---
title: "Electric Potential; Conductors and Capacitors"
book: "University Physics — Year 1"
subject: physics
language: en
chapter: 27
exercises: 12
source: https://one-course.com/books/physics/3/en/chapter/27-electric-potential-conductors-and-capacitors
---

# Chapter 27 — Electric Potential; Conductors and Capacitors

A field is three numbers at every point; a potential is one. Because the electrostatic force is [conservative](https://one-course.com/books/physics/3/en/chapter/13-work-energy-and-potential-energy#def-b1-work-and-energy-conservative), the whole field of a charge distribution can be folded into a single scalar function, the potential, from which it is recovered by differentiation — the same economy that turned forces into [potential energy](https://one-course.com/books/physics/3/en/chapter/13-work-energy-and-potential-energy#def-b1-work-and-energy-conservative) in mechanics. The potential is what a voltmeter reads, what a battery maintains, what accelerates electrons in a tube; its level surfaces map a field at a glance. This chapter builds it, uses it on the dipole (the first electric model of a molecule), and then turns to the two objects every circuit is made of: conductors, on which charge arranges itself so as to kill the field inside, and [capacitors](#def-b1-potential-capacitors-capacitance), which store charge and energy in a field.

![Charged to a few hundred kilovolts by a Van de Graaff generator, a person is an equipotential conductor whose hairs, all at the same potential and all charged alike, repel one another.](https://one-course.com/images/onecourse/chapters/physics-3/b1-potential-capacitors/img-b002bebad4b0.jpg)

*Charged to a few hundred kilovolts by a Van de Graaff generator, a person is an equipotential conductor whose hairs, all at the same potential and all charged alike, repel one another.*

## 27.1 The electrostatic potential

**Theorem 27.1 (The electrostatic field is conservative).**

There exists a scalar function $V(M)$, the *[electrostatic potential](#thm-b1-potential-capacitors-potential)* (volt, $\mathrm{V}$), such that

$$
\vect E = -\overrightarrow{\operatorname{grad}}\,V , \qquad
V_A - V_B = \int_A^B \vect E\cdot\dd\vect l
$$

along any path from $A$ to $B$ (the *circulation* of $\vect E$ from $A$ to $B$ is independent of the path, and zero round a closed loop). For a point charge $q$ at $O$, $V(M) = \dfrac{q}{4\pi\varepsilon_0\,OM}$ with the convention $V(\infty) = 0$; for several charges the potentials add. The [potential energy](https://one-course.com/books/physics/3/en/chapter/13-work-energy-and-potential-energy#def-b1-work-and-energy-conservative) of a charge $q'$ at $M$ is $E_p = q'V(M)$, and the work of the electric force from $A$ to $B$ is $q'(V_A - V_B)$.

**Proof.** The Coulomb force of a fixed charge $q$ on $q'$ has the form of Newton’s [gravitation](https://one-course.com/books/physics/3/en/chapter/16-central-forces-planets-and-satellites#def-b1-central-forces-gravitation) with $-Gm_1m_2$ replaced by $qq'/4\pi\varepsilon_0$; the same computation that gave $E_p = -Gm_1m_2/r$ ([Proposition 13.7](https://one-course.com/books/physics/3/en/chapter/13-work-energy-and-potential-energy#prop-b1-work-and-energy-eps)) gives $E_p = qq'/4\pi\varepsilon_0r$: dividing by $q'$ gives $V$, and $\vect E = \vect F/q' =
-\overrightarrow{\operatorname{grad}}(E_p)/q'$. Superposition carries the result to any distribution; the circulation formula is the definition of the gradient integrated along the path. ∎

**Proposition 27.2 (Reading a potential).**

1. The *[equipotential surfaces](#prop-b1-potential-capacitors-reading)* $V = \text{const}$ are everywhere perpendicular to the [field lines](https://one-course.com/books/physics/3/en/chapter/26-electrostatics-field-and-gausss-law#def-b1-electrostatics-gauss-field) , and the field points toward *decreasing* potential.
2. $|\vect E|$ is the rate of change of $V$ along the steepest direction: where equipotentials are close, the field is strong.
3. $V$ is defined up to an additive constant; only differences ( [voltages](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) ) are measurable. Ground or infinity is set to zero by convention.
4. In [spherical coordinates](https://one-course.com/books/physics/3/en/chapter/11-kinematics-of-a-point#def-b1-point-kinematics-spherical) with symmetry about $O$ , $E_r = -\dd V/  \dd r$ ; in [polar coordinates](https://one-course.com/books/physics/3/en/chapter/11-kinematics-of-a-point#def-b1-point-kinematics-cylindrical) $(r, \theta)$ in a plane, $E_r = -\partial V/\partial r$ and $E_\theta = -\dfrac1r\,\partial V/\partial\theta$ .

**Proof.** On an equipotential $\dd V = -\vect E\cdot\dd\vect l = 0$ for every displacement in the surface, so $\vect E$ is normal to it; moving along $\vect E$, $\dd V = -E\,\dd l < 0$. The polar components follow from $\dd V = (\partial V/\partial r)\dd r + (\partial V/\partial\theta)\dd\theta$ and $\dd\vect l = \dd r\,\vect e_r + r\,\dd\theta\,\vect e_\theta$. ∎

![Equipotentials (orange) and field lines (blue) of a point charge and of a plane capacitor. The lines cross the equipotentials at right angles and run from high to low potential; where the equipotentials crowd, the field is strong.](https://one-course.com/images/onecourse/chapters/physics-3/b1-potential-capacitors/fig-4a9144971422.svg)

*Equipotentials (orange) and [field lines](https://one-course.com/books/physics/3/en/chapter/26-electrostatics-field-and-gausss-law#def-b1-electrostatics-gauss-field) (blue) of a point charge and of a [plane capacitor](#prop-b1-potential-capacitors-three). The lines cross the equipotentials at right angles and run from high to low potential; where the equipotentials crowd, the field is strong.*

**Method 27.3 (Computing a potential).**

Either sum $\dd q/4\pi\varepsilon_0r$ over the distribution (when the sum is tractable and the distribution is finite, so that $V(\infty) = 0$ makes sense), or integrate $\dd V = -\vect E\cdot\dd\vect l$ along a convenient path from a point where $V$ is known, using the field found by [Gauss’s law](https://one-course.com/books/physics/3/en/chapter/26-electrostatics-field-and-gausss-law#thm-b1-electrostatics-gauss-gauss). For infinite distributions (wire, plane) only the second way works, and the zero of $V$ must be chosen at a finite point.

**Proposition 27.4 (Potentials of the classic distributions).**

1. Sphere of radius $R$ with $Q$ in its volume: $V(r) = \dfrac{Q}{4\pi\varepsilon_0r}$ outside; $V(r) = \dfrac{Q}{4\pi\varepsilon_0}\,\dfrac{3R^2 - r^2}{2R^3}$ inside, so $V(0) = \tfrac32\,Q/4\pi\varepsilon_0R$ . With $Q$ on the surface, $V$ is uniform inside, equal to $Q/4\pi\varepsilon_0R$ .
2. Infinite wire $\lambda$ : $V(r) = -\dfrac{\lambda}{2\pi\varepsilon_0}\ln\dfrac{r}{r_0}$ .
3. [Plane capacitor](#prop-b1-potential-capacitors-three) , field $E$ from plate 1 to plate 2 a distance $d$ apart: $V$ decreases linearly across the gap and $V_1 - V_2 = Ed$ .

**Proof.** Integrate $E_r = -\dd V/\dd r$ with the fields of [Proposition 26.12](https://one-course.com/books/physics/3/en/chapter/26-electrostatics-field-and-gausss-law#prop-b1-electrostatics-gauss-three), from infinity inward for the sphere (continuity at $R$ fixes the inside constant), from $r_0$ for the wire, across the gap for the [capacitor](#def-b1-potential-capacitors-capacitance). ∎

![Potential (solid) and field (dashed) of a uniformly charged sphere. The potential is continuous with a continuous slope; it is parabolic inside, where the field is linear, and in 1/r outside. Its maximum sits at the centre, where the field vanishes — a maximum of V is a point of zero field, not the reverse.](https://one-course.com/images/onecourse/chapters/physics-3/b1-potential-capacitors/fig-aaa178022d93.svg)

*Potential (solid) and field (dashed) of a uniformly charged sphere. The potential is continuous with a continuous slope; it is parabolic inside, where the field is linear, and in $1/r$ outside. Its maximum sits at the centre, where the field vanishes — a maximum of $V$ is a point of zero field, not the reverse.*

**Example 27.5 (The electron-volt, again).**

An electron crossing a gap of $100\,\mathrm{V}$ gains $eU = 100\,\mathrm{eV} =
1.6 \times 10^{-17}\,\mathrm{J}$, whatever the shape of the field between the electrodes: the potential makes the path irrelevant ([Definition 17.3](https://one-course.com/books/physics/3/en/chapter/17-charged-particles-in-e-and-b-fields#def-b1-charged-particles-eV)). A $1\,\text{µ}\mathrm{C}$ charge at $1\,\mathrm{m}$ from another has $V = 9\,\mathrm{kV}$ and the pair stores $9\,\mathrm{mJ}$; the potential at the centre of a gold nucleus is $24\,\mathrm{MV}$ ([Exercise 27.6](#exo-b1-potential-capacitors-6)).

## 27.2 The electric dipole

**Definition 27.6 (Electric dipole).**

Two opposite charges $-q$ at $N$ and $+q$ at $P$, a distance $a = NP$ apart, form an *electric dipole* of *dipole moment* $\vect p = q\,\vect{NP}$ ($\mathrm{C}\,\mathrm{m}$), pointing from $-$ to $+$. The dipole approximation is the study of its field at distances $r \gg a$. A molecule whose centres of positive and negative charge do not coincide (water, $p = 6.2 \times 10^{-30}\,\mathrm{C}\,\mathrm{m}$) is a dipole; a neutral atom in a field becomes one (induced dipole).

**Proposition 27.7 (Field of a dipole).**

At $M$ with $OM = r \gg a$ ($O$ the middle of $NP$) and $\theta$ the angle between $\vect p$ and $\vect{OM}$,

$$
V(M) = \frac{p\cos\theta}{4\pi\varepsilon_0 r^2}, \qquad
E_r = \frac{2p\cos\theta}{4\pi\varepsilon_0r^3}, \qquad
E_\theta = \frac{p\sin\theta}{4\pi\varepsilon_0r^3} .
$$

The field of a dipole falls as $1/r^3$, faster than a charge’s $1/r^2$: from afar the two charges almost cancel, and what remains is their separation.

**Proof.** $V = \dfrac{q}{4\pi\varepsilon_0}\Bigl(\dfrac1{PM} - \dfrac1{NM}\Bigr)$ with $PM \approx r - \tfrac a2\cos\theta$ and $NM \approx r + \tfrac a2\cos\theta$ (projection of $\pm\tfrac a2$ on $\vect{OM}$); to first order in $a/r$, $1/PM - 1/NM \approx a\cos\theta/r^2$. Then $E_r = -\partial V/\partial r$ and $E_\theta = -(1/r)\,\partial V/\partial\theta$. ∎

![Left: the geometry of the dipole approximation — M seen from the centre O at distance r and angle from p, with the radial and orthoradial field components. Right: field lines (blue, r 2) and equipotentials (orange, r √| |) of a dipole pointing right; the plane through O perpendicular to p is the equipotential V = 0.](https://one-course.com/images/onecourse/chapters/physics-3/b1-potential-capacitors/fig-f70f9c36fab9.svg)

*Left: the geometry of the dipole approximation — $M$ seen from the centre $O$ at distance $r$ and angle $\theta$ from $\vect p$, with the radial and orthoradial field components. Right: [field lines](https://one-course.com/books/physics/3/en/chapter/26-electrostatics-field-and-gausss-law#def-b1-electrostatics-gauss-field) (blue, $r \propto \sin^2\theta$) and equipotentials (orange, $r \propto
\sqrt{|\cos\theta|}$) of a dipole pointing right; the plane through $O$ perpendicular to $\vect p$ is the equipotential $V = 0$.*

**Proposition 27.8 (Dipole in an external field).**

A dipole $\vect p$ in a uniform field $\vect E$ feels no net force, a [torque](https://one-course.com/books/physics/3/en/chapter/15-angular-momentum#def-b1-angular-momentum-moment) $\vect\Gamma = \vect p\wedge\vect E$ that tends to align it with the field, and has the [potential energy](https://one-course.com/books/physics/3/en/chapter/13-work-energy-and-potential-energy#def-b1-work-and-energy-conservative) $E_p = -\vect p\cdot\vect E$, minimal when aligned. In a non-uniform field it is moreover pulled toward the regions of strong field (in one dimension, $F_x = p\,\dd E/\dd x$ for an aligned dipole).

**Proof.** Forces $\pm q\vect E$ cancel; their moment about $O$ is $\vect{OP}\wedge
q\vect E + \vect{ON}\wedge(-q\vect E) = q\vect{NP}\wedge\vect E$. Energy: $qV(P) - qV(N) = -q\,\vect{NP}\cdot\vect E$ for a uniform field (since $V(P)
- V(N) = -\vect E\cdot\vect{NP}$). Non-uniform, aligned along $x$: $F_x =
qE(x + a) - qE(x) \approx qa\,\dd E/\dd x$. ∎

**Example 27.9 (Why the comb attracts the paper).**

A charged comb creates a field that falls off with distance; a scrap of paper, neutral, is polarized by it (induced dipole along $\vect E$), and the force $p\,\dd E/\dd x$ on an aligned dipole pulls it toward the strong-field region: the comb. Sign of the comb’s charge irrelevant; the same mechanism holds the balloon on the wall and makes the water stream bend. For water molecules, $pE$ in a $1\,\mathrm{MV}/\mathrm{m}$ field is $6 \times 10^{-24}\,\mathrm{J}$, $600$ times less than $k_BT$ at room temperature: only a slight net alignment survives the thermal tumbling ([Exercise 27.7](#exo-b1-potential-capacitors-7)).

## 27.3 Conductors in electrostatic equilibrium

**Theorem 27.10 (Properties of a conductor in equilibrium).**

In a conductor where the free charges are at rest:

1. $\vect E = \vect 0$ inside, and the conductor is an equipotential volume: $V$ uniform in it and on its surface.
2. Any net charge sits on the surface, with a surface density $\sigma$ ; the inside is neutral.
3. Just outside, the field is normal to the surface and equals $\vect E = \dfrac{\sigma}{\varepsilon_0}\,\vect n$ ( *Coulomb’s theorem* ).
4. A cavity inside a conductor, empty of charge, has $\vect E =  \vect 0$ and uniform $V$ whatever happens outside ( *[Faraday cage](#thm-b1-potential-capacitors-conductor)* ).

**Proof.** (1) A field would move the free charges: equilibrium requires $\vect E =
\vect 0$, hence $\dd V = 0$ between any two inner points; the surface is the limit. (2) [Gauss’s law](https://one-course.com/books/physics/3/en/chapter/26-electrostatics-field-and-gausss-law#thm-b1-electrostatics-gauss-gauss) on any closed surface drawn inside the conductor: zero flux, zero enclosed charge. (3) The surface is an equipotential, so $\vect E$ just outside is normal to it; a pillbox straddling the surface has flux $E\,\dd S$ through its outer face only (inside $\vect E = \vect 0$) and charge $\sigma\,\dd S$. (4) The cavity wall is an equipotential; a [field line](https://one-course.com/books/physics/3/en/chapter/26-electrostatics-field-and-gausss-law#def-b1-electrostatics-gauss-field) inside the cavity would have to go from the wall to the wall across a potential drop — impossible on an equipotential; so no line, no field (admitted in this form). ∎

**Example 27.11 (Point effect and lightning rod).**

Two spheres of radii $R_1 > R_2$, far apart and joined by a wire, share the same $V = Q_i/4\pi\varepsilon_0R_i$, so $\sigma_i = Q_i/4\pi R_i^2 = \varepsilon_0V/R_i$: the *smaller* the [radius of curvature](https://one-course.com/books/physics/3/en/chapter/11-kinematics-of-a-point#def-b1-point-kinematics-frenet), the *larger* the surface charge and the field $\sigma/\varepsilon_0 = V/R$. At a sharp point the field exceeds the breakdown value of air ($3\,\mathrm{MV}/\mathrm{m}$) long before it does anywhere else: the air ionizes and charge leaks away quietly. A lightning rod works by this *[point effect](#ex-b1-potential-capacitors-point)* — it does not attract the stroke so much as offer it a path, and bleed the charge under a cloud before it builds up.

![Left: a hollow conductor in an external field — induced charges on its surface cancel the field inside and in the cavity (the Faraday cage). Right: the point effect — on a single equipotential conductor the surface charge and the outside field are largest where the curvature is sharpest.](https://one-course.com/images/onecourse/chapters/physics-3/b1-potential-capacitors/fig-0503400825d0.svg)

*Left: a hollow conductor in an external field — induced charges on its surface cancel the field inside and in the cavity (the [Faraday cage](#thm-b1-potential-capacitors-conductor)). Right: the [point effect](#ex-b1-potential-capacitors-point) — on a single equipotential conductor the surface charge and the outside field are largest where the curvature is sharpest.*

## 27.4 Capacitors

**Definition 27.12 (Capacitance).**

An isolated conductor at potential $V$ carries a charge $Q$ proportional to $V$: $Q = CV$, with $C$ its *capacitance* (farad, $\mathrm{F}$); for a sphere $C = 4\pi\varepsilon_0R$. A *capacitor* is a pair of conductors (*plates*) carrying opposite charges $\pm Q$ and facing each other so that the field is confined between them; its capacitance is $C = Q/U$ with $U = V_1 - V_2$ the [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) between the plates. The symbol and the circuit law $i = C\,\dd u/\dd t$ were used in [Chapter 7](https://one-course.com/books/physics/3/en/chapter/7-transient-regimes-first-and-second-order#ch-b1-transient-regimes).

**Proposition 27.13 (Three capacitors).**

1. Plane: plates of area $S$ a distance $d \ll \sqrt S$ apart: $C = \dfrac{\varepsilon_0S}{d}$ .
2. Spherical: concentric spheres of radii $R_1 < R_2$ : $C = \dfrac{4\pi\varepsilon_0R_1R_2}{R_2 - R_1}$ .
3. Cylindrical (coaxial), radii $a < b$ , length $L \gg b$ : $C = \dfrac{2\pi\varepsilon_0L}{\ln(b/a)}$ .

Filling the gap with an insulator of relative permittivity $\varepsilon_r$ multiplies $C$ by $\varepsilon_r$ (the dielectric’s molecules polarize and weaken the field, $\varepsilon_r = 2$ to $10$ for common plastics and glass, $80$ for water).

**Proof.** In each case the field between the plates is that of [Proposition 26.12](https://one-course.com/books/physics/3/en/chapter/26-electrostatics-field-and-gausss-law#prop-b1-electrostatics-gauss-three): $E = \sigma/\varepsilon_0 = Q/\varepsilon_0S$, $U = Ed$; $E = Q/4\pi\varepsilon_0r^2$, $U = \int_{R_1}^{R_2}E\,\dd r = \dfrac{Q}{4\pi\varepsilon_0}
\Bigl(\dfrac1{R_1} - \dfrac1{R_2}\Bigr)$; $E = Q/2\pi\varepsilon_0Lr$, $U =
\dfrac{Q}{2\pi\varepsilon_0L}\ln\dfrac ba$. Divide $Q$ by $U$. The dielectric factor is admitted. ∎

**Theorem 27.14 (Energy of a capacitor).**

A [capacitor](#def-b1-potential-capacitors-capacitance) charged to $Q$ under $U$ stores

$$
W = \tfrac12 CU^2 = \tfrac12 QU = \frac{Q^2}{2C} ,
$$

the work needed to carry the charge from one plate to the other against the growing [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage). In a [plane capacitor](#prop-b1-potential-capacitors-three) $W = \tfrac12\varepsilon_0E^2
\times Sd$: the energy is in the field, at a density $\tfrac12\varepsilon_0E^2$ per unit volume — a statement that holds for every electrostatic field.

**Proof.** Moving $\dd q$ from the negative to the positive plate when the charge is $q$ costs $\dd W = (q/C)\,\dd q$; integrate from $0$ to $Q$. Then substitute $C = \varepsilon_0S/d$ and $U = Ed$; the general density is admitted. ∎

**Example 27.15 (Three capacitors and their energies).**

A square metre of foil pair $1\,\mathrm{mm}$ apart: $C = 8.85 \times 10^{-12}/
10^{-3} = 8.9\,\mathrm{nF}$ — the [farad](#def-b1-potential-capacitors-capacitance) is a huge unit. A defibrillator’s $150\,\text{µ}\mathrm{F}$ at $2\,\mathrm{kV}$: $300\,\mathrm{J}$, delivered in milliseconds. A supercapacitor of $3000\,\mathrm{F}$ at $2.7\,\mathrm{V}$: $11\,\mathrm{kJ}$ in a can the size of a soda bottle — its [capacitance](#def-b1-potential-capacitors-capacitance) comes from a molecular gap $d$ and an enormous porous $S$. The energy density of a field at the breakdown of air, $\tfrac12\varepsilon_0(3 \times 10^6)^2 = 40\,\mathrm{J}/\mathrm{m}^{3}$, is why [capacitors](#def-b1-potential-capacitors-capacitance) store so little compared with batteries.

![Plane, spherical and dielectric-filled capacitors. The field (orange) is confined between the plates except for the fringing at the edges, neglected when d is small; the energy 1/2 _0E2 per unit volume lives in that field.](https://one-course.com/images/onecourse/chapters/physics-3/b1-potential-capacitors/fig-29a3450569b0.svg)

*Plane, spherical and dielectric-filled [capacitors](#def-b1-potential-capacitors-capacitance). The field (orange) is confined between the plates except for the fringing at the edges, neglected when $d$ is small; the energy $\tfrac12\varepsilon_0E^2$ per unit volume lives in that field.*

**Remark 27.16 (Capacitors in association).**

In parallel (same $U$) the charges add: $C = C_1 + C_2$; in series (same $Q$) the [voltages](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) add: $1/C = 1/C_1 + 1/C_2$ — the opposite of [resistors](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-resistor), because $C$ measures an ease (charge per volt) where $R$ measures a difficulty. These rules follow from $Q = CU$ and the circuit laws of [Chapter 6](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#ch-b1-dc-circuits).

## 27.5 Exercises

**Exercise 27.1 ★.**

Potential at $1.0\,\mathrm{m}$ from a $1.0\,\text{µ}\mathrm{C}$ charge; energy needed to bring a second $1.0\,\text{µ}\mathrm{C}$ from infinity to that point; speed with which it would fly off if released, if its mass is $1\,\mathrm{g}$.

**Solution of Exercise 27.1.**

$V = 8.99 \times 10^9 \times 10^{-6}/1 = 9.0\,\mathrm{kV}$; $W = qV = 9.0\,\mathrm{mJ}$; released, $\tfrac12mv^2 = 9\,\mathrm{mJ}$: $v = \sqrt{18} = 4.2\,\mathrm{m}/\mathrm{s}$.

**Exercise 27.2 ★.**

Plates $1.0\,\mathrm{mm}$ apart under $100\,\mathrm{V}$: field; [kinetic energy](https://one-course.com/books/physics/3/en/chapter/13-work-energy-and-potential-energy#thm-b1-work-and-energy-ke) and speed gained by an electron crossing the gap from rest; time of flight.

**Solution of Exercise 27.2.**

$E = 100/10^{-3} = 1 \times 10^{5}\,\mathrm{V}/\mathrm{m}$; $E_k = eU = 100\,\mathrm{eV} = 1.6 \times 10^{-17}\,\mathrm{J}$, so $v = \sqrt{2E_k/m_e} = 5.9 \times 10^{6}\,\mathrm{m}/\mathrm{s}$; uniformly accelerated, $t = 2d/v = 3.4 \times 10^{-10}\,\mathrm{s}$.

**Exercise 27.3 ★.**

On a map of equipotentials drawn every $10\,\mathrm{V}$, the lines are $2\,\mathrm{mm}$ apart near $A$ and $8\,\mathrm{mm}$ apart near $B$. Field at $A$ and at $B$; which way does the field point relative to the increasing potentials; which way does an electron accelerate?

**Solution of Exercise 27.3.**

$E_A = 10/0.002 = 5\,\mathrm{kV}/\mathrm{m}$, $E_B = 10/0.008 = 1.25\,\mathrm{kV}/\mathrm{m}$. The field points toward decreasing potential; an electron ($q < 0$) accelerates the other way, toward increasing $V$.

**Exercise 27.4 ★.**

A Van de Graaff sphere of radius $15\,\mathrm{cm}$ is charged to $200\,\mathrm{kV}$. [Capacitance](#def-b1-potential-capacitors-capacitance), charge, field at its surface. Maximum potential before the air around it breaks down ($3\,\mathrm{MV}/\mathrm{m}$).

**Solution of Exercise 27.4.**

$C = 4\pi\varepsilon_0R = 1.11 \times 10^{-10} \times 0.15 = 17\,\mathrm{pF}$; $Q = CV =
3.3\,\text{µ}\mathrm{C}$; $E = V/R = 2 \times 10^5/0.15 = 1.3\,\mathrm{MV}/\mathrm{m}$. Breakdown at $V = E_{\max}R = 3 \times 10^6 \times 0.15 = 450\,\mathrm{kV}$ — big machines use big spheres.

**Exercise 27.5 ★★.**

Potential on the axis of a ring of radius $R$ and charge $Q$; recover the axial field by differentiation. Same for the uniformly charged disk (use [Exercise 26.5](https://one-course.com/books/physics/3/en/chapter/26-electrostatics-field-and-gausss-law#exo-b1-electrostatics-gauss-5)): $V(z) = \dfrac{\sigma}{2\varepsilon_0}
\bigl(\sqrt{z^2 + R^2} - z\bigr)$.

**Solution of Exercise 27.5.**

Every element of the ring is at distance $\sqrt{R^2 + z^2}$: $V =
Q/4\pi\varepsilon_0\sqrt{R^2 + z^2}$; $E_z = -\dd V/\dd z = Qz/4\pi\varepsilon_0(R^2 +
z^2)^{3/2}$. Disk: $\dd q = 2\pi\sigma a\,\dd a$ at $\sqrt{a^2 + z^2}$: $V =
\dfrac{\sigma}{2\varepsilon_0}\displaystyle\int_0^R\dfrac{a\,\dd a}{\sqrt{a^2 + z^2}} =
\dfrac{\sigma}{2\varepsilon_0}\bigl(\sqrt{z^2 + R^2} - z\bigr)$, and $-\dd V/\dd z =
\dfrac{\sigma}{2\varepsilon_0}\Bigl(1 - \dfrac{z}{\sqrt{z^2 + R^2}}\Bigr)$.

**Exercise 27.6 ★★.**

Derive the potential inside a uniformly charged sphere from its field. Potential at the centre of a gold nucleus ($Z = 79$, $R = 7.0\,\mathrm{fm}$); energy a proton would need to reach the centre from far away, in $\mathrm{MeV}$; comment on Rutherford’s $5\,\mathrm{MeV}$ alpha particles.

**Solution of Exercise 27.6.**

$V(r) = V(R) + \displaystyle\int_r^R\dfrac{Qr'\,\dd r'}{4\pi\varepsilon_0R^3} =
\dfrac{Q}{4\pi\varepsilon_0R} + \dfrac{Q(R^2 - r^2)}{8\pi\varepsilon_0R^3} = \dfrac{Q}{4\pi\varepsilon_0}
\dfrac{3R^2 - r^2}{2R^3}$. Gold: $V(0) = 1.5 \times 8.99 \times 10^9 \times 1.26 \times
10^{-17}/7 \times 10^{-15} = 24\,\mathrm{MV}$: a proton needs $24\,\mathrm{MeV}$, an alpha ($2e$) $48\,\mathrm{MeV}$ to reach the centre and $32\,\mathrm{MeV}$ to touch the surface. Rutherford’s $5\,\mathrm{MeV}$ alphas turn back at $r$ where $2eV =
5\,\mathrm{MeV}$: $V = 2.5\,\mathrm{MV}$, $r = 45\,\mathrm{fm}$ — far outside the nucleus, which is why his point-charge analysis worked.

**Exercise 27.7 ★★.**

Water molecule, $p = 6.2 \times 10^{-30}\,\mathrm{C}\,\mathrm{m}$. Field it creates at $1.0\,\mathrm{nm}$ along its axis and perpendicular to it; [torque](https://one-course.com/books/physics/3/en/chapter/15-angular-momentum#def-b1-angular-momentum-moment) and energy difference between aligned and anti-aligned positions in $E = 1.0\,\mathrm{MV}/\mathrm{m}$; compare with $k_BT$ at $300\,\mathrm{K}$. Field of a sodium ion at $0.3\,\mathrm{nm}$ and the energy of the water dipole aligned in it (hydration).

**Solution of Exercise 27.7.**

Axis: $2p/4\pi\varepsilon_0r^3 = 2 \times 8.99 \times 10^9 \times 6.2 \times 10^{-30}/10^{-27} =
1.1 \times 10^{8}\,\mathrm{V}/\mathrm{m}$; perpendicular, half: $5.6 \times 10^{7}\,\mathrm{V}/\mathrm{m}$. [Torque](https://one-course.com/books/physics/3/en/chapter/15-angular-momentum#def-b1-angular-momentum-moment) $pE =
6.2 \times 10^{-24}\,\mathrm{N}\,\mathrm{m}$; energy gap $2pE = 1.2 \times 10^{-23}\,\mathrm{J}$ against $k_BT =
4.1 \times 10^{-21}\,\mathrm{J}$: $330$ times smaller — only a $0.1\%$ net alignment. Sodium ion at $0.3\,\mathrm{nm}$: $E = e/4\pi\varepsilon_0r^2 = 1.6 \times 10^{10}\,\mathrm{V}/\mathrm{m}$, $-pE
= -1.0 \times 10^{-19}\,\mathrm{J} = -0.6\,\mathrm{eV}$, $24\,k_BT$: the water molecules lock onto the ion — hydration.

**Exercise 27.8 ★★.**

A conducting sphere of radius $1.0\,\mathrm{cm}$ in air: maximum charge and potential before breakdown. Two conducting spheres, $R_1 = 10\,\mathrm{cm}$ and $R_2 = 1.0\,\mathrm{mm}$, joined by a thin wire and raised to $20\,\mathrm{kV}$: surface field on each; which one discharges into the air?

**Solution of Exercise 27.8.**

$Q = 4\pi\varepsilon_0R^2E = 10^{-4} \times 3 \times 10^6/8.99 \times 10^9 = 33\,\mathrm{nC}$, $V = ER = 30\,\mathrm{kV}$. Joined spheres: $E_1 = V/R_1 = 0.2\,\mathrm{MV}/\mathrm{m}$, $E_2 = V/R_2 = 20\,\mathrm{MV}/\mathrm{m} > 3\,\mathrm{MV}/\mathrm{m}$: the small sphere ionizes the air and leaks the charge — the [point effect](#ex-b1-potential-capacitors-point).

**Exercise 27.9 ★★.**

[Capacitances](#def-b1-potential-capacitors-capacitance): two $1.0\,\mathrm{m}^{2}$ foils $10\,\text{µ}\mathrm{m}$ apart with a polymer film $\varepsilon_r = 2.2$ (a film [capacitor](#def-b1-potential-capacitors-capacitance)); the same foils rolled up — does rolling change $C$? Energy at $400\,\mathrm{V}$. A $150\,\text{µ}\mathrm{F}$ [capacitor](#def-b1-potential-capacitors-capacitance) at $2.0\,\mathrm{kV}$ (defibrillator): charge, energy, mean power if delivered in $5\,\mathrm{ms}$.

**Solution of Exercise 27.9.**

$C = \varepsilon_r\varepsilon_0S/d = 2.2 \times 8.85 \times 10^{-12}/10^{-5} = 2.0\,\text{µ}\mathrm{F}$; rolling changes the shape, not the facing area: same $C$ (with a second film so that each foil faces the other on both sides, it doubles). $W = \tfrac12 \times 2 \times 10^{-6} \times 400^2 = 0.16\,\mathrm{J}$. Defibrillator: $Q = 150 \times 10^{-6} \times 2000 = 0.30\,\mathrm{C}$, $W = \tfrac12CU^2 = 300\,\mathrm{J}$, $P = 300/0.005 = 60\,\mathrm{kW}$.

**Exercise 27.10 ★★★.**

*Spherical and [cylindrical capacitors](#prop-b1-potential-capacitors-three).* (a) Derive $C$ for concentric spheres and check that for $R_2 - R_1 = d \ll R_1$ it reduces to the plane formula, and for $R_2 \to \infty$ to the isolated sphere. (b) Coaxial cable, $a = 0.5\,\mathrm{mm}$, $b = 2.5\,\mathrm{mm}$, polyethylene $\varepsilon_r = 2.3$: [capacitance](#def-b1-potential-capacitors-capacitance) per metre (compare with the $100\,\mathrm{pF}/\mathrm{m}$ of standard cables). (c) Field at the inner conductor under $1\,\mathrm{kV}$; safe?

**Solution of Exercise 27.10.**

(a) $E = Q/4\pi\varepsilon_0r^2$, $U = \dfrac{Q}{4\pi\varepsilon_0}\Bigl(\dfrac1{R_1} -
\dfrac1{R_2}\Bigr)$, $C = 4\pi\varepsilon_0R_1R_2/(R_2 - R_1)$; with $R_2 - R_1 = d \ll
R_1$, $R_1R_2 \approx R^2$ and $C \approx 4\pi R^2\varepsilon_0/d = \varepsilon_0S/d$; with $R_2 \to
\infty$, $C \to 4\pi\varepsilon_0R_1$. (b) $C/L = 2\pi\varepsilon_r\varepsilon_0/\ln(b/a) = 2\pi \times 2.3
\times 8.85 \times 10^{-12}/\ln5 = 79\,\mathrm{pF}/\mathrm{m}$, the right order. (c) $E(a) =
U/[a\ln(b/a)] = 1000/(5 \times 10^{-4} \times 1.61) = 1.2\,\mathrm{MV}/\mathrm{m}$: below the $20\,\mathrm{MV}/\mathrm{m}$ of polyethylene, safe; in air it would be marginal.

**Exercise 27.11 ★★★.**

*Energy in the field.* (a) Energy of a charged conducting sphere ($Q$, $R$) from $Q^2/2C$. (b) Set it equal to the electron’s rest energy $m_ec^2 = 0.511\,\mathrm{MeV}$ and solve for $R$: the “classical electron radius” (within a factor). (c) An isolated [plane capacitor](#prop-b1-potential-capacitors-three) charged to $Q$ has its plates pulled from $d$ to $2d$: change of stored energy, and the force between the plates deduced from it; check against [Exercise 26.8](https://one-course.com/books/physics/3/en/chapter/26-electrostatics-field-and-gausss-law#exo-b1-electrostatics-gauss-8). (d) Same with the [capacitor](#def-b1-potential-capacitors-capacitance) kept at fixed $U$ by a battery: energy change, and where the difference goes.

**Solution of Exercise 27.11.**

(a) $W = Q^2/2C = Q^2/8\pi\varepsilon_0R$. (b) $R = e^2/8\pi\varepsilon_0m_ec^2 = 2.31 \times
10^{-28}/(2 \times 8.19 \times 10^{-14}) = 1.4 \times 10^{-15}\,\mathrm{m}$ (the usual “classical radius” $e^2/4\pi\varepsilon_0m_ec^2 = 2.8\,\mathrm{fm}$ drops the $\tfrac12$). (c) $W =
Q^2/2C = Q^2d/2\varepsilon_0S$ doubles: $\Delta W = Q^2d/2\varepsilon_0S = F\,d$ with $F =
Q^2/2\varepsilon_0S = \sigma^2S/2\varepsilon_0$, as found directly. (d) At fixed $U$, $W =
\varepsilon_0SU^2/2d$ halves: $\Delta W = -W/2$; the puller still works $W/2$ (integrating $F = \varepsilon_0SU^2/2x^2$ from $d$ to $2d$); both halves, $W$ in all, flow back into the battery as the charge $CU/2$ returns to it at [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) $U$.

**Exercise 27.12 ★★★.**

*Coulomb energy of a nucleus.* (a) Build a uniformly charged sphere ($Q$, $R$) shell by shell and show that its electrostatic energy is $W = \dfrac{3Q^2}{20\pi\varepsilon_0R}$. (b) Uranium-238: $Z = 92$, $R =
7.4\,\mathrm{fm}$: $W$ in $\mathrm{MeV}$. (c) Split it into two equal fragments ($Z/2$, $A/2$, radius $R/2^{1/3}$) far apart: Coulomb energy released. (d) The measured energy of fission is about $200\,\mathrm{MeV}$: what opposes the Coulomb gain?

**Solution of Exercise 27.12.**

(a) The sphere of radius $r$ carries $q = Qr^3/R^3$ at surface potential $q/4\pi\varepsilon_0r$; adding the shell $\dd q = 3Qr^2\,\dd r/R^3$ costs $\dd W =
q\,\dd q/4\pi\varepsilon_0r = 3Q^2r^4\,\dd r/4\pi\varepsilon_0R^6$; integrate: $W =
3Q^2/20\pi\varepsilon_0R$. (b) $Q = 92e = 1.47 \times 10^{-17}\,\mathrm{C}$: $W = 3 \times 2.17 \times
10^{-34}/(20\pi \times 8.85 \times 10^{-12} \times 7.4 \times 10^{-15}) = 1.6 \times 10^{-10}\,\mathrm{J}
= 990\,\mathrm{MeV}$. (c) Each fragment: charge $Q/2$, radius $R/2^{1/3}$, energy $\tfrac14 2^{1/3}W$; two of them: $2^{1/3}W/2 = 0.63W$: released $0.37 \times 990 = 370\,\mathrm{MeV}$. (d) The nuclear “surface tension”: two fragments have more surface than one nucleus, which costs about $170\,\mathrm{MeV}$; the balance, near $200\,\mathrm{MeV}$, appears mostly as the [kinetic energy](https://one-course.com/books/physics/3/en/chapter/13-work-energy-and-potential-energy#thm-b1-work-and-energy-ke) of the two fragments flying apart under their mutual repulsion.

![Millikan’s oil-drop apparatus, about 1916 (Museum of Science and Industry, Chicago): the brass chamber holds the two horizontal plates between which the drops were watched.](https://one-course.com/images/onecourse/chapters/physics-3/b1-potential-capacitors/img-e7375e9d57d0.jpg)

*Millikan’s oil-drop apparatus, about 1916 (Museum of Science and Industry, Chicago): the brass chamber holds the two horizontal plates between which the drops were watched.*

## 27.6 Problem: Millikan’s oil drop

**Problem 27.1.**

Weekend problem — weighing the electron’s charge with a drop of oil: Stokes, the plane capacitor, and a table of numbers that refuse to be anything but multiples of one

Oil drops are sprayed between two horizontal plates $d = 16\,\mathrm{mm}$ apart, $20\,\mathrm{cm}$ in diameter, and watched through a [microscope](https://one-course.com/books/physics/3/en/chapter/4-optical-instruments-and-the-eye#def-b1-optical-instruments-microscope). Data: oil density $\rho = 900\,\mathrm{kg}/\mathrm{m}^{3}$, air density $1.2\,\mathrm{kg}/\mathrm{m}^{3}$, air viscosity $\eta = 1.8 \times 10^{-5}\,\mathrm{Pa}\,\mathrm{s}$, $g =
9.8\,\mathrm{m}/\mathrm{s}^{2}$, $\varepsilon_0 = 8.85 \times 10^{-12}\,\mathrm{F}/\mathrm{m}$. A small sphere of radius $r$ moving at $v$ in air feels the Stokes drag $6\pi\eta rv$.

**Part I — The falling drop.**

1. Forces on a drop falling with no [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) applied; show that it reaches a [terminal velocity](https://one-course.com/books/physics/3/en/chapter/12-newtons-laws-of-dynamics#prop-b1-newton-dynamics-terminal) $v_1 = 2\rho gr^2/9\eta$ (neglect the air’s buoyancy — justify).
2. A drop falls $1.0\,\mathrm{mm}$ in $10\,\mathrm{s}$ : its radius and mass.
3. [Time constant](https://one-course.com/books/physics/3/en/chapter/7-transient-regimes-first-and-second-order#def-b1-transient-regimes-firstorder) of the approach to $v_1$ ; distance covered meanwhile. Is the drop ever seen accelerating?
4. Reynolds number $\rho_{\text{air}}v_1r/\eta$ of the motion: is [Stokes’s law](https://one-course.com/books/physics/3/en/chapter/12-newtons-laws-of-dynamics#prop-b1-newton-dynamics-drag) legitimate?
5. Why must the drops be so small? (Compare the weight with the electric force on one [elementary charge](https://one-course.com/books/physics/3/en/chapter/26-electrostatics-field-and-gausss-law#def-b1-electrostatics-gauss-charge) in a field of $3 \times 10^{5}\,\mathrm{V}/\mathrm{m}$ .)
6. The drop is seen to jitter slightly around its mean fall: what is this, and why does it limit the precision?

**Part II — The field.** A [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) $U$ is applied, the upper plate positive.

7. Field between the plates for $U = 5.0\,\mathrm{kV}$ ; why is it uniform, and what are the equipotentials?
8. [Capacitance](#def-b1-potential-capacitors-capacitance) of the plates, charge on them at $5.0\,\mathrm{kV}$ , and stored energy.
9. The drop of question 2 is held stationary for $U_0 = 1080\,\mathrm{V}$ : sign and value of its charge, in coulombs and in units of $e$ .
10. With $U = 5.0\,\mathrm{kV}$ the same drop rises at $v_2$ : show that $q = 6\pi\eta r(v_1 + v_2)d/U$ and compute $v_2$ .
11. Why is timing a rise and a fall better than hunting for the balance [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) ?
12. [Potential energy](https://one-course.com/books/physics/3/en/chapter/13-work-energy-and-potential-energy#def-b1-work-and-energy-conservative) of the drop’s charge between the plates at $5\,\mathrm{kV}$ , and the energy dissipated by drag during a rise of $1\,\mathrm{cm}$ .

**Part III — The table of charges.** Exposed to a weak source of X-rays between measurements, the same drop changes its charge; successive measurements give (in $10^{-19}$ C): $4.80$, $8.01$, $6.42$, $3.19$, $9.63$.

13. Show that all five are close to integer multiples of one quantity, and give the integers.
14. Best estimate of the [elementary charge](https://one-course.com/books/physics/3/en/chapter/26-electrostatics-field-and-gausss-law#def-b1-electrostatics-gauss-charge) from the five values; [standard uncertainty](https://one-course.com/books/physics/3/en/chapter/1-units-dimensions-and-measurement#def-b1-units-dimensions-uncertainty) of the mean ( [Chapter 1](https://one-course.com/books/physics/3/en/chapter/1-units-dimensions-and-measurement#ch-b1-units-dimensions) ).
15. Why does the charge change by whole numbers of $e$ between measurements?
16. Show that, in this method, $q \propto \eta^{3/2}$ : by how much does a $1\%$ error on the viscosity shift $e$ ? (Millikan’s original value was $0.6\%$ low, for this very reason.)
17. Quarks carry $\pm e/3$ and $\pm 2e/3$ : why has no oil drop ever shown a third of $e$ ?
18. The Faraday constant, the charge of one [mole](https://one-course.com/books/physics/3/en/chapter/20-kinetic-theory-and-the-perfect-gas#def-b1-kinetic-theory-scales) of electrons, is $F = 96\,485\,\mathrm{C}/\mathrm{mol}$ : deduce Avogadro’s number.
19. Thomson had measured $e/m_e = 1.76 \times 10^{11}\,\mathrm{C}/\mathrm{kg}$ for the electron: deduce its mass, and compare with a hydrogen atom ( $1.67 \times 10^{-27}\,\mathrm{kg}$ ).

**Part IV — The apparatus, by the potential.**

20. Sketch the equipotentials and [field lines](https://one-course.com/books/physics/3/en/chapter/26-electrostatics-field-and-gausss-law#def-b1-electrostatics-gauss-field) between the plates, including the fringing near the edges; justify neglecting it.
21. A positively charged drop at height $z$ above the lower plate (at $0\,\mathrm{V}$ ): [potential energy](https://one-course.com/books/physics/3/en/chapter/13-work-energy-and-potential-energy#def-b1-work-and-energy-conservative) as a function of $z$ ; total [potential energy](https://one-course.com/books/physics/3/en/chapter/13-work-energy-and-potential-energy#def-b1-work-and-energy-conservative) including gravity; condition on $U$ for the stationary point.
22. Compare the energy of the drop’s charge (question 12) with the energy stored in the [capacitor](#def-b1-potential-capacitors-capacitance) : does the drop disturb the field?
23. The [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) is reversed (lower plate positive) with $U = 5.0\,\mathrm{kV}$ : what does the drop of question 9 do, and at what speed?
24. A dust grain of $1\,\text{µ}\mathrm{g}$ carrying $10^4e$ in the same field: ratio of electric force to weight; why such grains cannot be levitated and why Millikan chose oil mist.
25. Sum up: what was measured (three times), what was deduced, with which uncertainty, and what two constants followed from it.

**Solution of Problem 27.1.**

**1.** Weight $\tfrac43\pi r^3\rho g$ down, drag $6\pi\eta rv$ up, buoyancy $\tfrac43\pi r^3\rho_{\text{air}}g$ up — $1.2/900 = 0.13\%$ of the weight, negligible. Terminal: $\tfrac43\pi r^3\rho g = 6\pi\eta rv_1$, $v_1 =
2\rho gr^2/9\eta$.

**2.** $v_1 = 1.0 \times 10^{-4}\,\mathrm{m}/\mathrm{s}$: $r = \sqrt{9\eta v_1/2\rho g} = \sqrt{9 \times
1.8 \times 10^{-5} \times 10^{-4}/(2 \times 900 \times 9.8)} = 9.6 \times 10^{-7}\,\mathrm{m}$, a micrometre; $m = \tfrac43\pi r^3\rho = 3.3 \times 10^{-15}\,\mathrm{kg}$.

**3.** $\tau = m/6\pi\eta r = 3.3 \times 10^{-15}/3.3 \times 10^{-10} = 1 \times 10^{-5}\,\mathrm{s}$; distance $\sim v_1\tau = 1\,\mathrm{nm}$. Never.

**4.** $\mathrm{Re} = 1.2 \times 10^{-4} \times 9.6 \times 10^{-7}/1.8 \times 10^{-5} =
6 \times 10^{-6} \ll 1$: Stokes holds.

**5.** $eE = 1.6 \times 10^{-19} \times 3 \times 10^5 = 4.8 \times 10^{-14}\,\mathrm{N}$ against $mg = 3.2 \times 10^{-14}\,\mathrm{N}$: comparable only for micron drops. A $10\,\text{µ}\mathrm{m}$ drop weighs a thousand times more; one [elementary charge](https://one-course.com/books/physics/3/en/chapter/26-electrostatics-field-and-gausss-law#def-b1-electrostatics-gauss-charge) would change its speed by $0.1\%$, invisible.

**6.** Brownian motion: the bombardment by air molecules jostles the drop randomly; the timings scatter, limiting the precision and setting a lower bound on useful drop size.

**7.** $E = U/d = 5000/0.016 = 3.1 \times 10^{5}\,\mathrm{V}/\mathrm{m}$; plates much wider than $d$, so the two-plane superposition gives a uniform field; the equipotentials are horizontal planes, $V = Uz/d$.

**8.** $C = \varepsilon_0\pi(0.1)^2/0.016 = 17\,\mathrm{pF}$; $Q = CU = 87\,\mathrm{nC}$; $W = \tfrac12CU^2 = 0.22\,\mathrm{mJ}$.

**9.** Upper plate positive: $\vect E$ points down; the force must point up, so $q < 0$. $|q| = mgd/U_0 = 3.3 \times 10^{-15} \times 9.8 \times 0.016/
1080 = 4.8 \times 10^{-19}\,\mathrm{C} = 3e$: $q = -3e$.

**10.** Rising at constant speed: $|q|E = mg + 6\pi\eta rv_2$ and $mg =
6\pi\eta rv_1$, so $|q|E = 6\pi\eta r(v_1 + v_2)$ and $|q| = 6\pi\eta r(v_1 +
v_2)d/U$. $v_2 = (|q|E - mg)/6\pi\eta r = (1.5 \times 10^{-13} - 3.2 \times 10^{-14})/
3.3 \times 10^{-10} = 3.6 \times 10^{-4}\,\mathrm{m}/\mathrm{s}$: $1\,\mathrm{mm}$ in $2.8\,\mathrm{s}$.

**11.** Balance is a null judgment on a drifting, jittering speck whose charge may change meanwhile; timings over a fixed distance can be repeated at will, $r$ is fixed once for all by $v_1$, and one drop then yields a whole series of charges.

**12.** $|q|U = 3 \times 1.6 \times 10^{-19} \times 5000 = 2.4 \times 10^{-15}\,\mathrm{J}$ ($15\,\mathrm{keV}$) across the gap; drag over $1\,\mathrm{cm}$: $(|q|E - mg) \times 0.01 =
1.2 \times 10^{-15}\,\mathrm{J}$, turned into heat in the air.

**13.** Divide by $1.60$: $3.00$, $5.01$, $4.01$, $1.99$, $6.02$ — integers $3, 5, 4, 2, 6$.

**14.** $q/n$: $1.600, 1.602, 1.605, 1.595, 1.605$; mean $1.601$, standard deviation $0.004$, uncertainty of the mean $0.004/\sqrt5 =
0.002$: $e = (1.601 \pm 0.002) \times 10^{-19}$ C.

**15.** The X-rays ionize the air; the drop picks up ions one at a time, each carrying a whole number of $e$.

**16.** $r \propto \eta^{1/2}$ from $v_1$, $m \propto r^3 \propto \eta^{3/2}$, and $|q| = mgd/U_0 \propto \eta^{3/2}$: $1\%$ on $\eta$ gives $1.5\%$ on $e$. Millikan’s $\eta$ was $0.4\%$ low, hence his $e$ $0.6\%$ low.

**17.** Quarks never come alone: they are confined inside hadrons, and every free object carries an integer charge.

**18.** $N_A = F/e = 96485/1.602 \times 10^{-19} = 6.02 \times 10^{23}\,\mathrm{mol}^{-1}$.

**19.** $m_e = 1.602 \times 10^{-19}/1.76 \times 10^{11} = 9.1 \times 10^{-31}\,\mathrm{kg}$: $1840$ times lighter than the hydrogen atom.

**20.** Horizontal planes between the plates, bulging round the rims; vertical [field lines](https://one-course.com/books/physics/3/en/chapter/26-electrostatics-field-and-gausss-law#def-b1-electrostatics-gauss-field) fanning outward at the edges over a width of order $d = 1.6\,\mathrm{cm}$ — small against the $20\,\mathrm{cm}$ diameter, and the drop is watched at the centre.

**21.** $E_p = qUz/d + mgz$, linear in $z$: no minimum; a stationary drop requires the slope to vanish, $q = -mgd/U$, and the equilibrium is then indifferent — any residual drift persists, which is why the balance is hard to judge.

**22.** $2.4 \times 10^{-15}$ against $2.2 \times 10^{-4}$ J: $10^{-11}$. The drop does not disturb the field.

**23.** Field up, force on the negative drop down: it falls at $(mg
+ |q|E)/6\pi\eta r = (3.2 \times 10^{-14} + 1.5 \times 10^{-13})/3.3 \times 10^{-10} =
5.6 \times 10^{-4}\,\mathrm{m}/\mathrm{s}$.

**24.** $F = 10^4 \times 1.6 \times 10^{-19} \times 3.1 \times 10^5 = 5 \times 10^{-10}\,\mathrm{N}$ against $9.8 \times 10^{-9}\,\mathrm{N}$: $5\%$ of the weight — not liftable, and with $10^4$ charges a change of one $e$ is invisible. Oil mist gives drops that are small, non-evaporating and spherical.

**25.** Measured: $v_1$, $v_2$ and $U$ (with $d$, $\eta$, $\rho$ known); deduced $r$, $m$, then $q$, found to be always an integer multiple of $e = (1.601 \pm 0.002) \times 10^{-19}$ C ($0.1\%$); from it $N_A = F/e$ and $m_e = e/(e/m_e)$.