Chapter 01 · Electrostatics · Full book

Electrostatics

Two chapters of the source text, read as one arc: from the force between charges to the field, the flux law, the scalar potential, and the capacitor.

Part I · Electric Charges & Fields

Core intuition: Chapter Roadmap

This chapter builds the electric field from the ground up and gives you two complementary tools for computing it.

§ Why Fields Exist We replace “action at a distance” with a field that fills space — the idea that makes everything afterward work.
§ Charge The source: its two signs, its quantisation in units of $e$, and its strict conservation.
§ Coulomb's Law The force between point charges, in vector form, and the superposition principle that extends it to any collection.
§ The Electric Field Force per unit charge — separating what the sources do from what a test charge feels.
§ Continuous Distributions Integration as superposition: the field of a ring and of a line of charge.
§ Field Lines & Flux A picture of the field, then a number attached to that picture — electric flux.
§ Gauss's Law The first power tool: flux through any closed surface counts the charge inside. Exact, and often faster than integration.
§ Applications of Gauss's Law Line, sheet, and sphere fields in a few lines each — whenever symmetry permits.
§ The Electric Dipole Equal and opposite charges as a single object: its moment, its $1/r^3$ field, and the torque a field exerts on it.

Two audiences, one passage. Every major result is derived in full (the calculus form for JEE Advanced / AP-C E&M) and then restated as a ready-to-use formula with its conditions of validity (the algebra form for NEET / AP-2). The boxed result is what you apply; the derivation is why you may trust it.

Part II · Electrostatic Potential & Capacitance

Core intuition: Chapter Roadmap

This chapter builds the second language of electrostatics: the electric potential. Chapter 2 gave us the electric field—a vector at every point in space. The potential gives us a scalar instead, one that superposes by ordinary addition, encodes all energy bookkeeping directly, and turns the problem of finding forces into the gentler problem of reading a landscape. The chapter develops in six movements:

Electric Potential. The work-per-unit-charge idea; the point charge potential $V = kQ/r$; the scalar superposition principle.
Field from Potential. $\vec{E} = -\nabla V$: the field is the steepness of the potential landscape. We apply this to recover known fields and to establish the zero-$V$, zero-$E$ trap.
Equipotential Surfaces. The level curves of the potential; their perpendicularity to field lines; conductors as equipotential volumes.
The Dipole: Potential, Field, and Energy. The $1/r^2$ fall-off of the dipole potential; the full dipole field from $\vec{E} = -\nabla V$ in spherical coordinates; the energy $U = -\vec{p}\cdot\vec{E}$ and the equipotential structure.
Conductors in Electrostatic Equilibrium. Why $\vec{E} = 0$ inside; the surface field $E = \sigma/\varepsilon_0$ by a Gaussian pillbox; electrostatic shielding and the Faraday cage.
Capacitance and Energy Storage. The geometry-determined charge-to-voltage ratio; parallel-plate, spherical, series, and parallel configurations; dielectrics; the energy density $u = \tfrac{1}{2}\varepsilon_0 E^2$.

Perplexing Questions

Perplexing questions

Before we begin, let these questions sit with you. Do not reach for formulas. You will return to each one—with full answers—before this chapter ends.

The Gap Between Charges: Two charges sit in a perfect vacuum, separated by a metre of absolutely nothing. No air, no medium, no string, no contact of any kind. One charge exerts a force on the other. Through what, exactly, does this force travel?
The Speed of the Force: If you suddenly moved one charge, when would the other “feel” the change? Instantly? After some delay? If there is a delay, what carries the information, and how fast?
The 1/$r^2$ Coincidence: Coulomb's force weakens as the inverse square of distance—exactly. Not $1/r^{1.9}$, not $1/r^{2.1}$: precisely $1/r^2$. Gravity obeys exactly the same dependence. Is this a coincidence, or is there a deeper geometric reason that both forces must fall off this way?
Two Kinds of Charge: Mass comes in only one sign—two masses always attract. But charge comes in two signs, and opposite signs attract while like signs repel. Why does nature permit negative charge when it does not permit negative mass? Is there a deeper asymmetry at work?
The Stronger Force That Does Not Rule: The electric force between a proton and an electron is approximately $10^{36}$ times stronger than the gravitational force between them. Yet at the scale of planets, galaxies, and the universe, gravity dominates. A force forty orders of magnitude weaker wins cosmologically. How?

Each question targets a trap that catches students every year.
By the end of this chapter, every one of them will be transparent.

Perplexing questions

Sit with these before opening a formula. Every answer is earned before this chapter ends.

The Vanishing Interior. A charge is placed anywhere inside a hollow conducting shell—not at the centre, not symmetric, anywhere. The electric field inside the cavity is zero regardless of what charges or conductors exist outside. The shell appears to “know” it must cancel all exterior fields without being told where they are. How does a conductor manage this without any information about external geometry?
The Floating Potential. Two conducting spheres of very different radii are connected by a thin wire and charged together. The smaller sphere ends up with a much stronger electric field at its surface, even though it holds less total charge. How can less charge produce a stronger field?
The Battery Paradox. A charged capacitor has a dielectric inserted while it remains connected to a battery: the stored energy increases. The same capacitor is disconnected before the dielectric is inserted: the stored energy decreases. Same capacitor, same dielectric—opposite energy changes. What is different about the two situations?
The Zero-Potential Field. The electric potential is exactly zero at every point on the perpendicular bisector plane of an electric dipole. Yet the electric field is not zero anywhere on that plane. How can a region of zero potential carry a nonzero field?
The Image Charge. A positive charge is brought near an infinite grounded conductor. The conductor develops an induced charge distribution that creates, outside the conductor, a field identical to the field from a phantom “image charge” of opposite sign placed symmetrically on the other side of the surface. No one told the conductor about image charges. Why does the induced distribution turn out to be so precisely the right one?
The Energy in Empty Space. Pulling the dielectric out of a charged, isolated capacitor requires work. The capacitor gains energy. The charge on the plates has not changed; nothing has been added between the plates. Where does this extra energy come from and where does it live?

Each question exposes a gap that trips students on every major exam.
Return to them as you read—each is answered precisely, not just plausibly.

Part I · Electric Charges & Fields

Why Fields Exist

Newtonian mechanics, as you learned it in Volume 1, is built on forces. Two objects interact; each exerts a force on the other; Newton's second law tells us how each accelerates. The framework is powerful. But it carries a hidden assumption that Newton himself found troubling: the force between two separated objects is transmitted instantaneously across empty space.

Pull the Sun from its location, and Earth would, on this assumption, “know” immediately—its orbit would change at the moment of removal, not after light has had time to travel eight minutes to us. This is action at a distance, and it is conceptually incoherent.

The Problem With Action at a Distance

To see why, consider what a force is. A force is an interaction: object $A$ influences object $B$. If $A$ and $B$ are separated by empty space, then for the influence to reach $B$ from $A$, something must cross that space.

Three logical possibilities exist:

The influence crosses instantaneously (action at a distance).
The influence is carried by some material medium between them.
The influence is carried by a non-material entity—a field—that exists at every point in space, and propagates at a finite speed.

Option 1 was Newton's reluctant default. He wrote in a letter that he found the idea of action at a distance “so great an absurdity that I believe no man who has a competent faculty of thinking can ever fall into it.” Yet he used it because he had no alternative.

Option 2—a material medium called the “aether”—was proposed in the nineteenth century, and demolished by the Michelson–Morley experiment.

Option 3 is what electromagnetism, and later gravity in general relativity, actually requires. The field is the answer.

Core intuition: What a Field Is

A field is a physical quantity that has a value at every point in space (and at every moment in time). A scalar field assigns one number to each point: temperature inside a room, atmospheric pressure across a continent. A vector field assigns a vector to each point: wind velocity across a weather map, gravitational acceleration at every point above Earth's surface. The electric field $\vec{E}$ is a vector field. At every point in space, it specifies the force per unit positive charge that a test charge placed there would experience: \[ \vec{E} \equiv \frac{\vec{F}}{q_0} \qquad \text{(definition of electric field)}. \] The field exists whether or not a test charge is present. It is not a bookkeeping device—it carries energy, momentum, and information, and propagates at a finite speed.

Field as Physical Reality

There is a common temptation, especially when first learning electrostatics, to treat the electric field as a mathematical shorthand—a way of summarising the force that charge $A$ would exert on any other charge placed nearby. This view is insufficient, and it becomes positively wrong in later chapters.

Two facts establish the field as physically real:

1. The field carries energy. A charged capacitor stores energy. If you discharge it, that energy is released as heat or light or mechanical work. Where was the energy stored? Not on the plates—the plates are conductors in equilibrium. The energy is stored in the electric field between the plates. We will derive the energy density $u = \tfrac{1}{2}\varepsilon_0 E^2$ explicitly in the next chapter (Electrostatic Potential and Capacitance).

2. The field propagates at a finite speed. If you accelerate a charge, the disturbance in the electric field radiates outward at the speed of light. A distant charge does not respond instantly; it responds after a delay $t = r/c$. This delay is carried by the field, not by the source charge itself.

Both facts are consequences of Maxwell's equations, which we develop fully in Chapter 6 (Electromagnetic Waves). For now, the important lesson is:

Core intuition: The Field Comes First

In the modern picture of electromagnetism, the sequence of events is:

Source charge creates a field at every point in space.
The field exerts a force on any other charge placed at that point.
The source charge never “reaches out” to the test charge directly.

The interaction is always: charge $\to$ field $\to$ charge. This is not a semantic distinction. It changes what we mean by “force between charges.” Coulomb's law, which we derive in Section , is the static approximation—valid when charges are not accelerating and we need not worry about propagation delays. But even in this limit, the field is the correct intermediary.

Connection to Gravitational Field

You have already worked with a field in Volume 1. The gravitational field $\vec{g}$ at a point in space is the gravitational force per unit mass that a test mass would experience there: \[ \vec{g} = \frac{\vec{F}_{\text{grav}}}{m}. \] Near Earth's surface $|\vec{g}| \approx 9.8\,\mathrm{m/s^2}$, directed downward. At distance $r$ from a point mass $M$: \[ \vec{g} = -\frac{GM}{r^2}\,\hat{r}. \]

The electric field is the charge-world analogue of this. Every statement you made about gravitational fields in Volume 1's gravitation chapter has an electromagnetic counterpart—with two crucial differences:

Gravitational mass comes in one sign only; charge comes in two. This means the electric field can point toward or away from its source, while the gravitational field always points toward its source.
The electric force is vastly stronger than gravity for subatomic particles (Perplexing Question 5 will be answered in Section ).

The conceptual vocabulary—field lines, flux, source and sink, potential energy per unit “something”—transfers directly from gravitation to electrostatics. We will exploit this parallel throughout the chapter.

Misconception alert: “The Field Is Just Shorthand for the Force”

The misconception: The electric field at point $P$ means nothing on its own; it only becomes real when we place a charge at $P$ to feel the force. Why it is wrong:

A charged capacitor stores measurable energy even in vacuum with no test charge inside. The energy is in the field.
The field propagates as an electromagnetic wave even in the complete absence of matter. Light is an oscillating electric and magnetic field travelling through empty space.
In quantum electrodynamics, the field is the fundamental object; particles are excitations of the field.

JEE/NEET implication: Questions of the form “what is the electric field at the centre of a uniformly charged ring?” have a definite, non-zero answer even if no test charge is placed there. The field is a property of space itself, not of the test charge.

✓ Checkpoint · Why Fields Exist

A positive charge $Q$ sits in an otherwise empty room. A student says: “There is no electric field anywhere in the room because there is nothing for $Q$ to push on.” State precisely what is wrong with this claim.
The electric field is a property of space created by $Q$, independent of whether any other charge is present. At every point $P$ in the room, $\vec{E}(P) = kQ/r^2\,\hat{r}$ where $r$ is the distance from $Q$ to $P$. A test charge placed at $P$ would experience a force $q_0\vec{E}$, but the field exists whether or not $q_0$ is there.
A gravitational field $\vec{g}$ and an electric field $\vec{E}$ both exist at a point $P$ in space. State one structural similarity and one qualitative difference between the two fields.
Similarity: both are vector fields that assign a force-per-unit-source quantity to every point in space; both are created by their respective sources and exist independently of any test object. Difference: $\vec{g}$ always points toward its source (mass has one sign); $\vec{E}$ can point toward or away from its source depending on the sign of the charge. Consequently $\vec{E}$ can be shielded (by conductors) whereas $\vec{g}$ cannot.
If you could suddenly remove a charge from a distant location, would a nearby test charge feel the change immediately? Explain in terms of the field concept.
No. The field propagates at the speed of light $c$. The information that the source has been removed travels outward as a disturbance in the field at speed $c$. A test charge at distance $r$ does not respond until a time $t = r/c$ has elapsed. Action at a distance (instantaneous response) is incompatible with the field picture and with special relativity.
A charged capacitor is discharged through a resistor. After the discharge is complete, the electric field between the plates is zero. Where did the energy that was stored in the capacitor go, and where was it stored before the discharge?
The energy was stored in the electric field between the plates (energy density $u = \tfrac{1}{2}\varepsilon_0 E^2$). During discharge the field collapses; the energy is transferred to the resistor and dissipated as heat. This example establishes that the field is a physical carrier of energy, not merely a mathematical device.

Worked Examples

Example 1 · Feeling the Field After a Delay

A charged sphere is suddenly jolted sideways. A tiny test charge sits $3.0\,\mathrm{km}$ away in vacuum. (a) How long after the jolt does the test charge begin to respond? (b) What actually crosses the gap to carry the news?

Solution. Given. Separation $r = 3.0\,\mathrm{km} = 3.0\times10^{3}\,\mathrm{m}$; the intervening space is vacuum. Find. The response delay $t$, and the physical carrier of the change. Setup. A change in the source is not felt instantaneously (that would be action at a distance). The disturbance travels outward through the field at the speed of light, $c = 3.0\times10^{8}\,\mathrm{m/s}$, so $t = r/c$. Solve. \[ t = \frac{r}{c} = \frac{3.0\times10^{3}}{3.0\times10^{8}} = 1.0\times10^{-5}\,\mathrm{s} = 10\,\mu\mathrm{s}. \] Answer. The test charge stays put for $10\,\mu\mathrm{s}$, then begins to move. What crosses the gap is a disturbance in the electric field—not the source charge, and not any material substance. Check. Light itself covers $3\,\mathrm{km}$ in $10\,\mu\mathrm{s}$, and the field change propagates at exactly light speed. Had we assumed instantaneous action, $t$ would be $0$—the very absurdity Newton distrusted. ✓

Charge

The electric field is generated by charge. Before we can write down how it is generated, we need a precise account of what charge is.

Charge as an Intrinsic Property

Every known particle either carries electric charge or it does not. This is not a property that can be added or removed by processing the particle—it is an intrinsic attribute, as fundamental as mass.

An electron carries charge $-e = -1.6 \times 10^{-19}\,\mathrm{C}$. A proton carries charge $+e = +1.6 \times 10^{-19}\,\mathrm{C}$. A neutron carries charge zero. The photon carries charge zero.

These values are fixed constants of nature. No force, no chemical reaction, no temperature change alters the charge of an electron.

In ordinary matter, atoms are electrically neutral because they contain equal numbers of protons (in the nucleus) and electrons (in the shell). When we say an object is “charged,” we mean it has a net surplus or deficit of electrons relative to its proton count. No protons are moved; no electrons are created or destroyed. Electrons are transferred.

Core intuition: Charge: The Intrinsic Source of Electric Interaction

Electric charge $q$ is the intrinsic property of a particle that determines:

The strength of the electric force it exerts on others.
The strength of the electric force it experiences from others.
The sign of each interaction (attraction or repulsion).

SI unit: coulomb (C). The elementary charge is: \[ e = 1.602 \times 10^{-19}\,\mathrm{C}. \] A coulomb is an enormous macroscopic unit: $1\,\mathrm{C}$ corresponds to approximately $6.24 \times 10^{18}$ elementary charges.

Sign and Type

Charge comes in exactly two varieties, labelled by convention as positive and negative. The labelling is arbitrary—Benjamin Franklin's convention happened to assign “positive” to the carrier that we now know is the proton—but the distinction is real and non-negotiable.

The rule governing sign:

Like charges (both positive or both negative) repel.
Unlike charges (one positive, one negative) attract.

This is qualitatively different from gravity, where the analogue of charge is mass, and mass has only one sign. Gravity is always attractive. The electric force can be either.

The physical consequence is profound: large accumulations of matter are electrically neutral, because any slight excess of one sign attracts the opposite sign until neutrality is restored. Gravity has no such self-neutralising mechanism. This is the answer to Perplexing Question 5: electric forces neutralise at large scales, so gravity—though vastly weaker—wins by default.

Core intuition: Why Gravity Dominates the Universe Despite Being Weaker

All macroscopic matter is electrically neutral to extraordinary precision. The electric force between neutral objects is essentially zero at large separations.
Mass cannot be negative, so gravitational effects never cancel. Every kilogram of mass in the universe adds to the total gravitational influence.
Over cosmological distances and timescales, gravity accumulates without limit; electric forces have already averaged out.

Resolution of PQ 5. The electric force is not “weaker” at the cosmological scale—it simply ceases to exist there, because matter is neutral. Gravity, lacking any such cancellation, is all that is left.

Quantisation of Charge

Electric charge does not come in arbitrary amounts. It is quantised: every observable charge is an integer multiple of the elementary charge $e$.

If $q$ is any measurable charge: \[ q = ne, \qquad n \in \mathbb{Z}. \]

This was established by Millikan's oil-drop experiment (1909), in which charged oil droplets in a known electric field revealed that their charges were always integer multiples of a fixed minimum.

Two notes of caution:

Quarks carry fractional charges ($+\tfrac{2}{3}e$ or $-\tfrac{1}{3}e$), but they are permanently confined inside hadrons. Free quarks have never been isolated. The minimum observable free charge remains $e$.
The origin of charge quantisation is not explained within classical electromagnetism. It is a consequence of deep quantum-field theory considerations. We accept it as an experimental fact.

Misconception alert: Charge as a “Substance” vs. an Intrinsic Property

The misconception: Charge is a kind of fluid or substance that flows through wires and accumulates on surfaces. A “positively charged” body has “extra positive fluid” on it. Why it is wrong:

When you charge a conductor negatively, you have added electrons. No positive fluid has left. When you charge it positively, you have removed electrons. No positive fluid has arrived. Only electrons—discrete, massive particles—move.
The “fluid” model cannot account for quantisation. A true fluid can be divided into arbitrarily small amounts; charge cannot.
The model creates systematic errors in problems involving induction, grounding, and redistribution of charge on conductors.

Correct picture: A “charged” body has a non-zero integer surplus of one type of elementary carrier. Charge is not a substance; it is a count.

Conservation of Charge

Charge is conserved absolutely. In any isolated system, the total charge—counting signs—never changes.

Core intuition: Conservation of Charge

For an isolated system: \[ Q_{\text{total}} = \sum_i q_i = \text{constant}. \] This is a local conservation law: charge cannot disappear at one point and reappear elsewhere without a continuous flow connecting them. This is more powerful than a mere bookkeeping identity—it constrains field propagation and underlies Kirchhoff's current law in circuits (Chapter 2). Charge conservation holds in every known process:

Chemical reactions: rearrange electrons, total charge unchanged.
Nuclear reactions: proton and neutron numbers change, but total charge is conserved ($\beta^+$ and $\beta^-$ decays each produce a lepton that carries away the required charge).
Pair production ($\gamma \to e^+ + e^-$): total charge before and after is zero.

Charge conservation and Coulomb's law. Notice that conservation of charge constrains totals, not distributions. Charge can flow from one part of a conductor to another; it can be transferred between objects by contact; it can be induced on surfaces by proximity. None of these processes violate conservation—they only redistribute charge while keeping the algebraic sum fixed.

This distinction matters for problem solving: when you rub two identical conducting spheres together, or connect them briefly with a wire, or separate them after induction, you must track the total charge at each step.

✓ Checkpoint · Charge

A neutral copper sphere contains approximately $2.4 \times 10^{24}$ free electrons. You transfer $10^{12}$ electrons from the sphere to a rubber rod. What is the resulting charge on the sphere? Is the fractional change in the sphere's electron population significant?
Charge transferred away: $10^{12} \times 1.6\times10^{-19}\,\mathrm{C} = 1.6\times10^{-7}\,\mathrm{C} = 0.16\,\mu\mathrm{C}$. Sphere becomes positive: $q = +0.16\,\mu\mathrm{C}$. Fractional change: $10^{12}/2.4\times10^{24} \approx 4\times10^{-13}$— less than one part in a trillion. Macroscopic charging involves a negligible fraction of available electrons.
In a nuclear reaction, a neutron decays into a proton, an electron, and an antineutrino: $\mathrm{n} \to \mathrm{p} + \mathrm{e}^- + \bar{\nu}_e$. Verify that charge is conserved.
Before: neutron has charge $0$. After: proton $+e$, electron $-e$, antineutrino $0$. Total after: $+e + (-e) + 0 = 0$. ✓ Charge is conserved exactly.
Two identical conducting spheres carry charges $+7\,\mu\mathrm{C}$ and $-3\,\mu\mathrm{C}$. They are touched together and separated. A third identical sphere (uncharged) is then touched to one of them and removed. What are the final charges on all three spheres?
After first contact: both spheres carry $(7-3)/2 = +2\,\mu\mathrm{C}$. Call them $A$ and $B$, both at $+2\,\mu\mathrm{C}$. Third sphere $C$ (uncharged) touches $A$: $(2+0)/2 = +1\,\mu\mathrm{C}$ each. Final: $q_A = +1\,\mu\mathrm{C}$, $q_B = +2\,\mu\mathrm{C}$, $q_C = +1\,\mu\mathrm{C}$. Total: $1+2+1 = 4\,\mu\mathrm{C}$; original total was $7+(-3)=4\,\mu\mathrm{C}$. ✓
A student claims: “Charge quantisation is irrelevant in circuit problems because individual electrons are too small to matter.” Is this claim defensible? Identify one situation where quantisation has a measurable consequence.
The claim is defensible for macroscopic currents: a current of $1\,\mathrm{mA}$ involves $\sim 6\times10^{15}$ electrons per second, and discreteness is undetectable. However, in single-electron transistors (SET devices), charge quantisation is the operating principle—adding or removing a single electron switches the device. Quantisation matters whenever the number of charge carriers is small.

Worked Examples

Example 2 · Counting the Missing Electrons

A metal pith ball carries a net charge $q = -4.8\,\mu\mathrm{C}$. (a) How many excess electrons does it hold? (b) By how much has its mass changed because of them?

Solution. Given. $q = -4.8\,\mu\mathrm{C} = -4.8\times10^{-6}\,\mathrm{C}$; $e = 1.6\times10^{-19}\,\mathrm{C}$; $m_e = 9.11\times10^{-31}\,\mathrm{kg}$. Find. The number $n$ of excess electrons and the added mass $\Delta m$. Setup. Charge is quantised: $|q| = n e$, so $n = |q|/e$. The added mass is $\Delta m = n m_e$. Solve. \[ n = \frac{|q|}{e} = \frac{4.8\times10^{-6}}{1.6\times10^{-19}} = 3.0\times10^{13}, \qquad \Delta m = n m_e = (3.0\times10^{13})(9.11\times10^{-31}) \approx 2.7\times10^{-17}\,\mathrm{kg}. \] Answer. About $3.0\times10^{13}$ excess electrons; the mass rises by only $\approx 2.7\times10^{-17}\,\mathrm{kg}$. Check. A charge in microcoulombs divided by the tiny elementary charge must give a huge integer—and it does ($10^{13}$). The mass change is utterly negligible, confirming that “charging” rearranges an imperceptible sliver of the ball's electrons. ✓

Example 3 · Is This a Possible Charge?

Charge quantisation forbids arbitrary values. (a) Could a body carry $q = 8.0\times10^{-19}\,\mathrm{C}$? (b) Could it carry $q = 2.4\times10^{-19}\,\mathrm{C}$? (c) In the alpha decay $^{238}_{\ 92}\mathrm{U}\to{}^{234}_{\ 90}\mathrm{Th}+{}^{4}_{2}\mathrm{He}$, verify charge is conserved.

Solution. Given. $e = 1.6\times10^{-19}\,\mathrm{C}$; two candidate charges; one decay. Find. Whether each charge is an integer multiple of $e$; the charge balance. Setup. A free charge is allowed only if $q/e$ is a whole number. Solve. \[ \frac{8.0\times10^{-19}}{1.6\times10^{-19}} = 5 \ (\text{integer}), \qquad \frac{2.4\times10^{-19}}{1.6\times10^{-19}} = 1.5 \ (\text{not an integer}). \] Decay: before, $Z = 92$ (charge $+92e$). After, $90e + 2e = 92e$. Answer. (a) Allowed ($n=5$). (b) Forbidden—no free charge of $1.5e$ exists. (c) $+92e \to +90e + 2e$: charge conserved. ✓ Check. Only whole-number multiples of $e$ survive; $1.5e$ would require a free fractional charge, never observed (quarks are confined). The decay's proton count is unchanged in total, as conservation demands. ✓

Coulomb's Law

◈ Coulomb’s law bench

live bench

Slide the separation and flip or scale each charge — the force vector and the inverse-square verdict update live.

◈ loading the interactive bench…

↗ open in the Simulations page

With the field concept established and charge defined, we can now ask: how does a point charge generate the electric field around it? The answer at the electrostatic level is Coulomb's law.

The Force Between Two Point Charges

Experiment establishes the following facts about the force between two stationary point charges $q_1$ and $q_2$ separated by distance $r$:

The force is proportional to each charge: $F \propto |q_1|$ and $F \propto |q_2|$.
The force falls off as the inverse square of the separation: $F \propto 1/r^2$.
The force acts along the line joining the two charges.
Like charges repel; unlike charges attract.
The forces on the two charges are equal in magnitude and opposite in direction (Newton's Third Law).

Combining these: \[ F = k \frac{|q_1||q_2|}{r^2}, \qquad k = \frac{1}{4\pi\varepsilon_0} = 9.0 \times 10^9\,\mathrm{N\,m^2\,C^{-2}}. \]

The constant $\varepsilon_0 = 8.85 \times 10^{-12}\,\mathrm{C^2\,N^{-1}\,m^{-2}}$ is the permittivity of free space. It characterises how easily the vacuum supports electric fields.

Core intuition: The Geometric Origin of the $1/r^2$ Dependence

The inverse-square law is not a coincidence. It is a consequence of geometry in three-dimensional space. Imagine the electric “influence” of a point charge spreading outward uniformly in all directions, like ripples from a stone dropped in a pool— but in three dimensions. At distance $r$, this influence is spread over a sphere of area $4\pi r^2$. If the total influence is fixed (charge is conserved; it does not leak away), then the influence per unit area—the field strength—must be: \[ E \propto \frac{1}{4\pi r^2} \propto \frac{1}{r^2}. \] This argument is geometric, not empirical. Any quantity that spreads uniformly in three dimensions from a point source, with no absorption or generation, must obey an inverse-square law. Gravity does it. Light intensity does it. Sound intensity does it. Electric field does it. The $1/r^2$ law is how three-dimensional geometry looks. The formal version of this argument is Gauss's law, which we derive in Section .

Vector Form of Coulomb's Law

The scalar form $F = k|q_1||q_2|/r^2$ gives the magnitude but discards the direction. For vector calculations—superposition, equilibrium, field calculations— the full vector form is essential.

◇ Figure 1

Fig 1. Coulomb's law as an action–reaction pair. The forces are equal in magnitude and opposite in direction ($\vec F_{12}=-\vec F_{21}$); like charges repel, as shown.

Birth of the formula: Vector Formulation of Coulomb's Law

Setup. Let charge $q_1$ be at position $\vec{r}_1$ and charge $q_2$ at $\vec{r}_2$. Define: \[ \vec{r}_{12} = \vec{r}_2 - \vec{r}_1 \qquad \text{(vector from $q_1$ to $q_2$)}. \] \[ r = |\vec{r}_{12}| \qquad \text{(distance between them)}. \] \[ \hat{r}_{12} = \frac{\vec{r}_{12}}{r} \qquad \text{(unit vector from $q_1$ to $q_2$)}. \] Force on $q_2$ due to $q_1$: \[ \vec{F}_{21} = k \frac{q_1 q_2}{r^2}\,\hat{r}_{12}. \] Unpacking the sign logic:

If $q_1 q_2 \gt 0$ (like charges): $\vec{F}_{21}$ points in the direction $\hat{r}_{12}$, i.e. away from $q_1$. Repulsion. ✓
If $q_1 q_2 \lt 0$ (unlike charges): $\vec{F}_{21}$ points opposite to $\hat{r}_{12}$, i.e. toward $q_1$. Attraction. ✓

By Newton's Third Law: \[ \vec{F}_{12} = -\vec{F}_{21} = k \frac{q_1 q_2}{r^2}(-\hat{r}_{12}) = k \frac{q_1 q_2}{r^2}\,\hat{r}_{21}. \] Equivalent form using position vectors directly: \[ \vec{F}_{21} = k \frac{q_1 q_2}{|\vec{r}_2 - \vec{r}_1|^3} (\vec{r}_2 - \vec{r}_1). \] This form is computationally convenient when coordinates are given explicitly—no need to separately compute the unit vector. Critical notation check: $\vec{F}_{21}$ is the force on $q_2$ due to $q_1$. The subscript convention is: “$\text{on} \leftarrow \text{due to}$,” read right-to-left. Inconsistent notation is the single most common source of sign errors in multi-charge problems.

Comparison With Gravity

The structural similarity between Coulomb's law and Newton's law of gravitation is striking:

Property	Gravity	Electrostatics
Law	$F = G m_1 m_2 / r^2$	$F = k q_1 q_2 / r^2$
Source	Mass $m$	Charge $q$
Constant	$G = 6.67\times10^{-11}$ N m$^2$ kg$^{-2}$	$k = 9.0\times10^{9}$ N m$^2$ C$^{-2}$
Sign	Always attractive	Depends on sign of $q_1 q_2$
Distance dependence	$1/r^2$	$1/r^2$
Shielding possible?	No	Yes (by conductors)

The strength ratio. For a proton–electron pair ($m_p = 1.67\times10^{-27}$ kg, $m_e = 9.11\times10^{-31}$ kg, $q = e = 1.6\times10^{-19}$ C, $r$ cancels):

\[ \frac{F_{\text{elec}}}{F_{\text{grav}}} = \frac{ke^2}{G m_p m_e} = \frac{(9\times10^9)(1.6\times10^{-19})^2} {(6.67\times10^{-11})(1.67\times10^{-27})(9.11\times10^{-31})} \approx 2.3 \times 10^{39}. \]

The electric force between a proton and an electron is more than $10^{39}$ times stronger than the gravitational force between them. This number is independent of the separation $r$ because both forces scale as $1/r^2$.

Superposition Principle

When more than two charges are present, the force on any one charge is the vector sum of the forces exerted on it by each other charge individually. No pair of charges “knows” about the others. This is the superposition principle.

Core intuition: Superposition Principle for Electric Forces

For a system of $n$ point charges $q_1, q_2, \ldots, q_n$, the total force on charge $q_i$ is: \[ \vec{F}_i = \sum_{\substack{j=1 \\ j \neq i}}^{n} k \frac{q_i q_j}{r_{ij}^2}\,\hat{r}_{ji}, \] where $r_{ij}$ is the distance between $q_i$ and $q_j$, and $\hat{r}_{ji}$ is the unit vector from $q_j$ toward $q_i$. Superposition holds exactly in classical electromagnetism. It is not an approximation. Fields (and forces) from different sources add as vectors without any modification due to the presence of the other sources.

Why superposition matters. The superposition principle is what makes Gauss's law work, enables the definition of electric potential, and ultimately allows us to compute the field of any arbitrary charge distribution by integration. Every result we derive from here onward rests on this principle.

Misconception alert: Four Common Errors in Coulomb's Law Problems

Error 1: Using $r$ in centimetres. The SI unit of force is newtons. Coulomb's constant $k = 9\times10^9\,\mathrm{N\,m^2\,C^{-2}}$ requires $r$ in metres. Always convert before substituting. Error 2: Using $|q_1 q_2|$ in the vector form. The scalar magnitude uses $|q_1||q_2|$. The vector form uses $q_1 q_2$ with signs, so that the sign of the product automatically gives the correct direction (positive product $\Rightarrow$ repulsion, negative product $\Rightarrow$ attraction). Mixing the two conventions generates ghost sign errors. Error 3: Superposition by adding magnitudes. When three or more charges are present, the net force on any one charge is the vector sum of individual forces. The magnitudes can only be added when all forces are parallel. The general case requires component decomposition. Error 4: “Zero force means zero field.” If the net force on a test charge is zero at a point, the electric field there is also zero. But if a problem asks for the field at a point where no charge is placed, there is no force to be zero—yet the field can be nonzero. These are different questions.

✓ Checkpoint · Coulomb's Law

The distance between two point charges is halved, and simultaneously both charges are doubled. By what factor does the force between them change?
$F \propto q_1 q_2 / r^2$. New force: $(2q_1)(2q_2)/(r/2)^2 = 4q_1q_2/(r^2/4) = 16\,q_1q_2/r^2$. The force increases by a factor of $\mathbf{16}$.
Charge $q_1 = +3\,\mu\mathrm{C}$ is at the origin and $q_2 = +3\,\mu\mathrm{C}$ is at $x = 6\,\mathrm{cm}$. Where on the $x$-axis should a third charge $q_3$ be placed so that the net force on $q_3$ is zero? Does the answer depend on the sign or magnitude of $q_3$?
Since $q_1 = q_2$, the forces from each on $q_3$ are equal in magnitude when $q_3$ is equidistant from both, i.e. at $x = 3\,\mathrm{cm}$. At this midpoint the two forces on $q_3$ are equal and opposite regardless of the sign or magnitude of $q_3$ (the sign of $q_3$ reverses both forces simultaneously, maintaining cancellation). The equilibrium position depends only on the source charges, not on $q_3$.
Three charges $+q$, $+q$, and $-q$ are placed at the three corners of an equilateral triangle of side $a$. Without calculating, determine by symmetry arguments whether the net force on the $-q$ charge points toward, away from, or along the base of the triangle.
The two $+q$ charges attract the $-q$ charge. By symmetry these two attraction forces are equal in magnitude and make equal angles with the altitude from the $-q$ vertex. Their horizontal components cancel; their vertical components add along the altitude, pointing from the $-q$ vertex toward the midpoint of the base (toward the base). The net force on $-q$ points along the altitude, toward the base.
The electric force between a proton and an electron in a hydrogen atom ($r \approx 5.3\times10^{-11}\,\mathrm{m}$) is approximately $8.2\times10^{-8}\,\mathrm{N}$. Using only this value and $g = 9.8\,\mathrm{m/s^2}$, find the mass of an object whose weight equals this electric force. Comment on the result.
$m = F/g = 8.2\times10^{-8}/9.8 \approx 8.4\times10^{-9}\,\mathrm{kg} \approx 8.4\,\mathrm{ng}$ (nanograms). A force holding together an atom equals the weight of a speck of dust visible to the naked eye. At atomic scales, electric forces are enormous compared to any macroscopic intuition about “small” forces.

Worked Examples

Example 4 · Two Perpendicular Coulomb Forces

A charge $q_3 = +2\,\mu\mathrm{C}$ sits at the origin. A charge $q_1 = +5\,\mu\mathrm{C}$ is at $(-0.30,0)\,\mathrm{m}$ and $q_2 = +5\,\mu\mathrm{C}$ is at $(0,-0.40)\,\mathrm{m}$. Find the net force on $q_3$ (magnitude and direction).

Solution. Given. $q_3=+2\,\mu\mathrm{C}$ at origin; $q_1=+5\,\mu\mathrm{C}$ at $x=-0.30\,\mathrm{m}$; $q_2=+5\,\mu\mathrm{C}$ at $y=-0.40\,\mathrm{m}$; $k=9.0\times10^{9}$. Find. $\vec{F}_{\text{net}}$ on $q_3$. Setup. Each like-charge pair repels, so $q_1$ pushes $q_3$ along $+x$ and $q_2$ pushes it along $+y$. The two forces are perpendicular; add by Pythagoras. Solve. \[ F_1 = \frac{k q_1 q_3}{(0.30)^2} = \frac{(9\times10^9)(5\times10^{-6})(2\times10^{-6})}{0.090} = 1.00\,\mathrm{N}\ (+x), \] \[ F_2 = \frac{k q_2 q_3}{(0.40)^2} = \frac{0.090}{0.160} = 0.5625\,\mathrm{N}\ (+y). \] \[ F_{\text{net}} = \sqrt{F_1^2+F_2^2} = \sqrt{1.00^2+0.5625^2} = 1.15\,\mathrm{N}, \qquad \phi = \tan^{-1}\!\frac{0.5625}{1.00} = 29.4^\circ\ \text{above } +x. \] Answer. $F_{\text{net}}\approx 1.15\,\mathrm{N}$, directed $29.4^\circ$ above the $+x$-axis (into the first quadrant, away from both source charges). Check. The stronger, closer charge ($q_1$ at $0.30\,\mathrm{m}$) dominates, so the resultant leans toward $+x$ ($\phi\lt 45^\circ$). Both forces are repulsive and point away from their sources, consistent with three positive charges. ✓

Example 5 · Where the Net Force Vanishes

Charge $+Q$ is at the origin and $+4Q$ is at $x = d = 0.60\,\mathrm{m}$. Where on the line joining them can a third charge be placed so that it feels no net force? Does the answer depend on the third charge's sign?

Solution. Given. $+Q$ at $x=0$, $+4Q$ at $x=d=0.60\,\mathrm{m}$; test charge $q$ placed at $x$. Find. The null position $x$; its dependence (if any) on $q$. Setup. Between two like charges the two forces oppose. Set their magnitudes equal: $\dfrac{kQq}{x^2} = \dfrac{k(4Q)q}{(d-x)^2}$. The $q$ and $kQ$ cancel. Solve. \[ \frac{1}{x^2} = \frac{4}{(d-x)^2} \;\Rightarrow\; (d-x)^2 = 4x^2 \;\Rightarrow\; d-x = 2x \;\Rightarrow\; x = \frac{d}{3} = 0.20\,\mathrm{m}. \] Answer. At $x = d/3 = 0.20\,\mathrm{m}$ from the $+Q$ charge (nearer the weaker charge). The position is independent of the test charge's sign and magnitude—reversing $q$ reverses both forces together, preserving cancellation. Check. The null must lie closer to the smaller charge, and $0.20 \lt 0.30$ (the midpoint), so it sits on the $+Q$ side—correct. Taking the negative root $d-x=-2x$ gives $x=-d$, an exterior point where the forces add, not cancel; rejected. ✓

Solved examples

Five fully-worked problems from this chapter, free — solution and answer shown in full. The complete set of worked examples is in the full book.

Example 1 · Charge from a Number of Electrons

A neutral metal sphere gains $5.0\times10^{12}$ electrons. Find its charge, and state the sign.

Solution. Given: $N = 5.0\times10^{12}$ excess electrons; $e = 1.6\times10^{-19}\,\mathrm{C}$.
Find: the net charge $Q$.
Setup: charge is quantised, $Q = \pm Ne$; adding electrons makes the sphere negative.
Solve: \[ Q = -Ne = -(5.0\times10^{12})(1.6\times10^{-19}) = -8.0\times10^{-7}\,\mathrm{C}. \] Answer: $\boxed{Q = -0.80\,\mu\mathrm{C}}$ (negative). Check: $0.80\,\mu\mathrm{C}$ is a typical lab charge, yet it is only $5\times10^{12}$ of the sphere's $\sim10^{23}$ electrons — a vanishing fraction, as charging should be. ✓

Example 2 · Coulomb Force: Magnitude and Direction

Charge $q_1 = +4\,\mu\mathrm{C}$ is fixed at the origin and $q_2 = -6\,\mu\mathrm{C}$ at $x = 0.30\,\mathrm{m}$. Find the force on $q_2$ due to $q_1$.

Solution. Given: $q_1 = +4\,\mu\mathrm{C}$, $q_2 = -6\,\mu\mathrm{C}$, $r = 0.30\,\mathrm{m}$.
Find: $\vec F_{21}$.
Setup: Coulomb's law for magnitude; opposite signs $\Rightarrow$ attraction, so the force on $q_2$ points toward $q_1$ ($-\hat x$).
Solve: \[ F = k\frac{|q_1||q_2|}{r^2} = (9\times10^9)\frac{(4\times10^{-6})(6\times10^{-6})}{(0.30)^2} = 2.4\,\mathrm{N}. \] Answer: $\boxed{\vec F_{21} = -2.4\,\hat x\,\mathrm{N}}$ (toward $q_1$). Check: opposite signs give attraction, so the force on $q_2$ must point back toward $q_1$ ($-\hat x$) — the sign of the answer matches the physics. ✓

Example 3 · Field of a Point Charge

Find the magnitude of the electric field $0.40\,\mathrm{m}$ from a $+10\,\mathrm{nC}$ point charge.

Solution. Given: $Q = +10\,\mathrm{nC}$, $r = 0.40\,\mathrm{m}$.
Find: $E$.
Setup: $E = kQ/r^2$, directed radially outward (positive charge).
Solve: \[ E = \frac{(9\times10^9)(10\times10^{-9})}{(0.40)^2} = \frac{90}{0.16} = 562.5\,\mathrm{N/C}. \] Answer: $\boxed{E \approx 5.6\times10^{2}\,\mathrm{N/C}}$, pointing away from the charge. Check: the force on a $+1\,\mathrm{C}$ probe here would be $560\,\mathrm{N}$; units $\mathrm{N/C}$ and the outward direction (positive source) are both correct. ✓

Example 4 · Net Flux from Enclosed Charge

A closed surface encloses $+4\,\mu\mathrm{C}$ and $-1\,\mu\mathrm{C}$. A further $+10\,\mu\mathrm{C}$ sits outside it. Find the net outward flux.

Solution. Given: enclosed $+4$ and $-1\,\mu\mathrm{C}$; external $+10\,\mu\mathrm{C}$.
Find: $\Phi_E$.
Setup: Gauss's law — only enclosed charge counts; external charge contributes zero net flux.
Solve: \[ Q_{\text{enc}} = (+4-1)\,\mu\mathrm{C} = +3\,\mu\mathrm{C},\qquad \Phi_E = \frac{Q_{\text{enc}}}{\varepsilon_0} = \frac{3\times10^{-6}}{8.85\times10^{-12}}. \] Answer: $\boxed{\Phi_E \approx 3.4\times10^{5}\, \mathrm{N\,m^2\,C^{-1}}}$. The $+10\,\mu\mathrm{C}$ outside is irrelevant. Check: only $Q_{\text{enc}}=+3\,\mu\mathrm{C}$ enters $\Phi=Q_{\text{enc}}/\varepsilon_0$; the external charge changes the field on the surface but adds zero net flux. ✓

Example 5 · Flux Through a Tilted Square

A uniform field $\vec E = 500\,\hat x\,\mathrm{N/C}$ passes through a square of side $0.30\,\mathrm{m}$ whose normal makes $60^\circ$ with $\hat x$. Find the flux.

Solution. Given: $E = 500\,\mathrm{N/C}$, $a = 0.30\,\mathrm{m}$, $\theta = 60^\circ$.
Find: $\Phi_E$.
Setup: $\Phi_E = EA\cos\theta$ with $A = a^2$.
Solve: \[ \Phi_E = (500)(0.09)\cos 60^\circ = (500)(0.09)(0.5) = 22.5\,\mathrm{N\,m^2\,C^{-1}}. \] Answer: $\boxed{\Phi_E = 22.5\,\mathrm{N\,m^2\,C^{-1}}}$ — exactly half the head-on maximum ($45$), as $\cos 60^\circ = 0.5$. Check: tilting the normal to $90^\circ$ would kill the flux entirely; at $60^\circ$ we keep half, the expected fraction. ✓

Electrostatics

Perplexing Questions

Why Fields Exist

The Problem With Action at a Distance

Field as Physical Reality

Connection to Gravitational Field

Worked Examples

Charge

Charge as an Intrinsic Property

Sign and Type

Quantisation of Charge

Conservation of Charge

Worked Examples

Coulomb's Law

The Force Between Two Point Charges

Vector Form of Coulomb's Law

Comparison With Gravity

Superposition Principle

Worked Examples

Solved examples

Problem bank

Chapter test

The chapter continues.