Electricity-And-Magnetism · Unit 3 · Video 1 · Interactive Practice

Coulomb's Law: Force Between Two Point Charges

IKey Formulas

Formula	Name	What it says
$Q = ne, \quad n = 0, \pm 1, \pm 2, \ldots$ $e = 1.602 \times 10^{-19}\ \text{C}$	Charge is quantized (and conserved)	Every free charge is a whole multiple of $e$ ; in a closed system the total charge never changes, it is only moved
$\vec{F}_{12} = k_e \dfrac{q_1 q_2}{r^2}\,\hat{r}, \quad \hat{r} = \dfrac{\vec{r}}{r}$	Coulomb's law	Force on $q_2$ due to $q_1$ ; $\hat{r}$ points from $q_1$ toward $q_2$ , and the sign of $q_1 q_2$ sets the direction
$\vec{F}_{21} = -\vec{F}_{12}, \quad \|\vec{F}_{12}\| = \|\vec{F}_{21}\| = k_e \dfrac{\|q_1 q_2\|}{r^2}$	Newton's third law	Equal magnitudes and opposite directions, even when $\|q_1\| \neq \|q_2\|$
$k_e = \dfrac{1}{4\pi\varepsilon_0} = 10^{-7} c^2 \approx 8.99 \times 10^{9}\ \text{N} \cdot \text{m}^2/\text{C}^2$	Coulomb constant	$\varepsilon_0 = 1/(\mu_0 c^2) \approx 8.854 \times 10^{-12}\ \text{C}^2/(\text{N} \cdot \text{m}^2)$ , the permittivity of free space

Key Insight: No separate rule is needed for like and unlike charges: $q_1 q_2 > 0$ makes $\vec{F}_{12}$ point along $\hat{r}$ (repulsion), $q_1 q_2 < 0$ makes it point against $\hat{r}$ (attraction).

IIVisualization 1 — Rubbing Moves Charge, It Never Makes It

Rubbing glass with silk moves whole electrons, so each object ends at a multiple of $e$ and the total stays zero.

Electrons moved, n 0

💡 Quarks inside the proton carry $+\tfrac{2}{3}e$ and $-\tfrac{1}{3}e$ , but no quark has ever been isolated, so every free charge is still a whole multiple of $e$ .

IIIVisualization 2 — Equal and Opposite, Whatever the Charges

The sign of $q_1 q_2$ alone sets the direction of $\vec{F}_{12}$ , and $\vec{F}_{21} = -\vec{F}_{12}$ even for unequal charges.

q₁ (μC) +5 q₂ (μC) +1

IVVisualization 3 — The Inverse-Square Law

Because $|\vec{F}| \propto 1/r^2$ , doubling the separation cuts the force to one quarter and tripling it to one ninth.

Separation r / r₀ 1.00

VQuiz Questions

Problem 1 · Scaling the Force

Given: Two point charges a distance $r$ apart exert a force of magnitude $0.36\ \text{N}$ on each other. One charge is doubled and the separation is tripled to $3r$ . Find the new magnitude of the force.

✅ Correct! Doubling one charge doubles

q_1 q_2

, and tripling

r

divides the force by

9

0.36 \times \tfrac{2}{9} = 0.08\ \text{N}

❌ Close, but…

0.36/9 = 0.04\ \text{N}

accounts for the distance only. The doubled charge doubles the product

q_1 q_2

, so the force doubles as well.

❌ Not quite. That treats the force as

1/r

. Coulomb's law is inverse square: tripling

r

divides the force by

3^2 = 9

, not by

3

❌ Not quite.

0.36 \times \tfrac{4}{9}

would mean both charges were doubled. Only one charge changes, so

q_1 q_2

changes by a factor of

2

, not

4

Show solution

The magnitude is $F = k_e |q_1 q_2|/r^2$ . Replace $q_1 \to 2q_1$ and $r \to 3r$ :

F' = k_e \frac{|2q_1 \, q_2|}{(3r)^2} = \frac{2}{9}\, k_e \frac{|q_1 q_2|}{r^2} = \frac{2}{9}\,F

F' = \frac{2}{9} \times 0.36\ \text{N} = 0.08\ \text{N}

The charge factor enters linearly; the distance factor enters squared.

Problem 2 · Does the Bigger Charge Push Harder?

Given: $q_1 = +5.0\ \mu\text{C}$ sits at the origin and $q_2 = +1.0\ \mu\text{C}$ sits $0.30\ \text{m}$ away on the $+x$ axis. The force on $q_2$ is $\vec{F}_{12} = 0.50\ \text{N}$ in the $+x$ direction. Find the force $\vec{F}_{21}$ on $q_1$ .

✅ Correct!

\vec{F}_{21} = -\vec{F}_{12}

: same magnitude, opposite direction, so

q_1

is pushed away from

q_2

toward

-x

❌ Not quite. The larger charge does not push harder. Both magnitudes equal

k_e |q_1 q_2|/r^2

, and the product

|q_1 q_2|

is the same number for both forces.

❌ Close, but check the direction. The magnitude is right, but the charges repel, so the force on

q_1

points away from

q_2

, toward

-x

\vec{F}_{21} = -\vec{F}_{12}

Show solution

By Newton's third law, $\vec{F}_{21} = -\vec{F}_{12}$ . The magnitudes are both

|\vec{F}_{12}| = |\vec{F}_{21}| = k_e \frac{|q_1 q_2|}{r^2} = (8.99 \times 10^{9}) \frac{(5.0 \times 10^{-6})(1.0 \times 10^{-6})}{(0.30)^2} \approx 0.50\ \text{N}

Since $\vec{F}_{12}$ points along $+x$ , $\vec{F}_{21} = 0.50\ \text{N}$ along $-x$ . The two like charges push each other apart with equal strength, regardless of which charge is larger.

Problem 3 · The Force as a Vector

Given: $q_1 = -4.0\ \text{nC}$ at the origin and $q_2 = +2.0\ \text{nC}$ at $(0.30\ \text{m},\ 0.40\ \text{m})$ . Use $k_e = 8.99 \times 10^{9}\ \text{N} \cdot \text{m}^2/\text{C}^2$ . Find the magnitude and the components of $\vec{F}_{12}$ , the force on $q_2$ .

What is the magnitude of the force?

What is $\vec{F}_{12}$ in components?

✅ Correct!

q_1 q_2 < 0

, so

\vec{F}_{12} = -(2.88 \times 10^{-7}\ \text{N})\,\hat{r}

points from

q_2

back toward

q_1

: attraction.

❌ Not quite. You divided by

r = 0.50\ \text{m}

instead of

r^2 = 0.25\ \text{m}^2

. Coulomb's law is inverse square.

❌ Not quite. The separation is not

0.30 + 0.40 = 0.70\ \text{m}

. It is the length of

\vec{r} = (0.30, 0.40)\ \text{m}

r = \sqrt{0.30^2 + 0.40^2} = 0.50\ \text{m}

❌ Not quite. A nanocoulomb is

10^{-9}\ \text{C}

, not

10^{-6}\ \text{C}

; using micro for nano inflates the force by

10^{6}

❌ Check the magnitude. Compute

r

from both components, then use

k_e |q_1 q_2|/r^2

❌ Check the sign. That vector points along

\hat{r}

, away from

q_1

, which is repulsion. Here

q_1 q_2 < 0

, so

\vec{F}_{12}

points against

\hat{r}

❌ Check the components.

\hat{r} = (0.30, 0.40)/0.50 = (0.60, 0.80)

: the

y

-component is the larger one, so the components are swapped.

❌ Check the direction. Coulomb forces act along the line joining the charges, so

\vec{F}_{12}

must be a multiple of

\hat{r} = (0.60, 0.80)

Show solution

Step 1: separation and unit vector. $\vec{r} = (0.30, 0.40)\ \text{m}$ points from $q_1$ to $q_2$ .

r = \sqrt{0.30^2 + 0.40^2} = 0.50\ \text{m}, \qquad \hat{r} = \frac{\vec{r}}{r} = (0.60,\ 0.80)

Step 2: the scalar factor, keeping the sign of $q_1 q_2$ :

k_e \frac{q_1 q_2}{r^2} = (8.99 \times 10^{9}) \frac{(-4.0 \times 10^{-9})(2.0 \times 10^{-9})}{0.25} = -2.88 \times 10^{-7}\ \text{N}

Step 3: multiply by $\hat{r}$ .

\vec{F}_{12} = (-2.88 \times 10^{-7}\ \text{N})(0.60,\ 0.80) = (-1.73,\ -2.30) \times 10^{-7}\ \text{N}

The magnitude is $2.88 \times 10^{-7}\ \text{N}$ , and the negative factor turns $\vec{F}_{12}$ around to point toward $q_1$ : opposite charges attract.

Problem 4 · Glass, Silk and the Force Between Them

Given: Rubbing moves $2.0 \times 10^{10}$ electrons from a glass rod to a silk cloth, both neutral beforehand. Treat the rod and the cloth as point charges $5.0\ \text{cm}$ apart, with $e = 1.602 \times 10^{-19}\ \text{C}$ and $k_e = 8.99 \times 10^{9}\ \text{N} \cdot \text{m}^2/\text{C}^2$ . Find the force between them.

✅ Correct! The glass ends at

+Q

and the silk at

-Q

with

Q = 3.20 \times 10^{-9}\ \text{C}

; opposite charges attract with

k_e Q^2/r^2 = 3.69 \times 10^{-5}\ \text{N}

❌ Close, but check the direction. The glass loses electrons (

+Q

) and the silk gains them (

-Q

). Then

q_1 q_2 < 0

, which means attraction.

❌ Not quite. Conservation keeps the total at zero, but each object is charged:

+Q

and

-Q

sit

5.0\ \text{cm}

apart and attract each other.

❌ Not quite. That uses

Q/2

on each object. Every electron that leaves the glass lands on the silk, so the glass gains

+Q

and the silk gets the full

-Q

Show solution

Step 1: charge moved (a whole number of electrons, $Q = ne$ ):

Q = (2.0 \times 10^{10})(1.602 \times 10^{-19}\ \text{C}) = 3.20 \times 10^{-9}\ \text{C}

The glass is left with $+Q$ and the silk carries $-Q$ ; the total is still $0$ .

Step 2: Coulomb's law with $r = 0.050\ \text{m}$ :

F = k_e \frac{|(+Q)(-Q)|}{r^2} = (8.99 \times 10^{9}) \frac{(3.20 \times 10^{-9})^2}{(0.050)^2} \approx 3.69 \times 10^{-5}\ \text{N}

Since $q_1 q_2 = -Q^2 < 0$ , the force is attractive.

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