---
title: "Electric and Gravitational Fields"
book: "High School Physics"
subject: physics
language: en
chapter: 14
exercises: 15
source: https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields
---

# Chapter 14 — Electric and Gravitational Fields

How does the Earth pull an apple it never touches? Newton himself called action at a distance “so great an absurdity” that he refused to defend it. The modern answer: a mass, or a charge, modifies the space around it — it creates a *[field](#def-g11-electric-gravitational-fields-field)* — and whatever is placed there responds to the [field](#def-g11-electric-gravitational-fields-field) at its own location. We build the two great examples side by side, until the analogy becomes an identity of form.

## 14.1 From action at a distance to the field

**Definition 14.1 (Field).**

A *field* assigns to every point of space a vector. A source (a charge, a mass) creates its field everywhere around it, whether or not anything is there to feel it; a test object then feels a [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) determined by the field *at its own location* alone. This two-step picture replaces action at a distance.

**Remark 14.2.**

The [field](#def-g11-electric-gravitational-fields-field) is not a bookkeeping trick: it carries [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy), and changes in the source travel outward through it at finite speed — light itself is such a traveling [field](#def-g11-electric-gravitational-fields-field) ([Chapter 22](https://one-course.com/books/physics/2/en/chapter/22-light-as-a-wave-diffraction-and-interference#ch-g12-light-as-wave)). Here we study [fields](#def-g11-electric-gravitational-fields-field) that do not change in time.

## 14.2 The electric field

**Definition 14.3 (Electric field).**

If a small test charge $q$ placed at a point $M$ feels an electric [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) $\vect F$, the *electric field* at $M$ is

$$
\vect E = \frac{\vect F}{q}
\qquad \text{(in $\mathrm{N}/\mathrm{C}$, also written $\mathrm{V}/\mathrm{m}$)}.
$$

$\vect E$ depends only on the sources and on $M$, not on the test charge: doubling $q$ doubles $\vect F$, leaving the quotient unchanged.

**Proposition 14.4 (Force on a charge).**

A charge $q$ placed where the [field](#def-g11-electric-gravitational-fields-field) is $\vect E$ feels the [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) $\vect F = q \vect E$: along $\vect E$ if $q > 0$, opposite to $\vect E$ if $q < 0$, of magnitude $F = \abs{q}\,E$.

**Proof.** Rearrange the definition; the sign rule is the algebra of a negative multiple. ∎

**Example 14.5 (Reading a field).**

A test charge $q = 2.0 \times 10^{-8}\,\mathrm{C}$ at $M$ feels a [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) of $6.0 \times 10^{-4}\,\mathrm{N}$ pointing east. The [field](#def-g11-electric-gravitational-fields-field) at $M$ is $E = 6.0 \times 10^{-4} / 2.0 \times 10^{-8} = 3.0 \times 10^{4}\,\mathrm{N}/\mathrm{C}$, pointing east; $q' = -1.0 \times 10^{-8}\,\mathrm{C}$ placed at the same point feels $F = 1.0 \times 10^{-8} \times 3.0 \times 10^{4} = 3.0 \times 10^{-4}\,\mathrm{N}$ — pointing west.

**Proposition 14.6 (Field of a point charge).**

A point charge $Q$ creates, at distance $d$, a [field](#def-g11-electric-gravitational-fields-field) along the line joining charge to point — away from $Q$ if $Q > 0$, toward it if $Q < 0$ — of magnitude

$$
E = k\,\frac{\abs{Q}}{d^2},
\qquad k = 8.99 \times 10^{9}\,\mathrm{N}\,\mathrm{m}^{2}/\mathrm{C}^{2}.
$$

**Proof.** Coulomb’s law ([Chapter 13](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#ch-g11-fundamental-interactions)) gives the [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) on a test charge $q$ at distance $d$: magnitude $k \abs{qQ} / d^2$, along the joining line, repulsive for like signs. Divide by $q$ and apply [Proposition 14.4](#prop-g11-electric-gravitational-fields-force). ∎

**Example 14.7 (A charged sphere).**

A Van de Graaff sphere carries $Q = 5.0 \times 10^{-7}\,\mathrm{C}$ (a small charged sphere acts like a point charge at its center). At $d = 0.50\,\mathrm{m}$:

$$
E = 8.99 \times 10^{9} \times 5.0 \times 10^{-7} / 0.50^2 \approx 1.8 \times 10^{4}\,\mathrm{N}/\mathrm{C},
$$

pointing radially away from the sphere. At $1.0\,\mathrm{m}$ it has dropped to a quarter of this: the $1/d^2$ of the [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) survives in the [field](#def-g11-electric-gravitational-fields-field).

**Definition 14.8 (Field lines).**

A *field line* is a curve everywhere tangent to the [field](#def-g11-electric-gravitational-fields-field), an arrow giving the [field](#def-g11-electric-gravitational-fields-field)’s direction. Three rules:

- they leave positive charges and arrive at negative charges;
- where lines crowd the [field](#def-g11-electric-gravitational-fields-field) is strong; where they spread, weak;
- two lines never cross: the [field](#def-g11-electric-gravitational-fields-field) has one direction at each point.

![Field lines of a point charge: radially out of +, into -; the crowding near the charge pictures the 1/d2 growth of the field.](https://one-course.com/images/onecourse/chapters/physics-2/g11-electric-gravitational-fields/fig-fb95c245904e.svg)

![Field lines of a point charge: radially out of +, into -; the crowding near the charge pictures the 1/d2 growth of the field.](https://one-course.com/images/onecourse/chapters/physics-2/g11-electric-gravitational-fields/fig-2f2b1844cddd.svg)

*[Field lines](#def-g11-electric-gravitational-fields-lines) of a point charge: radially out of $+$, into $-$; the crowding near the charge pictures the $1/d^2$ growth of the [field](#def-g11-electric-gravitational-fields-field).*

## 14.3 The uniform field of a capacitor

**Definition 14.9 (Parallel-plate capacitor).**

Two facing parallel metal plates, close together and carrying opposite charges, form a *capacitor*. Between the plates the [field](#def-g11-electric-gravitational-fields-field) is *uniform*: same magnitude and direction everywhere — perpendicular to the plates, from the positive plate to the negative.

**Proposition 14.10 (Field between the plates).**

If the [voltage](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage) between the plates ([Chapter 12](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#ch-g11-circuits-and-power)) is $U$ and their separation is $d$, the [field](#def-g11-electric-gravitational-fields-field) between them has magnitude

$$
E = \frac{U}{d} \qquad \text{(in $\mathrm{V}/\mathrm{m}$)}.
$$

**Proof.** *Admitted at this level.* ∎

**Remark 14.11.**

Uniformity and $E = U/d$ are derived honestly in the Year 1 volume; the formula explains the [unit](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) $\mathrm{V}/\mathrm{m}$. [Thin](https://one-course.com/books/physics/2/en/chapter/10-lenses-images-and-the-eye#def-g11-lenses-and-eye-lens) gaps make strong [fields](#def-g11-electric-gravitational-fields-field): $100\,\mathrm{V}$ across a millimetre is already $10^{5}\,\mathrm{V}/\mathrm{m}$.

**Example 14.12 (Steering an electron beam).**

In an [oscilloscope](https://one-course.com/books/physics/2/en/chapter/8-signals-and-waves#def-g10-signals-and-waves-oscilloscope), the beam passes between plates with $U = 100\,\mathrm{V}$ and $d = 5.0\,\mathrm{mm}$: $E = 100 / 5.0 \times 10^{-3} = 2.0 \times 10^{4}\,\mathrm{V}/\mathrm{m}$, and each electron feels $F = 1.60 \times 10^{-19} \times 2.0 \times 10^{4} = 3.2 \times 10^{-15}\,\mathrm{N}$ toward the positive plate — $4 \times 10^{14}$ times its [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) of $8.9 \times 10^{-30}\,\mathrm{N}$, so the beam deflects visibly (the curved path is [Chapter 25](https://one-course.com/books/physics/2/en/chapter/25-newtons-laws#ch-g12-newtons-laws)’s business).

![The uniform field of a capacitor: parallel, equally spaced lines from the + plate to the - plate, E = U/d; a positive charge is pushed along the field, a negative one against it.](https://one-course.com/images/onecourse/chapters/physics-2/g11-electric-gravitational-fields/fig-aac0076074e6.svg)

*The [uniform field](#def-g11-electric-gravitational-fields-capacitor) of a [capacitor](#def-g11-electric-gravitational-fields-capacitor): parallel, equally spaced lines from the $+$ plate to the $-$ plate, $E = U/d$; a positive charge is pushed along the [field](#def-g11-electric-gravitational-fields-field), a negative one against it.*

## 14.4 The gravitational field

**Definition 14.13 (Gravitational field).**

If a test mass $m$ placed at a point feels a [gravitational force](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-force) (a [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight)) $\vect P$, the *gravitational field* there is

$$
\vect g = \frac{\vect P}{m} \qquad \text{(in $\mathrm{N}/\mathrm{kg}$)},
$$

and conversely $\vect P = m \vect g$. An earlier chapter introduced the number $g$; $\vect g$ adds the direction a dropped stone starts to move.

**Proposition 14.14 (Field of the Earth).**

At distance $d$ from the Earth’s center (mass $M$), the [gravitational field](#def-g11-electric-gravitational-fields-gfield) points toward the center and has magnitude

$$
g = G\,\frac{M}{d^2},
\qquad G = 6.67 \times 10^{-11}\,\mathrm{N}\,\mathrm{m}^{2}/\mathrm{kg}^{2}.
$$

**Proof.** Universal [gravitation](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-four) ([Chapter 4](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#ch-g10-universal-gravitation)) gives the [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) on a test mass $m$: magnitude $G M m / d^2$, toward the center; divide by $m$. At the surface, $d = R$ and this is the $g = GM/R^2 \approx 9.81\,\mathrm{N}/\mathrm{kg}$ of the earlier chapter. ∎

**Example 14.15 (The field where the Moon lives).**

At the Moon’s distance $d = 3.84 \times 10^{8}\,\mathrm{m}$:

$$
g = 6.67 \times 10^{-11} \times 5.97 \times 10^{24} / (3.84 \times 10^{8})^2
\approx 2.7 \times 10^{-3}\,\mathrm{N}/\mathrm{kg}.
$$

Earth’s [field](#def-g11-electric-gravitational-fields-field) never stops — it only [thins](https://one-course.com/books/physics/2/en/chapter/10-lenses-images-and-the-eye#def-g11-lenses-and-eye-lens) out as $1/d^2$; this is the [field](#def-g11-electric-gravitational-fields-field) that holds the Moon on its [orbit](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite).

**Remark 14.16 (Uniform near the surface).**

Over a laboratory or even a mountain, $d$ barely changes compared with $R = 6.37 \times 10^{6}\,\mathrm{m}$, so $\vect g$ is [uniform](#def-g11-electric-gravitational-fields-capacitor) to high accuracy: vertical lines, constant $9.81\,\mathrm{N}/\mathrm{kg}$ — the Earth’s [field](#def-g11-electric-gravitational-fields-field) looks like the [capacitor](#def-g11-electric-gravitational-fields-capacitor)’s, just as the round Earth looks flat from a garden.

![The Earth’s gravitational field: radial, pointing at the center, of magnitude GM/d2 — the exact portrait of the field of a negative point charge.](https://one-course.com/images/onecourse/chapters/physics-2/g11-electric-gravitational-fields/fig-a6821902517f.svg)

*The Earth’s [gravitational field](#def-g11-electric-gravitational-fields-gfield): radial, pointing at the center, of magnitude $GM/d^2$ — the exact portrait of the [field](#def-g11-electric-gravitational-fields-field) of a negative point charge.*

## 14.5 One idea, two forces

The two [fields](#def-g11-electric-gravitational-fields-field) of this chapter are the same mathematics in two costumes:

|  | **electric** | **gravitational** |
| --- | --- | --- |
| source | charge $Q$ | mass $M$ |
| [field](#def-g11-electric-gravitational-fields-field) | $\vect E = \vect F / q$ | $\vect g = \vect P / m$ |
| [unit](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) | $\mathrm{N}/\mathrm{C}$ = $\mathrm{V}/\mathrm{m}$ | $\mathrm{N}/\mathrm{kg}$ |
| point source | $E = k \abs{Q} / d^2$ | $g = G M / d^2$ |
| constant | $k = 8.99 \times 10^{9}\,\mathrm{N}\,\mathrm{m}^{2}/\mathrm{C}^{2}$ | $G = 6.67 \times 10^{-11}\,\mathrm{N}\,\mathrm{m}^{2}/\mathrm{kg}^{2}$ |
| [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) on a test | $\vect F = q \vect E$ | $\vect P = m \vect g$ |
| [field lines](#def-g11-electric-gravitational-fields-lines) | out of $+$, into $-$ | always into the mass |

Every formula on the left becomes the one on the right under the dictionary $q \leftrightarrow m$, $\vect E \leftrightarrow \vect g$, $k \leftrightarrow G$.

**Remark 14.17 (Where the analogy breaks).**

Charge comes in two signs; mass in one. So [electric fields](#def-g11-electric-gravitational-fields-efield) can be canceled: the rearranged charges of a metal box wipe out any external [field](#def-g11-electric-gravitational-fields-field) inside it (a *Faraday cage* — why an elevator kills your phone [signal](https://one-course.com/books/physics/2/en/chapter/8-signals-and-waves#def-g10-signals-and-waves-signal)), and bulk matter, with balanced $+$ and $-$, is electrically silent. Gravity can be neither screened nor neutralized: every [kilogram](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) adds its pull — which is why the [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) weaker by $10^{39}$ ([Exercise 14.10](#exo-g11-electric-gravitational-fields-10)) runs the universe at large scales.

## 14.6 Exercises

**Exercise 14.1 ★.**

A test charge $q = 2.0 \times 10^{-6}\,\mathrm{C}$ at a point $M$ feels a [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) of $5.0 \times 10^{-3}\,\mathrm{N}$ pointing north. Give the [field](#def-g11-electric-gravitational-fields-field) at $M$ (magnitude and direction), then the [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) on $q' = -4.0 \times 10^{-6}\,\mathrm{C}$ placed at $M$.

**Solution of Exercise 14.1.**

$E = 5.0 \times 10^{-3} / 2.0 \times 10^{-6} = 2.5 \times 10^{3}\,\mathrm{N}/\mathrm{C}$, pointing north. On $q'$: $F = 4.0 \times 10^{-6} \times 2.5 \times 10^{3} = 1.0 \times 10^{-2}\,\mathrm{N}$, pointing *south* ($q' < 0$).

**Exercise 14.2 ★.**

A point charge $Q = 5.0\,\mathrm{nC}$ sits at $O$.

1. Compute the [field](#def-g11-electric-gravitational-fields-field) at $d = 30\,\mathrm{cm}$ : magnitude, direction.
2. At what distance is the [field](#def-g11-electric-gravitational-fields-field) half as strong?

**Solution of Exercise 14.2.**

*1.* $E = 8.99 \times 10^{9} \times 5.0 \times 10^{-9} / 0.30^2
\approx 5.0 \times 10^{2}\,\mathrm{N}/\mathrm{C}$, pointing away from $O$ ($Q > 0$).

*2.* Halving $k\abs{Q}/d^2$ needs $d^2$ doubled: $d' = 0.30\sqrt{2} \approx 0.42\,\mathrm{m}$.

**Exercise 14.3 ★.**

A [capacitor](#def-g11-electric-gravitational-fields-capacitor) has $U = 600\,\mathrm{V}$ across plates $3.0\,\mathrm{mm}$ apart. Compute the [field](#def-g11-electric-gravitational-fields-field), then the [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) on one electron there ($e = 1.60 \times 10^{-19}\,\mathrm{C}$).

**Solution of Exercise 14.3.**

$E = 600 / 3.0 \times 10^{-3} = 2.0 \times 10^{5}\,\mathrm{V}/\mathrm{m}$; $F = eE = 1.60 \times 10^{-19} \times 2.0 \times 10^{5} = 3.2 \times 10^{-14}\,\mathrm{N}$.

**Exercise 14.4 ★.**

On Mars, a $250\,\mathrm{kg}$ probe weighs $930\,\mathrm{N}$. Compute the Martian surface [field](#def-g11-electric-gravitational-fields-field) $g_{\text{Mars}}$, then the [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) of an $80\,\mathrm{kg}$ astronaut there.

**Solution of Exercise 14.4.**

$g_{\text{Mars}} = P/m = 930/250 = 3.7\,\mathrm{N}/\mathrm{kg}$. Astronaut: $P = 80 \times 3.72 \approx 3.0 \times 10^{2}\,\mathrm{N}$ (about $780\,\mathrm{N}$ on Earth).

**Exercise 14.5 ★.**

True or false, with one reason each: (a) two [field lines](#def-g11-electric-gravitational-fields-lines) may cross where the [field](#def-g11-electric-gravitational-fields-field) is strong; (b) [field lines](#def-g11-electric-gravitational-fields-lines) leave negative charges and arrive at positive ones; (c) bunched [field lines](#def-g11-electric-gravitational-fields-lines) [signal](https://one-course.com/books/physics/2/en/chapter/8-signals-and-waves#def-g10-signals-and-waves-signal) a strong [field](#def-g11-electric-gravitational-fields-field).

**Solution of Exercise 14.5.**

(a) False: the [field](#def-g11-electric-gravitational-fields-field) has one direction at each point, so lines never cross. (b) False: they leave $+$ and arrive at $-$. (c) True: crowding codes strength.

**Exercise 14.6 ★★.**

The International Space Station [orbits](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) at altitude $420\,\mathrm{km}$ ($R = 6.37 \times 10^{6}\,\mathrm{m}$, $M = 5.97 \times 10^{24}\,\mathrm{kg}$).

1. Compute $g$ at the station; what fraction of the surface value?
2. Astronauts float. Reconcile this with your answer in one sentence.

**Solution of Exercise 14.6.**

*1.* $d = 6.37 \times 10^{6} + 4.2 \times 10^{5} = 6.79 \times 10^{6}\,\mathrm{m}$, so $g = 6.67 \times 10^{-11} \times 5.97 \times 10^{24} / (6.79 \times 10^{6})^2
\approx 8.6\,\mathrm{N}/\mathrm{kg}$ — $88\%$ of the surface value.

*2.* They float because station and astronaut fall together (free fall), not because gravity is absent.

**Exercise 14.7 ★★.**

What [field](#def-g11-electric-gravitational-fields-field) magnitude would balance an electron’s [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) ($m_e = 9.11 \times 10^{-31}\,\mathrm{kg}$)? Compare with the $2.0 \times 10^{4}\,\mathrm{V}/\mathrm{m}$ of [Example 14.12](#ex-g11-electric-gravitational-fields-beam); conclude about the place of gravity in electron physics.

**Solution of Exercise 14.7.**

$E = m_e g / e = 9.11 \times 10^{-31} \times 9.81 / 1.60 \times 10^{-19}
\approx 5.6 \times 10^{-11}\,\mathrm{V}/\mathrm{m}$: some $4 \times 10^{14}$ times weaker than the [oscilloscope](https://one-course.com/books/physics/2/en/chapter/8-signals-and-waves#def-g10-signals-and-waves-oscilloscope)’s [field](#def-g11-electric-gravitational-fields-field). Gravity is utterly negligible in electron physics.

**Exercise 14.8 ★★.**

A dust grain of mass $2.0 \times 10^{-6}\,\mathrm{kg}$ floats at rest between horizontal plates $2.0\,\mathrm{cm}$ apart with $U = 5.0\,\mathrm{kV}$ across them.

1. Compute the [field](#def-g11-electric-gravitational-fields-field) , then the magnitude of the grain’s charge.
2. The grain’s charge is positive: which plate is the positive one?
3. How many [elementary charges](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-elementary) does the grain carry?

**Solution of Exercise 14.8.**

*1.* $E = 5.0 \times 10^{3} / 2.0 \times 10^{-2} = 2.5 \times 10^{5}\,\mathrm{V}/\mathrm{m}$; at rest $qE = mg$, so $q = 2.0 \times 10^{-6} \times 9.81 / 2.5 \times 10^{5} \approx 7.8 \times 10^{-11}\,\mathrm{C}$.

*2.* The [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) on the positive grain must point up, so $\vect E$ points up: the *lower* plate is positive.

*3.* $n = 7.85 \times 10^{-11} / 1.60 \times 10^{-19} \approx 4.9 \times 10^{8}$ [elementary charges](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-elementary).

**Exercise 14.9 ★★.**

Charges $+Q$ and $+4Q$ sit $30\,\mathrm{cm}$ apart. Where on the segment joining them is the total [field](#def-g11-electric-gravitational-fields-field) zero? Explain first why the point must lie between the charges, closer to the smaller one.

**Solution of Exercise 14.9.**

Outside the segment both [fields](#def-g11-electric-gravitational-fields-field) point the same way; between, they oppose — and balance needs the point nearer the weaker source $+Q$. With $x$ the distance to $+Q$: $kQ/x^2 = 4kQ/(0.30 - x)^2$, so $0.30 - x = 2x$ and $x = 0.10\,\mathrm{m}$ from $+Q$.

**Exercise 14.10 ★★.**

In a hydrogen [atom](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder), the proton ($m_p = 1.67 \times 10^{-27}\,\mathrm{kg}$, charge $+e$) and the electron ($m_e = 9.11 \times 10^{-31}\,\mathrm{kg}$, charge $-e$) are $d = 5.3 \times 10^{-11}\,\mathrm{m}$ apart. Compute the electric and the [gravitational forces](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-force) between them, and their ratio.

**Solution of Exercise 14.10.**

$F_e = ke^2/d^2 = 8.99 \times 10^{9} \times (1.60 \times 10^{-19})^2 /
(5.3 \times 10^{-11})^2 \approx 8.2 \times 10^{-8}\,\mathrm{N}$; $F_g = G m_p m_e / d^2 = 6.67 \times 10^{-11} \times 1.67 \times 10^{-27} \times
9.11 \times 10^{-31} / (5.3 \times 10^{-11})^2 \approx 3.6 \times 10^{-47}\,\mathrm{N}$. Ratio $F_e / F_g \approx 2.3 \times 10^{39}$.

**Exercise 14.11 ★★.**

At a point $M$, one source alone would create a [field](#def-g11-electric-gravitational-fields-field) of $300\,\mathrm{N}/\mathrm{C}$ pointing east; a second, $400\,\mathrm{N}/\mathrm{C}$ pointing north. [Fields](#def-g11-electric-gravitational-fields-field) add as vectors: give the total [field](#def-g11-electric-gravitational-fields-field) at $M$ (magnitude and direction), then the [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) on $q = -2.0 \times 10^{-6}\,\mathrm{C}$ placed there.

**Solution of Exercise 14.11.**

$E = \sqrt{300^2 + 400^2} = 500\,\mathrm{N}/\mathrm{C}$, at $\tan^{-1}(400/300) \approx 53{}^{\circ}$ north of east. On $q < 0$: $F = 2.0 \times 10^{-6} \times 500 = 1.0 \times 10^{-3}\,\mathrm{N}$, opposite the [field](#def-g11-electric-gravitational-fields-field) — $53{}^{\circ}$ south of west.

**Exercise 14.12 ★★★.**

At what altitude has the Earth’s [field](#def-g11-electric-gravitational-fields-field) dropped to half its surface value? And at altitude $h = R$, to what fraction?

**Solution of Exercise 14.12.**

$GM/(R+h)^2 = \tfrac12\,GM/R^2$ gives $R + h = R\sqrt{2}$, so $h = (\sqrt{2} - 1)R \approx 2.6 \times 10^{6}\,\mathrm{m}$ (about $2600\,\mathrm{km}$). At $h = R$, $d = 2R$: one quarter of the surface value.

**Exercise 14.13 ★★★.**

Charges of $2.0\,\mathrm{nC}$ sit at $A$ and $B$, $6.0\,\mathrm{cm}$ apart. $M$ is on the perpendicular bisector of $[AB]$, $4.0\,\mathrm{cm}$ from its midpoint.

1. Compute the distance $AM$ , then the magnitude of the [field](#def-g11-electric-gravitational-fields-field) each charge creates at $M$ .
2. Add the two [fields](#def-g11-electric-gravitational-fields-field) as vectors (use the symmetry) and give the total [field](#def-g11-electric-gravitational-fields-field) at $M$ , magnitude and direction.

**Solution of Exercise 14.13.**

*1.* $AM = \sqrt{3.0^2 + 4.0^2} = 5.0\,\mathrm{cm}$; each charge creates $E_1 = 8.99 \times 10^{9} \times 2.0 \times 10^{-9} / 0.050^2
\approx 7.2 \times 10^{3}\,\mathrm{N}/\mathrm{C}$.

*2.* The components along $AB$ cancel by symmetry; each [field](#def-g11-electric-gravitational-fields-field) contributes $E_1 \times 4/5$ along the bisector, away from the segment: $E = 2 \times 7.19 \times 10^{3} \times 0.80 \approx 1.2 \times 10^{4}\,\mathrm{N}/\mathrm{C}$, directed along the bisector, away from $[AB]$.

**Exercise 14.14 ★★★.**

Explain, using the two signs of charge, how the free charges of a metal box cancel an external [electric field](#def-g11-electric-gravitational-fields-efield) everywhere inside it — and why no arrangement of masses can do the same for gravity. What impossible ingredient would a gravity shield require?

**Solution of Exercise 14.14.**

The box’s free charges move under the external [field](#def-g11-electric-gravitational-fields-field): $+$ piles up on one face, $-$ on the other, and these displaced charges create an internal [field](#def-g11-electric-gravitational-fields-field) opposing the external one. They keep moving until the total [field](#def-g11-electric-gravitational-fields-field) inside is exactly zero — then equilibrium. Gravity offers only one sign of source: every mass *adds* an attracting [field](#def-g11-electric-gravitational-fields-field), none can oppose it. A gravity shield would need negative mass, which does not exist.

**Exercise 14.15 ★★★.**

A small ball ($m = 0.50\,\mathrm{g}$, $q = 2.0 \times 10^{-7}\,\mathrm{C}$) hangs from a thread in a horizontal [uniform field](#def-g11-electric-gravitational-fields-capacitor) $E = 1.0 \times 10^{4}\,\mathrm{V}/\mathrm{m}$. At rest, the thread makes an angle $\theta$ with the vertical.

1. List the three [forces](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) ; explain why $\tan\theta = qE / (mg)$ .
2. Compute $\theta$ , then the [tension](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) of the thread.

**Solution of Exercise 14.15.**

*1.* [Weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) $m\vect g$ (down), [tension](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) $\vect T$ (along the thread), electric [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) $q\vect E$ (horizontal). At rest, $T\sin\theta = qE$ and $T\cos\theta = mg$; divide: $\tan\theta = qE/(mg)$.

*2.* $\tan\theta = 2.0 \times 10^{-7} \times 1.0 \times 10^{4} /
(5.0 \times 10^{-4} \times 9.81) \approx 0.41$, so $\theta \approx 22{}^{\circ}$; $T = mg/\cos\theta \approx 5.3 \times 10^{-3}\,\mathrm{N}$.

## 14.7 Problem: Millikan’s droplet

**Problem 14.1.**

Weekend problem — weighing the electron’s charge: an oil droplet parked in mid-air between two plates reveals that charge comes in indivisible steps of $1.60 \times 10^{-19}$ [coulombs](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-charge)

In 1909, Robert Millikan sprayed oil droplets between the horizontal plates of a [capacitor](#def-g11-electric-gravitational-fields-capacitor). By tuning the [voltage](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage) until a chosen droplet hung motionless — electric [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) exactly balancing [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) — he could weigh its charge. Our droplet has mass $m = 4.90 \times 10^{-15}\,\mathrm{kg}$ (measured from its slow fall with the [field](#def-g11-electric-gravitational-fields-field) off; we take it as given); the plates are $d = 8.0\,\mathrm{mm}$ apart, the *upper* one positive; the droplet’s charge is negative.

**Part I — The stage.**

1. Describe the [field](#def-g11-electric-gravitational-fields-field) between the plates; what does “ [uniform](#def-g11-electric-gravitational-fields-capacitor) ” buy?
2. For $U = 800\,\mathrm{V}$ , compute $E$ .
3. In which direction is the electric [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) on the droplet? Why did the *upper* plate have to be the positive one?
4. Compute the droplet’s [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) .
5. Write the at-rest condition as an equality of two [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) magnitudes.

**Part II — The balance.**

6. From it, express $\abs{q}$ in terms of $m$ , $g$ , $d$ and $U$ .
7. The droplet hangs at rest for $U = 800\,\mathrm{V}$ : compute $\abs{q}$ .
8. Divide by $e = 1.60 \times 10^{-19}\,\mathrm{C}$ . How many excess electrons does the droplet carry?
9. The [voltage](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage) is nudged up to $900\,\mathrm{V}$ . In which direction does the net [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) now point?
10. Show that the balancing [voltage](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage) is $U = mgd/\abs{q}$ : the bigger the charge, the smaller the [voltage](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage) .

**Part III — The steps.** A burst of radiation can knock electrons onto or off the droplet; after each change, Millikan re-tunes the [voltage](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage) to balance.

11. After one burst, balance requires $U = 1200\,\mathrm{V}$ . Compute the new $\abs{q}$ . Did the droplet gain or lose an electron?
12. Successive balances are found at $U = 1200$ , $800$ , $600$ , $480$ and $400\,\mathrm{V}$ . Compute the five charges.
13. Show that all five are integer multiples of one quantity; give its value.
14. Between consecutive balances, by how much does the charge change? Interpret.
15. Explain why this droplet can never hang at rest at $U = 700\,\mathrm{V}$ , however patiently one tunes.
16. Millikan repeated this on hundreds of droplets: every measured charge was an integer multiple of the same value. State the conclusion, and name the quantity measured.

**Part IV — Why there is no gravitational Millikan.**

17. In this chapter’s dictionary, what plays the roles of $q$ , $E$ and $qE$ in the balance of Part II?
18. Could a mass overhead hold the droplet up instead? Compute the [field](#def-g11-electric-gravitational-fields-field) of a $1000\,\mathrm{kg}$ lead sphere $1.0\,\mathrm{m}$ above the droplet, the [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) on the droplet, and compare with its [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) .
19. The balance needs an upward pull strong enough to fight the whole Earth. Why can a plate of charge provide it and no arrangement of masses can?
20. The droplet’s *mass* also changes in steps (one oil [molecule](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) : about $5 \times 10^{-25}\,\mathrm{kg}$ ). Compare the relative jumps in $m$ and in $\abs{q}$ when one electron lands, and conclude: why does this balance see the [atoms](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) of electricity but not the [molecules](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) of oil? Punchline: charge is quantized, and its [atom](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) is $e = 1.60 \times 10^{-19}\,\mathrm{C}$ .

**Solution of Problem 14.1.**

**1.** [Uniform](#def-g11-electric-gravitational-fields-capacitor), perpendicular to the plates, pointing down (from the $+$ upper plate to the $-$ lower one); uniformity means the [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) on the droplet is the same wherever it drifts. **2.** $E = U/d = 800 / 8.0 \times 10^{-3} = 1.0 \times 10^{5}\,\mathrm{V}/\mathrm{m}$. **3.** The charge is negative, so the [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) is *opposite* $\vect E$: upward. Only with the $+$ plate on top does $\vect E$ point down and the electric [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) fight the [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight). **4.** $P = mg = 4.90 \times 10^{-15} \times 9.81 \approx
4.8 \times 10^{-14}\,\mathrm{N}$. **5.** $\abs{q}E = mg$. **6.** $\abs{q} = mg/E = mgd/U$. **7.** $\abs{q} = 4.81 \times 10^{-14} \times 8.0 \times 10^{-3} / 800
\approx 4.8 \times 10^{-19}\,\mathrm{C}$. **8.** $n = 4.8 \times 10^{-19} / 1.60 \times 10^{-19} = 3$ excess electrons. **9.** $E$ grows, so $\abs{q}E > mg$: the net [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) points up. **10.** From $\abs{q}\,U/d = mg$, $U = mgd/\abs{q}$ — inversely proportional to the charge. **11.** $\abs{q} = mgd/U = 3.85 \times 10^{-16}/1200 \approx
3.2 \times 10^{-19}\,\mathrm{C} = 2e$: down from $3e$, the droplet *lost* one electron. **12.** $\abs{q} = 3.85 \times 10^{-16}/U$: $3.2 \times 10^{-19}\,\mathrm{C}$, $4.8 \times 10^{-19}\,\mathrm{C}$, $6.4 \times 10^{-19}\,\mathrm{C}$, $8.0 \times 10^{-19}\,\mathrm{C}$, $9.6 \times 10^{-19}\,\mathrm{C}$. **13.** They are $2e, 3e, 4e, 5e, 6e$ with $e = 1.60 \times 10^{-19}\,\mathrm{C}$. **14.** Each step changes $\abs{q}$ by exactly $1.6 \times 10^{-19}\,\mathrm{C}$: one electron lands at a time. **15.** It would need $\abs{q} = 3.85 \times 10^{-16}/700 \approx
5.5 \times 10^{-19}\,\mathrm{C} = 3.4\,e$ — not a whole number of electrons, so no burst of radiation can ever produce it. **16.** Charge is *quantized*: it comes in integer multiples of the [elementary charge](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-elementary) $e = 1.60 \times 10^{-19}\,\mathrm{C}$ — the quantity Millikan measured. **17.** $q \leftrightarrow m$, $E \leftrightarrow g$, and $\abs{q}E \leftrightarrow mg$: the [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) itself. **18.** $g_{\text{sphere}} = GM/d^2 = 6.67 \times 10^{-11} \times 1000 /
1.0^2 = 6.7 \times 10^{-8}\,\mathrm{N}/\mathrm{kg}$; [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) on the droplet $4.90 \times 10^{-15} \times 6.7 \times 10^{-8} \approx 3.3 \times 10^{-22}\,\mathrm{N}$ — about $1.5 \times 10^{8}$ times smaller than its [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight). Hopeless. **19.** Because $G$ is so small, only a planet-sized mass overhead could rival the Earth’s pull — and no mass repels. The enormous $k$ and the two signs of charge let a bench-top plate out-pull the planet in either direction. **20.** One electron: $\Delta\abs{q}/\abs{q} = e/3e \approx 33\%$; one oil [molecule](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder): $\Delta m/m = 5 \times 10^{-25} / 4.90 \times 10^{-15}
\approx 10^{-10}$. The charge staircase moves the balancing [voltage](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage) by hundreds of [volts](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage); the mass staircase is invisible at any [voltage](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage). The balance therefore resolves the [atom](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) of electricity — $e = 1.60 \times 10^{-19}\,\mathrm{C}$ — while the [molecules](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-ladder) of oil blur into a continuum.