Electric Charge and Static Electricity

Beginner

Pull a wool jumper over your head in a dark room and you may see tiny sparks and hear them crackle. That is static electricity: electric charge that has built up on an object and stays put.

Step 1: two kinds of charge. Everything is made of atoms. In the middle of each atom sit protons, which carry positive charge (+); around them move much lighter electrons, which carry negative charge (−). An ordinary object has exactly as many electrons as protons, so it is neutral.

Step 2: the rule of the game. Like charges push each other away; opposite charges pull each other together.

Step 3: charging by rubbing. Some materials hold on to their electrons more tightly than others. Rub a rubber balloon on a wool scarf and electrons leave the wool for the balloon. The balloon now has extra electrons, so it is negative; the wool is short of the same number, so it is positive. No charge is made or destroyed: it only moves, and the total stays the same.

Step 4: conductors and insulators. In metals some electrons roam freely, so extra charge spreads over the whole object; these are conductors. In rubber, plastic, paper and wool the electrons stay put, so charge stays where it was put; these are insulators.

Step 5: pulling on neutral things. A charged balloon nudges the charges inside nearby paper, so the side facing it becomes slightly positive. That side is closer, so the pull wins and the paper jumps up.

Check yourself: a plastic comb gains electrons from your hair. Is the comb positive or negative? Will it attract or repel your hair? Answer: negative; attract, because your hair is now positive.

In the animation, blue electrons hop from the wool onto the balloon, leaving red plus signs behind; then bits of paper leap up to the balloon.

Intermediate

What charge is, and how it is counted

Electric charge is a property of the particles that atoms are made of. It comes in two kinds, named positive and negative by Benjamin Franklin: protons in the nucleus carry positive charge, the much lighter electrons around it negative charge, and neutrons none. Like charges repel, unlike charges attract. An atom has as many electrons as protons, so it is neutral: its net charge is zero, but it holds plenty of both kinds.

Charge is measured in coulombs (C), and static charges are usually nanocoulombs (1 nC is $10^{-9}\mathrm{C}$) or microcoulombs. It comes in whole numbers of a smallest portion, the elementary charge: a proton carries $+e$ and an electron $-e$, where
$e=1.602\times 10^{-19}\mathrm{C}$
so the charge of any object is
$q=\pm ne,n=0,1,2,\dots$

In solids the nuclei stay locked in place and only electrons move, so $n$ counts the electrons gained (sign $-$) or lost (sign $+$). One coulomb is $6.24\times 10^{18}$ elementary charges.

Rubbing separates charge; the total is conserved

When two different materials touch, some surface electrons move to the one that binds them more tightly; rubbing multiplies the points of contact. Rubber, PVC and polythene take electrons from wool, hair and skin, while glass loses some to silk. Such rankings, called triboelectric series, are rough guides: humidity and dirt can change them.

Rubbing does not make charge; it separates charge that was already there. In an isolated system, one that no charge enters or leaves, the total never changes:
$\sum q=\mathrm{const}\text{(isolated system)}$
The sum is taken with signs: a negative balloon and an equally positive scarf add up to zero, as they did before they touched. No experiment has ever found this law broken.

Worked example: the balloon in the animation

In the animation the balloon is rubbed with four up-and-down strokes, and the counter climbs to $n=6.0\times 10^{10}$ electrons; each blue minus sign that hops across stands for $10^{10}$ of them. The balloon has gained electrons, so
$q=-ne=-6.0\times 10^{10}\cdot 1.602\times 10^{-19}\mathrm{C}$
$q=-9.6\times 10^{-9}\mathrm{C}=-9.6\mathrm{nC}$

The wool has lost exactly those electrons, so it carries $+9.6\mathrm{nC}$. These are the final values on the blue and red readouts; at every moment of the rubbing the two are equal and opposite, as every red plus sign left on the wool matches a blue minus sign on the balloon.

Is that a lot? A 3 g balloon contains about $10^{24}$ electrons, so only one in 17 trillion is extra. Electric forces are so strong that even this imbalance lifts paper.

Conductors and insulators

In a metal, one or two electrons per atom roam freely through the whole piece, so extra charge spreads out at once and settles on the outer surface. Such materials are conductors: metals, salty water and the human body, whose charge carriers are ions.

In insulators such as rubber, plastic, glass and dry paper, electrons stay bound to their atoms, so charge stays where it was put. That is why the animation turns the balloon's rubbed side towards the paper. A metal spoon held in your bare hand cannot be charged by rubbing: the charge escapes through you to the ground.

Connecting a conductor to the ground, which can give or take any amount of charge, is called earthing. In damp weather a water film on surfaces also leaks charge, though more slowly, so static tricks work best on dry days.

Worked example: sharing charge between spheres

When two conductors touch, electrons flow until nothing pushes them further. Identical conductors must, by symmetry, end with equal charges, and conservation fixes the total:
$q^{\prime }=\frac{q_1+q_2}{2}\text{(identical conductors in contact)}$

Three identical metal spheres stand on insulating stands: A carries $-12\mathrm{nC}$, B is neutral and C carries $+4\mathrm{nC}$. A touches B, then B touches C. What does each sphere end with?

First contact: the total is $-12\mathrm{nC}$, so A and B end with $-6\mathrm{nC}$ each, after $6\mathrm{nC}/e=3.7\times 10^{10}$ electrons have moved from A to B. Second contact: the total is $-6+4=-2\mathrm{nC}$, so B and C end with $-1\mathrm{nC}$ each. Check: $-6-1-1=-8\mathrm{nC}$, the starting total $-12+0+4$. In the second contact $5\mathrm{nC}/e=3.1\times 10^{10}$ electrons crossed from B to C, cancelling its $+4\mathrm{nC}$ and leaving it slightly negative; no positive charge travelled.

Attraction without contact: polarisation and induction

A charged object can pull on a neutral one without touching it. Bring a negative rod near a metal ball: the ball's free electrons are pushed to the far side, leaving the near side positive. This is electrostatic induction. The ball stays neutral, but its positive side is closer to the rod, and electric forces weaken with distance, so the attraction wins.

In paper, an insulator, electrons cannot travel; instead each molecule is stretched slightly into a tiny dipole, a pair of equal and opposite charges a small distance apart. The paper is polarised: positive on the face towards the balloon and negative on the far face, as the close-up shows with its + and − signs drawing apart.

The arrows on the middle bit show the contest. The $mg$ arrow, the bit's weight, stays the same, while the green $F$ arrow, the electric pull, grows faster and faster as the balloon comes down. When $F$ is as long as $mg$, the bit takes off; the middle bit, right under the balloon, goes first. An electroscope uses induction too: a charged rod near its cap drives charge of the rod's sign into two thin leaves, which repel and spread apart.

Common mistakes

• Rubbing does not create charge: the balloon's $-9.6\mathrm{nC}$ and the wool's $+9.6\mathrm{nC}$ were already there in neutral matter, and they still add up to zero.
• A positive object has lost electrons, not gained protons: the wool's $+9.6\mathrm{nC}$ means $6.0\times 10^{10}$ missing electrons.
• Neutral does not mean uncharged: a neutral 3 g balloon holds about 160,000 C of negative charge in its electrons, balanced exactly by its nuclei.
• Attraction does not prove that a body is charged: a neutral bit of paper is attracted by either sign. Only repulsion shows that two bodies carry charges of the same sign.
• Multiply by $e$, do not divide: $6.0\times 10^{10}$ electrons carry $9.6\times 10^{-9}\mathrm{C}$, while dividing gives an absurd $3.7\times 10^{29}\mathrm{C}$.

Where you meet it

Static electricity works in laser printers and photocopiers, whose charged drum picks up toner in the pattern of the page, in spray painting and in electrostatic precipitators, which remove over 99 % of the ash from power-station smoke. It is also a hazard: walking across a carpet on a dry day can leave about a microcoulomb on you, enough for a spark at a door handle, and fuel tankers are earthed before unloading. A lightning stroke carries some 5 to 30 C.

The next lesson, Coulomb's Law, measures the force between two charges. Electric Field and Potential, Capacitors, and Electric Current and Resistance follow charge into the space around it, into storage and into motion.

Advanced

A force that almost cancels itself

Electric forces run the everyday world even though most matter looks uncharged. Two small charges $q_1$ and $q_2$ a distance $r$ apart attract or repel each other with a force of size
$F=\frac{k|q_1{q}_2|}{r^2}$
(Coulomb's law), where $k=8.99\times 10^9\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{C}}^2$. Like gravity, this force falls off as the inverse square of the distance. Unlike gravity, it is incomparably stronger.

Compare the two forces between two protons, each of charge $e$ and mass $m_p$, with $G$ the gravitational constant. Both forces carry the same $1/r^2$, so $r$ cancels and the ratio is the same at every distance:
$\frac{F_e}{F_g}=\frac{k{e}^2}{G{m}_p^2}\approx 1.2\times 10^{36}$

Why, then, do we so rarely notice it? Because charge comes in two signs, and matter balances them almost perfectly: the charge of an atom or molecule is zero to within one part in $10^{21}$ of $e$.

A rough estimate shows why the balance has to be that fine. A 1 g drop of water contains $3.3\times 10^{23}$ electrons, a charge of about $5.4\times 10^4\mathrm{C}$, matched by the same positive charge in its nuclei. Take one electron in a million out of each of two such drops, and place them 1 m apart. Each drop is left with $q=+0.054\mathrm{C}$, and they repel with
$F=\frac{k{q}^2}{r^2}\approx 2.6\times 10^7\mathrm{N}$
the weight of some 2600 tonnes.

What we do feel are leftovers. Neighbouring atoms are neutral, but at atomic distances their positive and negative charges do not sit exactly on top of each other. That small mismatch is the push of a table, the tension in a rope and the bonds of chemistry.

Quantised, invariant and conserved

Three experimental facts define what charge is. Every theory of matter has to respect all three.

First, charge is quantised: it comes only in whole multiples of the elementary charge $e$. Between 1909 and 1913 Robert Millikan showed this by balancing tiny charged oil drops against gravity in an electric field; every charge he measured was a whole number times $e$. Since 2019 the SI fixes $e=1.602176634\times 10^{-19}\mathrm{C}$ exactly, and this value defines the coulomb. Quarks carry fractions, $+\frac{2}{3}e$ or $-\frac{1}{3}e$, but they are never found alone. They are always bound into particles with whole charges: the proton (uud) has $+e$, the neutron (udd) zero.

Second, charge is invariant: it does not depend on how fast the charged particle moves. Inside a helium nucleus the protons move at a sizeable fraction of the speed of light; in a hydrogen molecule they are nearly at rest. If a proton's charge changed with its speed, one of the two gases would carry a net charge, yet both are neutral to about one part in $10^{20}$. Energy, momentum, lengths and time intervals all depend on the observer; charge does not.

Third, charge is conserved: in every process ever observed, the total charge after equals the total before. A gamma ray passing near a nucleus can turn into an electron and a positron, $0\to (-e)+(+e)$. A free neutron decays into a proton, an electron and an antineutrino, $0\to (+e)+(-e)+0$. Particles appear and disappear, but the net charge does not change.

Local conservation: the continuity equation

Legend

$S$	Closed surface that bounds the volume $V$.
$V$	The volume inside $S$, held fixed.	${\mathrm{m}}^3$
$\rho$	Charge per unit volume inside $V$ (the red dots).	$C/{\mathrm{m}}^3$
$\vec{J}$	Current density: current per unit area, pointing the way positive charge flows.	$A/{\mathrm{m}}^2$
$d\vec{A}$	One small patch of $S$ as a vector: its length is the patch's area $dA$, and it points out of $V$.	${\mathrm{m}}^2$
$\theta$	Angle between $\vec{J}$ and $d\vec{A}$; above $90^{\circ }$ where charge flows in.	$\mathrm{rad}$

Conservation says more than "the total stays the same". Charge cannot vanish in one place and reappear in another; it has to travel there. This section turns that idea into an equation, and Figure 1 sets up the bookkeeping.

Figure 1 shows a fixed volume $V$ bounded by a closed surface $S$. The charge inside is described by its density $\rho$, the charge per unit volume (the red dots). Moving charge is described by the current density $\vec{J}$, the current per unit area, pointing the way positive charge flows. The arrows $\vec{J}$ cross $S$ in both directions: most leave $V$, but the two on the left enter it.

To count what crosses $S$, cut it into small patches. The enlarged patch on the right has area $dA$, and we turn it into a vector $d\vec{A}$ of length $dA$, pointing out of $V$. Only the part of $\vec{J}$ along $d\vec{A}$ carries charge through the patch; the part parallel to the patch slides past it. So the current out through the patch is
$\vec{J}\cdot d\vec{A}=JdA\cos \theta$
where $\theta$ is the angle between the two arrows of the enlarged patch. Where charge flows out, $\theta$ is below $90^{\circ }$ and this term is positive; where charge flows in, $\theta$ is above $90^{\circ }$ and it is negative. Summed over all patches, ${\oint }_S\vec{J}\cdot d\vec{A}$ is the net current out of $V$.

Now the physics. Because $V$ is fixed, the charge inside it can change only by crossing $S$. So the charge inside falls exactly as fast as the net current flows out:
$\frac{d}{dt}{\int }_V\rho dV=-{\oint }_S\vec{J}\cdot d\vec{A}$
This is the integral form of local conservation. To turn it into a statement about single points, use the divergence theorem, which rewrites the flux out of a closed surface as a volume integral of the divergence: ${\oint }_S\vec{J}\cdot d\vec{A}={\int }_V\nabla \cdot \vec{J}dV$. Since $V$ does not move, the time derivative can go inside the integral on the left, and both sides become integrals over the same volume:
${\int }_V\left(\frac{\partial \rho }{\partial t}+\nabla \cdot \vec{J}\right)dV=0$
This holds for every fixed volume, however small. An integral that vanishes over every small region can do so only if the integrand is zero at every point, so
$\frac{\partial \rho }{\partial t}+\nabla \cdot \vec{J}=0$
This is the continuity equation. In words: wherever the charge density is falling, current is spreading out from that point.

Why insist on the local law, when the global one would keep the total just as well? Imagine charge vanishing on Earth and reappearing at the same moment on the Moon. The total would never change. But in relativity, two distant events that are simultaneous for one observer are not simultaneous for another observer moving past. That second observer would see the charge missing for a while, or present in both places at once. Only charge that flows continuously across boundaries is conserved for every observer.

The animation's $q(t)$ graph shows the integral form at work. Draw $S$ around the balloon. Charge crosses it only where balloon and wool touch, so the slope of the blue balloon curve is the current into the balloon. The red wool curve has the opposite slope at every instant: what one gains, the other loses. The charge changes by 9.6 nC in 3.2 s, an average current of 3.0 nA. The transfer follows the distance rubbed, so the current peaks at about 4.7 nA in mid-stroke and drops to zero at each end of a stroke, where the balloon stops and reverses.

Why a neutral body is pulled

Legend

$Q$	Positive source charge.	$\mathrm{C}$
$d$	Distance from $Q$ to the centre of the body.	$\mathrm{m}$
$q$	Size of each of the two charges that the field separates in the body.	$\mathrm{C}$
$\vec{s}$	Separation of the charges, from $-q$ to $+q$; the dipole moment is $\vec{p}=q\vec{s}$.	$\mathrm{m}$
$\vec{E}(\vec{r})$	Field of $Q$ at the $-q$ end, the nearer one, at position $\vec{r}$ from $Q$.	$N/C$
$\vec{E}(\vec{r}+\vec{s})$	Field of $Q$ at the $+q$ end: weaker, because that end is farther from $Q$.	$N/C$
${\vec{F}}_{-}$	Force on $-q$, towards $Q$.	$\mathrm{N}$
${\vec{F}}_{+}$	Force on $+q$, away from $Q$ and smaller.	$\mathrm{N}$
$\vec{F}$	Net force on the body, ${\vec{F}}_{-}+{\vec{F}}_{+}$: towards $Q$.	$\mathrm{N}$

A charged balloon picks up bits of paper that carry no net charge. How does it pull them, and how quickly does the pull grow as the balloon comes closer?

Figure 2 shows a neutral body at a distance $d$ from a positive charge $Q$. The body's charges are not rigidly fixed; they can shift a little. The field of $Q$ points away from $Q$, so it pushes positive charge away and draws negative charge closer. A charge $-q$ gathers on the near side and $+q$ on the far side, separated by the small vector $\vec{s}$, drawn from $-q$ to $+q$. The body has become an induced dipole, with dipole moment $\vec{p}=q\vec{s}$.

How large $\vec{p}$ becomes depends on how easily the charges shift. In a field that is weak compared with the fields inside atoms, it is proportional to the field,
$\vec{p}=\alpha \vec{E}$
where the constant $\alpha$, the polarisability, measures that ease.

Now look at the forces. Each end of the dipole feels the field where it sits. Let $\vec{r}$ be the position of the near end, measured from $Q$. The $-q$ end at $\vec{r}$ sits in the field $\vec{E}(\vec{r})$ and is pulled towards $Q$ by ${\vec{F}}_{-}$. The $+q$ end at $\vec{r}+\vec{s}$ sits farther out, in the weaker field $\vec{E}(\vec{r}+\vec{s})$, and is pushed away by the smaller force ${\vec{F}}_{+}$. The two charges are equal and opposite, but the fields they sit in are not equal, so the forces do not cancel. Their sum, the net force $\vec{F}$ drawn below the body, points towards $Q$.

To find its size, add the two forces and expand the field across the short step $\vec{s}$: $\vec{E}(\vec{r}+\vec{s})\approx \vec{E}(\vec{r})+(\vec{s}\cdot \nabla )\vec{E}$, where $(\vec{s}\cdot \nabla )\vec{E}$ is the change in $\vec{E}$ over that step. Then
$\vec{F}=q[\vec{E}(\vec{r}+\vec{s})-\vec{E}(\vec{r})]\approx (\vec{p}\cdot \nabla )\vec{E}$
So a dipole feels a net force only where the field changes from place to place.

Next, put in $\vec{p}=\alpha \vec{E}$. For any field, $(\vec{E}\cdot \nabla )\vec{E}=\frac{1}{2}\nabla (E^2)-\vec{E}\times (\nabla \times \vec{E})$, and an electrostatic field has no curl, $\nabla \times \vec{E}=0$. Therefore
$\vec{F}=\frac{\alpha }{2}\nabla (E^2)$
Near a point charge, $E=kQ/r^2$, so $E^2=k^2{Q}^2/r^4$. Its gradient points towards $Q$, with size $4k^2{Q}^2/r^5$. At the body, $r=d$, and the pull is
$F=\frac{2\alpha k^2{Q}^2}{d^5}\text{(point charge)}$

Three consequences follow. First, the pull always points towards stronger field, whatever the sign of the source, because $E^2$ ignores signs. Make $Q$ in Figure 2 negative: $\vec{E}$ and $\vec{p}$ both reverse and the $+q$ end becomes the nearer one, yet $\vec{F}$ still points towards $Q$. Second, in a uniform field $E^2$ is the same everywhere, so there is no net force. An elongated body still turns, because it polarises most easily along its length; that is why the paper bits swing round in flight and land on one end. Third, the pull is very steep: halving $d$ multiplies it by $2^5=32$.

The animation's $F/mg$ graph for the middle bit shows this steepness. When the balloon starts down at 5.6 s, the pull is 6 % of the bit's weight. It reaches a quarter at 7.0 s and more than half at 7.4 s, and it crosses the dashed line at 1 at about 7.6 s, when the bit lifts.

These readings can be compared with the point-charge law. Over the same time the distance from the balloon's centre to the bit shrinks from 80 to 52 units. For a charge at the centre, the $1/d^5$ law would raise the pull $(80/52{)}^5\approx 8.6$ times; it actually grows 17 times. The reason is that the charge sits on the balloon's lower surface, whose nearest part comes almost twice as close. Once the bit lifts, the gap closes further and the pull runs away.

Conductors in equilibrium

In a conductor some charges are free to move: in a metal, electrons. They keep moving as long as any field pushes them, so once everything has settled the field inside the material is zero, $\vec{E}=0$. Gauss's law, developed in Electric Field and Potential, then says there is no net charge anywhere inside: all excess or induced charge sits on the surface. The field just outside is perpendicular to the surface, since a component along it would still move the surface charges. Its size is $E=\sigma /{\epsilon }_0$, where $\sigma$ is the charge per unit area of the surface and ${\epsilon }_0=1/(4\pi k)$ is the permittivity of free space.

The same mobility explains charging by induction. Bring a charged rod near a conductor, and charge of the opposite sign gathers on the side facing it. Connect the conductor to earth: charge flows through the connection until the conductor carries a net charge opposite to the rod's, held there by the rod's attraction. Break the earth connection first, then remove the rod, and the conductor keeps that charge.

Charge crowds where the surface curves

Legend

$V$	Potential of the wire and of both spheres: the same everywhere on the conductor.	$\mathrm{V}$
$R_1$	Radius of the larger sphere.	$\mathrm{m}$
$R_2$	Radius of the smaller sphere.	$\mathrm{m}$
$q_1$	Charge of the larger sphere.	$\mathrm{C}$
$q_2$	Charge of the smaller sphere.	$\mathrm{C}$
${\vec{E}}_1$	Field just outside the larger sphere, perpendicular to its surface.	$N/C$
${\vec{E}}_2$	Field just outside the smaller sphere: stronger, since $E=V/R$.	$N/C$

On a charged conductor the charge is not spread evenly: it piles up where the surface curves most sharply. Figure 3 shows the simplest case that makes this precise.

Figure 3 shows two conducting spheres of radii $R_1$ and $R_2$, the first the larger, far apart and joined by a thin wire. Together they form one conductor, so charge flows through the wire until all of it, wire and both spheres, is at one potential $V$. Because the spheres are far apart, each one's potential comes from its own charge alone, and a sphere of radius $R$ holding charge $q$ has $V=kq/R$. Setting the two potentials equal, $k{q}_1/R_1=k{q}_2/R_2$, so
$\frac{q_1}{q_2}=\frac{R_1}{R_2}$
The larger sphere holds the larger charge.

Now look at the field arrows. The field at a sphere's surface is
$E=\frac{kq}{R^2}=\frac{V}{R}$
and since $V$ is shared, $E$ goes as $1/R$. So ${\vec{E}}_2$, on the small sphere, is longer than ${\vec{E}}_1$ in the ratio $R_1:R_2$; the arrows are drawn this way for $R_1=2R_2$. The smaller the sphere, the stronger the field at its surface, and, since $E=\sigma /{\epsilon }_0$, the more charge it carries per unit area.

The same holds on a single conductor of any shape: a sharp point behaves like a small sphere and a flat region like a large one. A tip with a radius of 0.1 mm on a body of radius 10 cm has a surface field about a thousand times that of the body. That is why sparks and the faint glow of corona discharge start at points and edges.

Worked problem: when does a bead jump?

This problem applies the induced-dipole force to a real object. Put the balloon's charge, $Q=9.6\mathrm{nC}$, on a metal sphere of radius 1 cm on an insulating stand. Outside the sphere, its field is exactly that of a point charge $Q$ at the centre. Below the sphere lies a light conducting bead of graphite-coated polystyrene foam, 2 mm across, with mass density ${\rho }_m=20kg/{\mathrm{m}}^3$. How close must the sphere come before the bead jumps up?

Step 1, the pull. A small conducting ball of radius $a$ has polarisability $\alpha =4\pi {\epsilon }_0{a}^3=a^3/k$. Putting this into $F=2\alpha k^2{Q}^2/d^5$ gives
$F=\frac{2k{Q}^2{a}^3}{d^5}$
where $d$ is the distance between the centres of the sphere and the bead.

Step 2, the weight. The bead's mass is its volume times its density, so $mg=\frac{4}{3}\pi a^3{\rho }_{m}g$.

Step 3, the condition. The bead lifts when $F>mg$. Both sides contain $a^3$, which cancels:
$\frac{2k{Q}^2}{d^5}>\frac{4}{3}\pi {\rho }_{m}g$
Solving for $d^5$:
$d^5<\frac{3k{Q}^2}{2\pi {\rho }_{m}g}$

Step 4, the numbers. In SI units $3k{Q}^2=2.5\times 10^{-6}$ and $2\pi {\rho }_{m}g=1.23\times 10^3$, so $d^5<2.0\times 10^{-9}{\mathrm{m}}^5$ and $d<1.8\mathrm{cm}$. The bead jumps when its centre comes within 1.8 cm of the centre of the sphere. Taking away the sphere's radius of 1 cm and the bead's of 1 mm leaves a gap of about 7 mm between the surfaces.

How the lifting distance scales

Legend

$d$	Distance from the centre of the charged sphere to the centre of the bead.	$\mathrm{m}$
$F$	Upward pull on the bead, $F=2k{Q}^2{a}^3/d^5$, where $a$ is the bead's radius.	$\mathrm{N}$
$mg$	Weight of the bead (dashed line).	$\mathrm{N}$
$Q$	Charge on the sphere; the left curve is for $Q$, the right one for $2Q$.	$\mathrm{C}$
$d_Q$	Distance at which the pull equals the weight for the charge $Q$: 1.8 cm.	$\mathrm{m}$
$d_{2Q}$	Distance at which the pull equals the weight for the charge $2Q$: 2.4 cm.	$\mathrm{m}$

The worked problem gave one number. Figure 4 shows how that number moves when the charge or the bead changes, and why the bead's size drops out.

Figure 4 plots the pull $F=2k{Q}^2{a}^3/d^5$ on the bead against the distance $d$ from the sphere's centre. The pull climbs steeply as $d$ shrinks. The weight $mg$ does not depend on $d$, so it is the dashed horizontal line. The bead lifts wherever a curve runs above that line, that is, at any distance inside the crossing point. For the charge $Q$, the left curve, the crossing is at $d_Q$, the 1.8 cm found above.

Now double the charge. The pull goes as $Q^2$, so it is 4 times larger at every distance, and the right curve, for $2Q$, crosses the line farther out. At the new crossing the pull must still equal $mg$, so $4/d_{2Q}^5=1/d_Q^5$ and
$d_{2Q}=4^{1/5}{d}_Q=2^{2/5}{d}_Q\approx 1.32d_Q$
which is 2.4 cm. Because of the fifth power, doubling the charge gains only about a third more distance.

The bead's size does not matter, only its density, as long as the bead stays small compared with $d$, because $a^3$ cancelled. A bead 32 times denser would raise the dashed line 32-fold and pull the crossing in by a factor $32^{1/5}=2$, to 0.9 cm. That is inside the sphere, whose radius is 1 cm, so this bead could never lift before touching.

What happens on contact? The conducting bead takes a share of the sphere's charge, of the same sign, so it is now repelled and flung away. Paper conducts so poorly that it charges only slowly, which is why a bit of paper clings to the balloon instead.

Limits of validity

Each model in this lesson works only within a certain range.

Air insulates only up to about 3 MV/m. Above that, stray free electrons gain enough energy between collisions with molecules to knock out further electrons, and the growing avalanche is a spark.

The balloon is far from this. Its charge, spread over a rubbed patch of about 100 cm², gives a field of roughly $\sigma /(2{\epsilon }_0)\approx 5\times 10^{4}V/m$, some 55 times too weak; the factor 2 appears because the field of a charged insulating sheet points out on both sides. The worked problem's sphere comes much closer: its surface field, $kQ/R^2=8.6\times 10^{5}V/m$ with $R=1\mathrm{cm}$, is within a factor of four of breakdown. At the other extreme, a lightning stroke carries 5 to 30 C, up to $2\times 10^{20}$ electrons, across hundreds of megavolts.

Quantisation is invisible at the balloon's scale, where one electron changes the charge by one part in $6\times 10^{10}$. That is why charge can be treated as a continuous fluid of density $\rho$, as in the continuity equation.

The other models have limits of their own. The law $\vec{p}=\alpha \vec{E}$ needs fields far below those inside atoms, about $5\times 10^{11}V/m$, and the $1/d^5$ law needs a body small compared with $d$. The triboelectric series is empirical: it is still debated whether electrons, ions or tiny bits of material carry the charge between rubbed surfaces. Finally, classical electrostatics cannot say why an electron does not fall onto a proton. That needs quantum mechanics, which begins in Atomic Models: the Bohr Atom.

Where this leads

Coulomb's Law treats the force between point charges in full. Electric Field and Potential turns fields, Gauss's law and equipotential conductors into tools; Capacitors stores separated charge; Electric Current and Resistance studies the current density $\vec{J}$ itself. Moving charges also feel magnetic forces (Magnetic Force on Moving Charges), and the continuity equation is built into Maxwell's equations (Electromagnetic Waves). Semiconductors, Diodes and LEDs treats a missing electron as a positive carrier, and Particles and Forces: the Standard Model derives charge conservation from a symmetry of nature.