Physics · Book 2 · Grades 10–12

High School Physics

High School Physics · Grades 10–12

14Electric 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 — and whatever is placed there responds to the 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 determined by the field at its own location alone. This two-step picture replaces action at a distance.

Remark 14.2

The field is not a bookkeeping trick: it carries energy, and changes in the source travel outward through it at finite speed — light itself is such a traveling field (Chapter 22). Here we study fields that do not change in time.

14.2 The electric field

Definition 14.3 (Electric field)

If a small test charge qq placed at a point MM feels an electric force F→\vect F, the electric field at MM is

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

E→\vect E depends only on the sources and on MM, not on the test charge: doubling qq doubles F→\vect F, leaving the quotient unchanged.

Proposition 14.4 (Force on a charge)

A charge qq placed where the field is E→\vect E feels the force F→=qE→\vect F = q \vect E: along E→\vect E if q>0q > 0, opposite to E→\vect E if q<0q < 0, of magnitude F=∣q∣ EF = \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=2.0×10−8 Cq = 2.0 \times 10^{-8}\,\mathrm{C} at MM feels a force of 6.0×10−4 N6.0 \times 10^{-4}\,\mathrm{N} pointing east. The field at MM is E=6.0×10−4/2.0×10−8=3.0×104 N/CE = 6.0 \times 10^{-4} / 2.0 \times 10^{-8} = 3.0 \times 10^{4}\,\mathrm{N}/\mathrm{C}, pointing east; q′=−1.0×10−8 Cq' = -1.0 \times 10^{-8}\,\mathrm{C} placed at the same point feels F=1.0×10−8×3.0×104=3.0×10−4 NF = 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 QQ creates, at distance dd, a field along the line joining charge to point — away from QQ if Q>0Q > 0, toward it if Q<0Q < 0 — of magnitude

E=k ∣Q∣d2,k=8.99×109 N m2/C2.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) gives the force on a test charge qq at distance dd: magnitude k∣qQ∣/d2k \abs{qQ} / d^2, along the joining line, repulsive for like signs. Divide by qq and apply Proposition 14.4. ∎

Example 14.7 (A charged sphere)

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

E=8.99×109×5.0×10−7/0.502≈1.8×104 N/C,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 m1.0\,\mathrm{m} it has dropped to a quarter of this: the 1/d21/d^2 of the force survives in the field.

Definition 14.8 (Field lines)

A field line is a curve everywhere tangent to the field, an arrow giving the field’s direction. Three rules:

  • they leave positive charges and arrive at negative charges;
  • where lines crowd the field is strong; where they spread, weak;
  • two lines never cross: the 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. Field lines of a point charge: radially out of +, into -; the crowding near the charge pictures the 1/d2 growth of the field.
Field lines of a point charge: radially out of ++, into −-; the crowding near the charge pictures the 1/d21/d^2 growth of the 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 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 between the plates (Chapter 12) is UU and their separation is dd, the field between them has magnitude

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

Proof. Admitted at this level. ∎

Remark 14.11

Uniformity and E=U/dE = U/d are derived honestly in the Year 1 volume; the formula explains the unit V/m\mathrm{V}/\mathrm{m}. Thin gaps make strong fields: 100 V100\,\mathrm{V} across a millimetre is already 105 V/m10^{5}\,\mathrm{V}/\mathrm{m}.

Example 14.12 (Steering an electron beam)

In an oscilloscope, the beam passes between plates with U=100 VU = 100\,\mathrm{V} and d=5.0 mmd = 5.0\,\mathrm{mm}: E=100/5.0×10−3=2.0×104 V/mE = 100 / 5.0 \times 10^{-3} = 2.0 \times 10^{4}\,\mathrm{V}/\mathrm{m}, and each electron feels F=1.60×10−19×2.0×104=3.2×10−15 NF = 1.60 \times 10^{-19} \times 2.0 \times 10^{4} = 3.2 \times 10^{-15}\,\mathrm{N} toward the positive plate — 4×10144 \times 10^{14} times its weight of 8.9×10−30 N8.9 \times 10^{-30}\,\mathrm{N}, so the beam deflects visibly (the curved path is Chapter 25’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.
The uniform field of a capacitor: parallel, equally spaced lines from the ++ plate to the −- plate, E=U/dE = U/d; a positive charge is pushed along the field, a negative one against it.

14.4 The gravitational field

Definition 14.13 (Gravitational field)

If a test mass mm placed at a point feels a gravitational force (a weight) P→\vect P, the gravitational field there is

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

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

Proposition 14.14 (Field of the Earth)

At distance dd from the Earth’s center (mass MM), the gravitational field points toward the center and has magnitude

g=G Md2,G=6.67×10−11 N m2/kg2.g = G\,\frac{M}{d^2}, \qquad G = 6.67 \times 10^{-11}\,\mathrm{N}\,\mathrm{m}^{2}/\mathrm{kg}^{2}.

Proof. Universal gravitation (Chapter 4) gives the force on a test mass mm: magnitude GMm/d2G M m / d^2, toward the center; divide by mm. At the surface, d=Rd = R and this is the g=GM/R2≈9.81 N/kgg = 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=3.84×108 md = 3.84 \times 10^{8}\,\mathrm{m}:

g=6.67×10−11×5.97×1024/(3.84×108)2≈2.7×10−3 N/kg.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 never stops — it only thins out as 1/d21/d^2; this is the field that holds the Moon on its orbit.

Remark 14.16 (Uniform near the surface)

Over a laboratory or even a mountain, dd barely changes compared with R=6.37×106 mR = 6.37 \times 10^{6}\,\mathrm{m}, so g→\vect g is uniform to high accuracy: vertical lines, constant 9.81 N/kg9.81\,\mathrm{N}/\mathrm{kg} — the Earth’s field looks like the 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.
The Earth’s gravitational field: radial, pointing at the center, of magnitude GM/d2GM/d^2 — the exact portrait of the field of a negative point charge.

14.5 One idea, two forces

The two fields of this chapter are the same mathematics in two costumes:

electricgravitational
sourcecharge QQmass MM
fieldE→=F→/q\vect E = \vect F / qg→=P→/m\vect g = \vect P / m
unitN/C\mathrm{N}/\mathrm{C} = V/m\mathrm{V}/\mathrm{m}N/kg\mathrm{N}/\mathrm{kg}
point sourceE=k∣Q∣/d2E = k \abs{Q} / d^2g=GM/d2g = G M / d^2
constantk=8.99×109 N m2/C2k = 8.99 \times 10^{9}\,\mathrm{N}\,\mathrm{m}^{2}/\mathrm{C}^{2}G=6.67×10−11 N m2/kg2G = 6.67 \times 10^{-11}\,\mathrm{N}\,\mathrm{m}^{2}/\mathrm{kg}^{2}
force on a testF→=qE→\vect F = q \vect EP→=mg→\vect P = m \vect g
field linesout of ++, into −-always into the mass

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

Remark 14.17 (Where the analogy breaks)

Charge comes in two signs; mass in one. So electric fields can be canceled: the rearranged charges of a metal box wipe out any external field inside it (a Faraday cage — why an elevator kills your phone signal), and bulk matter, with balanced ++ and −-, is electrically silent. Gravity can be neither screened nor neutralized: every kilogram adds its pull — which is why the force weaker by 103910^{39} (Exercise 14.10) runs the universe at large scales.

14.6 Exercises

Exercise 14.1 ★

A test charge q=2.0×10−6 Cq = 2.0 \times 10^{-6}\,\mathrm{C} at a point MM feels a force of 5.0×10−3 N5.0 \times 10^{-3}\,\mathrm{N} pointing north. Give the field at MM (magnitude and direction), then the force on q′=−4.0×10−6 Cq' = -4.0 \times 10^{-6}\,\mathrm{C} placed at MM.

Solution

Solution of Exercise 14.1.

E=5.0×10−3/2.0×10−6=2.5×103 N/CE = 5.0 \times 10^{-3} / 2.0 \times 10^{-6} = 2.5 \times 10^{3}\,\mathrm{N}/\mathrm{C}, pointing north. On q′q': F=4.0×10−6×2.5×103=1.0×10−2 NF = 4.0 \times 10^{-6} \times 2.5 \times 10^{3} = 1.0 \times 10^{-2}\,\mathrm{N}, pointing south (q′<0q' < 0).

Exercise 14.2 ★

A point charge Q=5.0 nCQ = 5.0\,\mathrm{nC} sits at OO.

  1. Compute the field at d=30 cmd = 30\,\mathrm{cm}: magnitude, direction.
  2. At what distance is the field half as strong?
Solution

Solution of Exercise 14.2.

1. E=8.99×109×5.0×10−9/0.302≈5.0×102 N/CE = 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 OO (Q>0Q > 0).

2. Halving k∣Q∣/d2k\abs{Q}/d^2 needs d2d^2 doubled: d′=0.302≈0.42 md' = 0.30\sqrt{2} \approx 0.42\,\mathrm{m}.

Exercise 14.3 ★

A capacitor has U=600 VU = 600\,\mathrm{V} across plates 3.0 mm3.0\,\mathrm{mm} apart. Compute the field, then the force on one electron there (e=1.60×10−19 Ce = 1.60 \times 10^{-19}\,\mathrm{C}).

Solution

Solution of Exercise 14.3.

E=600/3.0×10−3=2.0×105 V/mE = 600 / 3.0 \times 10^{-3} = 2.0 \times 10^{5}\,\mathrm{V}/\mathrm{m}; F=eE=1.60×10−19×2.0×105=3.2×10−14 NF = 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 kg250\,\mathrm{kg} probe weighs 930 N930\,\mathrm{N}. Compute the Martian surface field gMarsg_{\text{Mars}}, then the weight of an 80 kg80\,\mathrm{kg} astronaut there.

Solution

Solution of Exercise 14.4.

gMars=P/m=930/250=3.7 N/kgg_{\text{Mars}} = P/m = 930/250 = 3.7\,\mathrm{N}/\mathrm{kg}. Astronaut: P=80×3.72≈3.0×102 NP = 80 \times 3.72 \approx 3.0 \times 10^{2}\,\mathrm{N} (about 780 N780\,\mathrm{N} on Earth).

Exercise 14.5 ★

True or false, with one reason each: (a) two field lines may cross where the field is strong; (b) field lines leave negative charges and arrive at positive ones; (c) bunched field lines signal a strong field.

Solution

Solution of Exercise 14.5.

(a) False: the 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 at altitude 420 km420\,\mathrm{km} (R=6.37×106 mR = 6.37 \times 10^{6}\,\mathrm{m}, M=5.97×1024 kgM = 5.97 \times 10^{24}\,\mathrm{kg}).

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

Solution of Exercise 14.6.

1. d=6.37×106+4.2×105=6.79×106 md = 6.37 \times 10^{6} + 4.2 \times 10^{5} = 6.79 \times 10^{6}\,\mathrm{m}, so g=6.67×10−11×5.97×1024/(6.79×106)2≈8.6 N/kgg = 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%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 magnitude would balance an electron’s weight (me=9.11×10−31 kgm_e = 9.11 \times 10^{-31}\,\mathrm{kg})? Compare with the 2.0×104 V/m2.0 \times 10^{4}\,\mathrm{V}/\mathrm{m} of Example 14.12; conclude about the place of gravity in electron physics.

Solution

Solution of Exercise 14.7.

E=meg/e=9.11×10−31×9.81/1.60×10−19≈5.6×10−11 V/mE = 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 4×10144 \times 10^{14} times weaker than the oscilloscope’s field. Gravity is utterly negligible in electron physics.

Exercise 14.8 ★★

A dust grain of mass 2.0×10−6 kg2.0 \times 10^{-6}\,\mathrm{kg} floats at rest between horizontal plates 2.0 cm2.0\,\mathrm{cm} apart with U=5.0 kVU = 5.0\,\mathrm{kV} across them.

  1. Compute the 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 does the grain carry?
Solution

Solution of Exercise 14.8.

1. E=5.0×103/2.0×10−2=2.5×105 V/mE = 5.0 \times 10^{3} / 2.0 \times 10^{-2} = 2.5 \times 10^{5}\,\mathrm{V}/\mathrm{m}; at rest qE=mgqE = mg, so q=2.0×10−6×9.81/2.5×105≈7.8×10−11 Cq = 2.0 \times 10^{-6} \times 9.81 / 2.5 \times 10^{5} \approx 7.8 \times 10^{-11}\,\mathrm{C}.

2. The force on the positive grain must point up, so E→\vect E points up: the lower plate is positive.

3. n=7.85×10−11/1.60×10−19≈4.9×108n = 7.85 \times 10^{-11} / 1.60 \times 10^{-19} \approx 4.9 \times 10^{8} elementary charges.

Exercise 14.9 ★★

Charges +Q+Q and +4Q+4Q sit 30 cm30\,\mathrm{cm} apart. Where on the segment joining them is the total field zero? Explain first why the point must lie between the charges, closer to the smaller one.

Solution

Solution of Exercise 14.9.

Outside the segment both fields point the same way; between, they oppose — and balance needs the point nearer the weaker source +Q+Q. With xx the distance to +Q+Q: kQ/x2=4kQ/(0.30−x)2kQ/x^2 = 4kQ/(0.30 - x)^2, so 0.30−x=2x0.30 - x = 2x and x=0.10 mx = 0.10\,\mathrm{m} from +Q+Q.

Exercise 14.10 ★★

In a hydrogen atom, the proton (mp=1.67×10−27 kgm_p = 1.67 \times 10^{-27}\,\mathrm{kg}, charge +e+e) and the electron (me=9.11×10−31 kgm_e = 9.11 \times 10^{-31}\,\mathrm{kg}, charge −e-e) are d=5.3×10−11 md = 5.3 \times 10^{-11}\,\mathrm{m} apart. Compute the electric and the gravitational forces between them, and their ratio.

Solution

Solution of Exercise 14.10.

Fe=ke2/d2=8.99×109×(1.60×10−19)2/(5.3×10−11)2≈8.2×10−8 NF_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}; Fg=Gmpme/d2=6.67×10−11×1.67×10−27×9.11×10−31/(5.3×10−11)2≈3.6×10−47 NF_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 Fe/Fg≈2.3×1039F_e / F_g \approx 2.3 \times 10^{39}.

Exercise 14.11 ★★

At a point MM, one source alone would create a field of 300 N/C300\,\mathrm{N}/\mathrm{C} pointing east; a second, 400 N/C400\,\mathrm{N}/\mathrm{C} pointing north. Fields add as vectors: give the total field at MM (magnitude and direction), then the force on q=−2.0×10−6 Cq = -2.0 \times 10^{-6}\,\mathrm{C} placed there.

Solution

Solution of Exercise 14.11.

E=3002+4002=500 N/CE = \sqrt{300^2 + 400^2} = 500\,\mathrm{N}/\mathrm{C}, at tan⁡−1(400/300)≈53∘\tan^{-1}(400/300) \approx 53{}^{\circ} north of east. On q<0q < 0: F=2.0×10−6×500=1.0×10−3 NF = 2.0 \times 10^{-6} \times 500 = 1.0 \times 10^{-3}\,\mathrm{N}, opposite the field — 53∘53{}^{\circ} south of west.

Exercise 14.12 ★★★

At what altitude has the Earth’s field dropped to half its surface value? And at altitude h=Rh = R, to what fraction?

Solution

Solution of Exercise 14.12.

GM/(R+h)2=12 GM/R2GM/(R+h)^2 = \tfrac12\,GM/R^2 gives R+h=R2R + h = R\sqrt{2}, so h=(2−1)R≈2.6×106 mh = (\sqrt{2} - 1)R \approx 2.6 \times 10^{6}\,\mathrm{m} (about 2600 km2600\,\mathrm{km}). At h=Rh = R, d=2Rd = 2R: one quarter of the surface value.

Exercise 14.13 ★★★

Charges of 2.0 nC2.0\,\mathrm{nC} sit at AA and BB, 6.0 cm6.0\,\mathrm{cm} apart. MM is on the perpendicular bisector of [AB][AB], 4.0 cm4.0\,\mathrm{cm} from its midpoint.

  1. Compute the distance AMAM, then the magnitude of the field each charge creates at MM.
  2. Add the two fields as vectors (use the symmetry) and give the total field at MM, magnitude and direction.
Solution

Solution of Exercise 14.13.

1. AM=3.02+4.02=5.0 cmAM = \sqrt{3.0^2 + 4.0^2} = 5.0\,\mathrm{cm}; each charge creates E1=8.99×109×2.0×10−9/0.0502≈7.2×103 N/CE_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 ABAB cancel by symmetry; each field contributes E1×4/5E_1 \times 4/5 along the bisector, away from the segment: E=2×7.19×103×0.80≈1.2×104 N/CE = 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][AB].

Exercise 14.14 ★★★

Explain, using the two signs of charge, how the free charges of a metal box cancel an external electric field everywhere inside it — and why no arrangement of masses can do the same for gravity. What impossible ingredient would a gravity shield require?

Solution

Solution of Exercise 14.14.

The box’s free charges move under the external field: ++ piles up on one face, −- on the other, and these displaced charges create an internal field opposing the external one. They keep moving until the total field inside is exactly zero — then equilibrium. Gravity offers only one sign of source: every mass adds an attracting 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 gm = 0.50\,\mathrm{g}, q=2.0×10−7 Cq = 2.0 \times 10^{-7}\,\mathrm{C}) hangs from a thread in a horizontal uniform field E=1.0×104 V/mE = 1.0 \times 10^{4}\,\mathrm{V}/\mathrm{m}. At rest, the thread makes an angle θ\theta with the vertical.

  1. List the three forces; explain why tan⁡θ=qE/(mg)\tan\theta = qE / (mg).
  2. Compute θ\theta, then the tension of the thread.
Solution

Solution of Exercise 14.15.

1. Weight mg→m\vect g (down), tension T→\vect T (along the thread), electric force qE→q\vect E (horizontal). At rest, Tsin⁡θ=qET\sin\theta = qE and Tcos⁡θ=mgT\cos\theta = mg; divide: tan⁡θ=qE/(mg)\tan\theta = qE/(mg).

2. tan⁡θ=2.0×10−7×1.0×104/(5.0×10−4×9.81)≈0.41\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 θ≈22∘\theta \approx 22{}^{\circ}; T=mg/cos⁡θ≈5.3×10−3 NT = 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 1.60×10−191.60 \times 10^{-19} coulombs

In 1909, Robert Millikan sprayed oil droplets between the horizontal plates of a capacitor. By tuning the voltage until a chosen droplet hung motionless — electric force exactly balancing weight — he could weigh its charge. Our droplet has mass m=4.90×10−15 kgm = 4.90 \times 10^{-15}\,\mathrm{kg} (measured from its slow fall with the field off; we take it as given); the plates are d=8.0 mmd = 8.0\,\mathrm{mm} apart, the upper one positive; the droplet’s charge is negative.

Part I — The stage.

  1. Describe the field between the plates; what does “uniform” buy?
  2. For U=800 VU = 800\,\mathrm{V}, compute EE.
  3. In which direction is the electric force on the droplet? Why did the upper plate have to be the positive one?
  4. Compute the droplet’s weight.
  5. Write the at-rest condition as an equality of two force magnitudes.

Part II — The balance.

  1. From it, express ∣q∣\abs{q} in terms of mm, gg, dd and UU.
  2. The droplet hangs at rest for U=800 VU = 800\,\mathrm{V}: compute ∣q∣\abs{q}.
  3. Divide by e=1.60×10−19 Ce = 1.60 \times 10^{-19}\,\mathrm{C}. How many excess electrons does the droplet carry?
  4. The voltage is nudged up to 900 V900\,\mathrm{V}. In which direction does the net force now point?
  5. Show that the balancing voltage is U=mgd/∣q∣U = mgd/\abs{q}: the bigger the charge, the smaller the 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 to balance.

  1. After one burst, balance requires U=1200 VU = 1200\,\mathrm{V}. Compute the new ∣q∣\abs{q}. Did the droplet gain or lose an electron?
  2. Successive balances are found at U=1200U = 1200, 800800, 600600, 480480 and 400 V400\,\mathrm{V}. Compute the five charges.
  3. Show that all five are integer multiples of one quantity; give its value.
  4. Between consecutive balances, by how much does the charge change? Interpret.
  5. Explain why this droplet can never hang at rest at U=700 VU = 700\,\mathrm{V}, however patiently one tunes.
  6. 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.

  1. In this chapter’s dictionary, what plays the roles of qq, EE and qEqE in the balance of Part II?
  2. Could a mass overhead hold the droplet up instead? Compute the field of a 1000 kg1000\,\mathrm{kg} lead sphere 1.0 m1.0\,\mathrm{m} above the droplet, the force on the droplet, and compare with its weight.
  3. 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?
  4. The droplet’s mass also changes in steps (one oil molecule: about 5×10−25 kg5 \times 10^{-25}\,\mathrm{kg}). Compare the relative jumps in mm and in ∣q∣\abs{q} when one electron lands, and conclude: why does this balance see the atoms of electricity but not the molecules of oil? Punchline: charge is quantized, and its atom is e=1.60×10−19 Ce = 1.60 \times 10^{-19}\,\mathrm{C}.
Solution

Solution of Problem 14.1.

1. Uniform, perpendicular to the plates, pointing down (from the ++ upper plate to the −- lower one); uniformity means the force on the droplet is the same wherever it drifts. 2. E=U/d=800/8.0×10−3=1.0×105 V/mE = 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 is opposite E→\vect E: upward. Only with the ++ plate on top does E→\vect E point down and the electric force fight the weight. 4. P=mg=4.90×10−15×9.81≈4.8×10−14 NP = mg = 4.90 \times 10^{-15} \times 9.81 \approx 4.8 \times 10^{-14}\,\mathrm{N}. 5. ∣q∣E=mg\abs{q}E = mg. 6. ∣q∣=mg/E=mgd/U\abs{q} = mg/E = mgd/U. 7. ∣q∣=4.81×10−14×8.0×10−3/800≈4.8×10−19 C\abs{q} = 4.81 \times 10^{-14} \times 8.0 \times 10^{-3} / 800 \approx 4.8 \times 10^{-19}\,\mathrm{C}. 8. n=4.8×10−19/1.60×10−19=3n = 4.8 \times 10^{-19} / 1.60 \times 10^{-19} = 3 excess electrons. 9. EE grows, so ∣q∣E>mg\abs{q}E > mg: the net force points up. 10. From ∣q∣ U/d=mg\abs{q}\,U/d = mg, U=mgd/∣q∣U = mgd/\abs{q} — inversely proportional to the charge. 11. ∣q∣=mgd/U=3.85×10−16/1200≈3.2×10−19 C=2e\abs{q} = mgd/U = 3.85 \times 10^{-16}/1200 \approx 3.2 \times 10^{-19}\,\mathrm{C} = 2e: down from 3e3e, the droplet lost one electron. 12. ∣q∣=3.85×10−16/U\abs{q} = 3.85 \times 10^{-16}/U: 3.2×10−19 C3.2 \times 10^{-19}\,\mathrm{C}, 4.8×10−19 C4.8 \times 10^{-19}\,\mathrm{C}, 6.4×10−19 C6.4 \times 10^{-19}\,\mathrm{C}, 8.0×10−19 C8.0 \times 10^{-19}\,\mathrm{C}, 9.6×10−19 C9.6 \times 10^{-19}\,\mathrm{C}. 13. They are 2e,3e,4e,5e,6e2e, 3e, 4e, 5e, 6e with e=1.60×10−19 Ce = 1.60 \times 10^{-19}\,\mathrm{C}. 14. Each step changes ∣q∣\abs{q} by exactly 1.6×10−19 C1.6 \times 10^{-19}\,\mathrm{C}: one electron lands at a time. 15. It would need ∣q∣=3.85×10−16/700≈5.5×10−19 C=3.4 e\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 e=1.60×10−19 Ce = 1.60 \times 10^{-19}\,\mathrm{C} — the quantity Millikan measured. 17. q↔mq \leftrightarrow m, E↔gE \leftrightarrow g, and ∣q∣E↔mg\abs{q}E \leftrightarrow mg: the weight itself. 18. gsphere=GM/d2=6.67×10−11×1000/1.02=6.7×10−8 N/kgg_{\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 on the droplet 4.90×10−15×6.7×10−8≈3.3×10−22 N4.90 \times 10^{-15} \times 6.7 \times 10^{-8} \approx 3.3 \times 10^{-22}\,\mathrm{N} — about 1.5×1081.5 \times 10^{8} times smaller than its weight. Hopeless. 19. Because GG is so small, only a planet-sized mass overhead could rival the Earth’s pull — and no mass repels. The enormous kk and the two signs of charge let a bench-top plate out-pull the planet in either direction. 20. One electron: Δ∣q∣/∣q∣=e/3e≈33%\Delta\abs{q}/\abs{q} = e/3e \approx 33\%; one oil molecule: Δm/m=5×10−25/4.90×10−15≈10−10\Delta m/m = 5 \times 10^{-25} / 4.90 \times 10^{-15} \approx 10^{-10}. The charge staircase moves the balancing voltage by hundreds of volts; the mass staircase is invisible at any voltage. The balance therefore resolves the atom of electricity — e=1.60×10−19 Ce = 1.60 \times 10^{-19}\,\mathrm{C} — while the molecules of oil blur into a continuum.