Case Study 2 — The Field of a Point Charge: Inverse-Square Law as Vector Calculus

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Case Study 2 — The Field of a Point Charge: Inverse-Square Law as Vector Calculus

Field: Physics, electrostatics, gravitation

A single proton sits at the origin. Around it, in empty space, there is no visible thing — and yet a second charge placed anywhere nearby feels a push. That push, drawn at every point at once, is a vector field: the electric field $\mathbf{E}$ of a point charge. This case study takes the most important field in classical physics — the inverse-square field, shared by electrostatics and gravity — and reads it through the two operators of this chapter. We will find that its divergence is zero everywhere except at the source, that it is irrotational, and that it is conservative with a potential you already know as voltage. Each of these is a §34.4–34.7 idea made physical.

The field

Coulomb's law says the electric field of a point charge $q$ at the origin is

$$\mathbf{E}(\mathbf{r}) = \frac{q}{4\pi\varepsilon_0}\,\frac{\mathbf{r}}{r^3}, \qquad \mathbf{r} = \langle x, y, z\rangle, \quad r = \|\mathbf{r}\| = \sqrt{x^2 + y^2 + z^2}.$$

Strip the constants and write $\mathbf{F} = \mathbf{r}/r^3$; we restore the physics at the end. The direction $\mathbf{r}/r$ is the outward unit radial, and the magnitude is

$$\|\mathbf{F}\| = \frac{\|\mathbf{r}\|}{r^3} = \frac{r}{r^3} = \frac{1}{r^2}.$$

There it is — the famous inverse-square decay (§34.2). At twice the distance the field is one-quarter as strong. The arrows point straight out from the origin (for positive $q$) and shrink as $1/r^2$. The same formula, with gravity's constants, is the gravitational field of a point mass; only an overall sign and the constant differ. Everything we prove about $\mathbf{F}$ holds for both.

Divergence: where are the sources?

Intuition from §34.4 says a field of arrows fleeing the origin should have positive divergence — it is the very picture of a source. Let us compute and watch a subtlety appear. Take the $x$-component $P = x/r^3 = x(x^2+y^2+z^2)^{-3/2}$. By the product and chain rules,

$$P_x = (x^2+y^2+z^2)^{-3/2} + x\cdot\left(-\tfrac{3}{2}\right)(x^2+y^2+z^2)^{-5/2}\cdot 2x = \frac{1}{r^3} - \frac{3x^2}{r^5}.$$

By symmetry the other two diagonal derivatives are $Q_y = \dfrac{1}{r^3} - \dfrac{3y^2}{r^5}$ and $R_z = \dfrac{1}{r^3} - \dfrac{3z^2}{r^5}$. Add them:

$$\nabla\cdot\mathbf{F} = \frac{3}{r^3} - \frac{3(x^2 + y^2 + z^2)}{r^5} = \frac{3}{r^3} - \frac{3r^2}{r^5} = \frac{3}{r^3} - \frac{3}{r^3} = 0.$$

The divergence is zero everywhere $r > 0$. This is one of the most important small calculations in physics, and at first it seems to contradict intuition: how can a field of outward-fleeing arrows be sourceless? The resolution (foreshadowed in §34.4) is that the source is not spread through space — it is concentrated entirely at the single point $r = 0$, where the formula blows up and the calculation above does not apply. Away from the charge, empty space holds no charge, so the field has no source there. All the "outflow" originates at the proton itself.

The deeper meaning (Chapter 37). That the divergence vanishes off the source while a nonzero charge sits at the center is precisely Gauss's law in disguise. The Divergence Theorem of Chapter 37 will convert "$\nabla\cdot\mathbf{E} = 0$ away from the charge" plus "flux through any enclosing surface $= q/\varepsilon_0$" into the statement $\nabla\cdot\mathbf{E} = \rho/\varepsilon_0$ with $\rho$ a point concentration. The chapter's operators are defined here; their integral meaning waits three chapters.

Curl: is the field irrotational?

Drop a paddle wheel anywhere in the field of a point charge. The arrows all point radially; nothing pushes one vane harder than its opposite, so the wheel should not spin. Curl should be zero. We can verify it without grinding through the determinant, by recognizing the field as a gradient — which by the master identity $\nabla\times(\nabla f) = \mathbf{0}$ (§34.6) is automatically irrotational. So we turn to the potential.

Conservative: the potential is the voltage

Guess the potential $\varphi = -\dfrac{1}{r} = -(x^2+y^2+z^2)^{-1/2}$ and check $\nabla\varphi = \mathbf{F}$. Differentiating,

$$\varphi_x = -\left(-\tfrac{1}{2}\right)(x^2+y^2+z^2)^{-3/2}\cdot 2x = \frac{x}{r^3},$$

and by symmetry $\varphi_y = y/r^3$, $\varphi_z = z/r^3$. Therefore

$$\nabla\varphi = \left\langle \frac{x}{r^3}, \frac{y}{r^3}, \frac{z}{r^3}\right\rangle = \frac{\mathbf{r}}{r^3} = \mathbf{F}. \qquad\checkmark$$

So $\mathbf{F}$ is conservative, with potential $\varphi = -1/r$. Restoring the constants, the physical potential is

$$V(\mathbf{r}) = \frac{q}{4\pi\varepsilon_0}\,\frac{1}{r}, \qquad \mathbf{E} = -\nabla V.$$

The minus sign is the §34.7 convention: the force points downhill in potential. And $V$ is not an abstraction — it is voltage, the quantity a voltmeter reads. The "potential function" of this chapter, the scalar whose gradient is the field, is literally electric potential. Because the field is a gradient, the curl test (§34.6) guarantees $\nabla\times\mathbf{E} = \mathbf{0}$ without any further computation. The single scalar $V$ encodes the entire vector field $\mathbf{E}$.

Energy and path-independence

Conservative fields earned their name from energy (§34.7–34.8). The work done moving a test charge $q_0$ from point $A$ to point $B$ through the field is, by the Fundamental Theorem for Line Integrals previewed in §34.8,

$$W = \int_C \mathbf{F}\cdot d\mathbf{r} = q_0\big(V(A) - V(B)\big),$$

depending only on the endpoints — not the path. This is why a battery's voltage difference, not the wire's route, sets the energy delivered. Carry the charge around any closed loop and the work is exactly $0$: start and end coincide, so $V(A) - V(B) = 0$. That zero-loop-work property is the experimental signature of a conservative field, and it is why electrostatic energy is stored, never lost going around a circuit of ideal wire.

Contrast: the magnetic field is different. The magnetic field around a current-carrying wire, $\mathbf{B}\propto\dfrac{1}{x^2+y^2}\langle -y, x, 0\rangle$ (§34.2), circulates rather than radiates. It is divergence-free (no magnetic monopoles, $\nabla\cdot\mathbf{B}=0$) but its global circulation is nonzero, and it is not the gradient of a single-valued potential — the same punctured-plane obstruction as the vortex of §34.6. Electric fields from static charges are conservative; magnetic fields are not gradients. The two operators tell them apart at a glance: $\mathbf{E}$ has divergence at its source and zero curl; $\mathbf{B}$ has zero divergence and circulation.

Why this one field matters so much

Three computations — a vanishing divergence, an automatic zero curl, a recovered potential — turned Coulomb's law into a complete vector-calculus object. The divergence located the source (and previews Gauss's law). The conservative structure handed us voltage and energy conservation for free. The same field, reinterpreted with gravitational constants, governs planetary orbits and is exactly the potential $\Phi = -GM/r$ that mission planners exploit (§34.7). One inverse-square field, read through divergence, curl, and potential, underlies both the force that holds atoms together and the force that holds the solar system in orbit. That is the reach of two derivatives.

Discussion Questions

We found $\nabla\cdot\mathbf{F} = 0$ for $r > 0$, yet the charge is obviously a source. Explain in your own words how a sourceless-looking divergence is consistent with a real source at the origin. What goes wrong with the computation exactly at $r = 0$?
Verify the curl is zero directly (compute $\nabla\times(\mathbf{r}/r^3)$) for at least one component, and confirm it matches the prediction from $\mathbf{F} = \nabla\varphi$.
The gravitational field of a point mass is $\mathbf{g} = -GM\,\mathbf{r}/r^3$ with potential $\Phi = -GM/r$. How do the signs differ from the electric case, and why (think about whether like masses attract or repel)?
Using $W = q_0(V(A) - V(B))$, explain why moving a charge along a complicated zig-zag path between two points does the same work as moving it straight. Which §34.8 property is this?

Annotated Further Reading

Stewart, Calculus: Early Transcendentals, §16.1 and §16.3 — vector fields and the Fundamental Theorem for Line Integrals; §16.1 includes the gravitational/electric inverse-square field as a worked example.
OpenStax Calculus Volume 3, §6.1 and §6.3 — "Vector Fields," then "Conservative Vector Fields" and finding potential functions, mirroring the potential recovery above.
Griffiths, Introduction to Electrodynamics, ch. 2 — the point-charge field, Gauss's law, and electric potential; the physics behind why $\nabla\cdot\mathbf{E}=\rho/\varepsilon_0$ and $\mathbf{E}=-\nabla V$.
Feynman Lectures on Physics, Vol. II, ch. 4 — an intuitive tour of electrostatics as the divergence and curl of $\mathbf{E}$, complementary to this chapter's operator view.