance. Which of the geometric factors in C is your finger changing?
These four show the section was not an abstract exercise. Each card is a direct application of one of the five properties.
The Faraday cage is the shielding result. The lightning rod is the sharp-point result. The camera flash is the energy result — a capacitor charged over several seconds and discharged in milliseconds, which is why the power is enormous even though the energy is modest. And the touchscreen is capacitance used as a measurement, where your finger changes C and the phone detects the change.
The flash is worth a moment because it separates energy from power. The same energy delivered in a thousandth of the time is a thousand times the power.
Matching
Match the pairs
Why: Zero interior field is what makes shielding work — no outside arrangement can produce a field inside a cavity. The equipotential property is what makes conductors so useful in Chapter 3: 'this whole surface sits at V₀' is exactly the kind of boundary condition Laplace's equation needs. Perpendicularity plus the boundary condition turns a field measurement just outside into a direct reading of the local surface charge.
The middle one is the bridge to the next deck: every problem in Chapter 3 is Laplace's equation plus a set of conductors at stated potentials.
Three properties, three consequences, and the mapping between them is the structure of the whole section.
Zero interior field gives shielding. Perpendicularity at the surface gives a direct read on the local surface charge. And the equipotential property is what makes a conductor usable as a boundary condition.
That middle one is the bridge to the next deck. Every problem in Chapter 3 is Laplace's equation plus a set of conductors held at stated potentials. Without the equipotential property there would be no boundary condition to impose and the whole method would be unavailable.
Comparison
Comparison matrix
| E | V | |
|---|---|---|
| type | vector (3 numbers) | scalar (1 number) |
| units | N/C = V/m | V = J/C |
| superposition | add as vectors | add as signed numbers |
| across a surface charge | jumps by σ/ε₀ | continuous, with a kink |
| inside a conductor | zero | constant, generally not zero |
| falls off from a point charge as | 1/r² | 1/r |
The last row explains a lot of confusion: at large distances the potential dies away much more slowly than the field.
Use this grid as the summary of the first two-thirds of the deck. Filling the blanks unaided is a good sign the vector-versus-scalar distinction has landed.
The superposition row is what justifies the entire strategy: vectors need diagrams and components, signed numbers need neither.
The row about behaviour across a surface charge is the one used for induced charge: E jumps, V merely kinks.
And the last row explains a persistent confusion. Far from a charge, the potential dies away much more slowly than the field — one over r against one over r squared. Expecting them to behave alike leads to being surprised by how far potentials reach.
Pattern
If Q does not cancel in that last step, you have made an algebra error — capacitance can never depend on the charge you imagined putting there.
This is the procedure whenever a problem mentions a conductor, and most of the work happens before any calculation.
Write down the five properties first, because in most problems a large part of the answer is already contained in them. Then, to find induced charges, draw a Gaussian surface inside the metal where the field is known to be zero — that single move solves nearly every induced-charge question.
For capacitance, follow the fixed recipe: put plus and minus Q on the conductors, find the field, integrate to get V, form Q over V. Watch Q cancel, and treat its failure to cancel as a signal that something has gone wrong rather than a result to report.
Reading surface charge off the local field is the last tool, and it works point by point on surfaces of any shape.
Check
Final question.
Check your understanding
A parallel-plate capacitor is charged to voltage V and then disconnected from the battery. The plates are then pulled twice as far apart. What happens to the stored energy?
Answer: A
Why: Disconnected means Q is fixed. Use W = Q²/2C: doubling d halves C = ε₀A/d, so W doubles. Physically, the plates attract each other, so pulling them apart takes work — and that work is exactly the extra stored energy. Note the field between the plates never changed, but there is now twice as much space filled with it.
This question separates memorising formulas from tracking what is being held fixed, so all four options are worth thinking about.
The critical word is 'disconnected'. That fixes the charge, not the voltage. With Q fixed, use the form of the energy that has Q in it: Q squared over 2C. Doubling the separation halves the capacitance, which doubles the energy.
The physical story confirms it and is more memorable than the algebra. The plates carry opposite charges, so they attract. Pulling them apart therefore takes work, and that work is exactly the increase in stored energy.
There is a nice detail here: the field between the plates does not change at all. What changes is the volume of space filled with that field, which doubles — and since energy density times volume is the energy, that alone accounts for the factor of two.
Exit ticket
Discussion prompt
Two sentences, no formulas: what is the potential, and why is a conductor an equipotential?
If those hold up, Chapter 3 is just techniques for solving Laplace's equation with conductors as boundary conditions.
A genuine check rather than a formality. Producing these two sentences without notation is the difference between being ready for the next deck and struggling through it regardless of how many formulas you have memorised.
The potential is the work per unit charge needed to bring a charge from the reference point to where you are — a landscape whose downhill direction is the field.
A conductor is an equipotential because any potential difference within it would be a slope, a slope is a field, and a field would push the free charges — so they move until no slope is left. That argument, in that order, is the one to be able to reproduce.
Recap
| the result | the formula |
|---|---|
| potential of a point charge | q/(4πε₀r) |
| potential inside a charged shell | q/(4πε₀R), constant |
| Poisson / Laplace | ∇²V = −ρ/ε₀ ; ∇²V = 0 |
| boundary conditions | E jumps by σ/ε₀ ; V continuous |
| energy density | u = (ε₀/2)E² |
| energy of a charged shell | q²/(8πε₀R) |
| parallel-plate capacitance | ε₀A/d |
| stored energy | ½CV² = Q²/2C |
Next deck: Chapter 3. Given the conductors and their potentials, solve Laplace's equation — with images, separation of variables, and the multipole expansion.
Take stock of what has changed in your toolkit. Before this deck, finding a field meant a vector integral or a lucky symmetry. Now there is a third route: solve for one scalar function and differentiate.
Go down the results table and make sure each entry is attached to a picture rather than floating free. The point-charge potential, the constant potential inside a shell, the boundary conditions, the energy density, the parallel-plate capacitance — each one has a diagram earlier in the deck that explains it.
Then look ahead. Everything so far assumed you know where the charge is. The next deck handles the far more common situation where the charge is induced and depends on the field you are trying to compute — which is what Laplace's equation, images and separation of variables are for.
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