Electricity-And-Magnetism · Unit 5 · Video 2 · Interactive Practice

Hydrogen Atom and Millikan's Oil Drop: Electric Forces by the Numbers

IKey Formulas

Formula	Name	What it says
$F_e = \dfrac{1}{4\pi\varepsilon_0}\dfrac{e^2}{r^2},\qquad E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{e}{r^2}$	Force and field in hydrogen	At $r = 5.3\times10^{-11}\ \text{m}$ : $F_e = 8.2\times10^{-8}\ \text{N}$ , $E = 5.1\times10^{11}\ \text{N/C}$
$\gamma = \dfrac{F_e}{F_g} = \dfrac{e^2/(4\pi\varepsilon_0)}{G\,m_p m_e} \approx 2.2\times10^{39}$	Electric over gravitational	Both forces fall as $1/r^2$ , so $r$ cancels and the ratio is the same at every distance
$M\vec{\mathbf{g}} + Q\vec{\mathbf{E}} = \vec{\mathbf{0}}\ \Rightarrow\ Q = -\dfrac{Mg}{E_y}$	Millikan balance	Field $-E_y\hat{\mathbf{j}}$ (down), force up, so the charge is negative
$M = \rho_{\text{oil}}\cdot\tfrac{4}{3}\pi R^3,\qquad n = \|Q\|/e$	Drop mass and electron count	Measured charges come out as whole multiples of $e$

Key Insight: The electric force is $10^{39}$ times stronger than gravity, yet matter is neutral to about one part in $10^{24}$ , so the attractions and repulsions between planets cancel and gravity, which has only one sign, steers them.

IIVisualization 1 — Thirty-Nine Decades at Every Distance

On a log scale both forces fall with slope $-2$ , so the gap between them never changes.

IIIVisualization 2 — Balance the Drop, Count the Electrons

A drop of mass $1.57\times10^{-14}\ \text{kg}$ hangs still only when $|Q|E_y = Mg$ , and that fixes $Q$ .

Drop Field magnitude E_y (×10⁵ N/C) 1.00

IVQuiz Questions

Problem 1 · The Proton's Field Farther Out

Given: $e = 1.6\times10^{-19}\ \text{C}$ and $\frac{1}{4\pi\varepsilon_0} = 9.0\times10^{9}\ \text{N}\,\text{m}^2/\text{C}^2$ . Find the magnitude of the proton's electric field at $r = 1.06\times10^{-10}\ \text{m}$ , twice the Bohr radius.

✅ Correct! Doubling

r

quarters the field:

5.1\times10^{11}/4 \approx 1.3\times10^{11}\ \text{N/C}

❌ Not quite. That halves the field; it falls as

1/r^2

, so doubling

r

divides it by

4

❌ Not quite.

e^2/(4\pi\varepsilon_0 r^2)

is the force on the electron in newtons; the field carries

e

to the first power.

❌ Not quite.

5.1\times10^{11}\ \text{N/C}

is the field at the Bohr radius itself.

❌ Not quite. Use

E = \frac{1}{4\pi\varepsilon_0}\frac{e}{r^2}

Show solution

E = \frac{(9.0\times10^{9})(1.6\times10^{-19})}{(1.06\times10^{-10})^2} = \frac{1.44\times10^{-9}}{1.124\times10^{-20}} = 1.3\times10^{11}\ \text{N/C},

pointing radially away from the proton.

Problem 2 · The Ratio at a New Distance

Given: at the Bohr radius $F_e/F_g = 2.2\times10^{39}$ for the proton and electron. Find $F_e/F_g$ when they are twice as far apart.

✅ Correct! Both forces scale as

1/r^2

, so the

r^2

cancels in the ratio.

❌ Not quite. The electric force does drop, but the gravitational force drops by the same factor of

4

❌ Not quite. Neither force grows with distance; both fall by

4

, leaving the ratio unchanged.

❌ Not quite. Write out

F_e/F_g

and look for

r

Show solution

\gamma = \frac{\dfrac{1}{4\pi\varepsilon_0}\dfrac{e^2}{r^2}}{\dfrac{G m_p m_e}{r^2}} = \frac{e^2/(4\pi\varepsilon_0)}{G m_p m_e},

which contains no $r$ , so the ratio is still $2.2\times10^{39}$ .

Problem 3 · Counting Electrons on a Smaller Drop

Given: An oil drop of radius $R = 1.0\times10^{-6}\ \text{m}$ and density $900\ \text{kg/m}^3$ hangs still in a downward field of magnitude $E_y = 7.7\times10^{4}\ \text{N/C}$ . Use $g = 9.8\ \text{m/s}^2$ , $e = 1.6\times10^{-19}\ \text{C}$ . Find the number of excess electrons.

✅ Correct!

|Q| = 4.8\times10^{-19}\ \text{C} = 3.0\,e

❌ Not quite. The sphere's volume is

\tfrac{4}{3}\pi R^3

; leaving out the

\tfrac{1}{3}

triples the mass and the count.

❌ Not quite. That count comes from a radius of

2.0\times10^{-6}\ \text{m}

; the given value is already the radius, and the mass grows as

R^3

❌ Not quite. Five electrons was the video's drop; this one is smaller and its field is different.

❌ Not quite. Find

M

, then

|Q| = Mg/E_y

, then divide by

e

Show solution

M = 900\cdot\tfrac{4}{3}\pi(1.0\times10^{-6})^3 = 3.77\times10^{-15}\ \text{kg},\qquad Mg = 3.69\times10^{-14}\ \text{N}.

|Q| = \frac{Mg}{E_y} = \frac{3.69\times10^{-14}}{7.7\times10^{4}} = 4.8\times10^{-19}\ \text{C} = 3.0\,e.

The force is up, against the downward field, so $Q = -3e$ : three excess electrons.

Problem 4 · Reversing the Field

Given: A charged drop hangs still in a uniform field that points up. The field is then reversed, with the same magnitude. Find the drop's acceleration just after the reversal (ignore air drag).

✅ Correct! The electric force flips from

Mg

up to

Mg

down and now adds to gravity.

❌ Not quite. The electric force reverses with the field, so the two forces no longer cancel.

❌ Not quite. The electric force does not vanish; it reverses, so it adds another

Mg

downward.

❌ Not quite. Before the reversal the force along the upward field held the drop up, so

Q > 0

; after it,

Q\vec{\mathbf{E}}

points down.

❌ Not quite. Find the sign of

Q

first, then the new net force.

Show solution

Balance with an upward field: $QE = Mg$ with the force along $\vec{\mathbf{E}}$ , so $Q > 0$ . After the reversal the electric force is $Mg$ downward, and the net force is

\vec{\mathbf{F}} = -Mg\,\hat{\mathbf{j}} - Mg\,\hat{\mathbf{j}} = -2Mg\,\hat{\mathbf{j}}\quad\Rightarrow\quad \vec{\mathbf{a}} = -2g\,\hat{\mathbf{j}}.

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