Classical-Mechanics · Unit 7 · Video 3 · Interactive Practice

Charge, Coulomb's Law, and Newton's Third Law

IKey Formulas

Formula	Name	What it says
$\vec{F}^{\,E}_{1,2} = k_e\,\dfrac{q_1 q_2}{r_{1,2}^{2}}\,\hat{r}_{1,2}$	Coulomb's Law	Electric force on charge 2 from charge 1
$k_e = 8.9875\times10^{9}\ \mathrm{N\,m^2/C^2}$	Coulomb constant	Sets the strength of the electric force
$Q = n\,e,\quad e = 1.602\times10^{-19}\ \mathrm{C}$	Charge quantization	Charge comes in whole multiples of $e$ ; total charge is conserved
$\hat{r}_{2,1} = -\hat{r}_{1,2}$	Third-law symmetry	Swapping labels only reverses the unit vector

Key Insight: Coulomb's Law has no minus sign out front — the direction lives entirely in the product $q_1 q_2$ : like charges ( $q_1 q_2 > 0$ ) repel, opposite charges ( $q_1 q_2 < 0$ ) attract.

IIQuantization and Conservation

Transferring electrons changes each object's charge in whole units of $e$ , while their total stays fixed.

IIICoulomb's Law: Sign and Separation

The sign of $q_1 q_2$ sets the force's direction; the separation $r$ sets its magnitude.

|q₁| (µC) 3 |q₂| (µC) 3

💡 Challenge: with the two signs fixed, find how far you must shrink $r$ to quadruple $|F|$ .

IVWhy Every Force Comes in an Equal-and-Opposite Pair

Relabelling $1 \leftrightarrow 2$ changes only one quantity — the unit vector reverses.

Step 1

Step 1 — Force on object 2

\vec{F}^{\,E}_{1,2} = k_e\,\frac{q_1 q_2}{r_{1,2}^{2}}\,\hat{r}_{1,2}

Step 2 — Swap labels $1 \leftrightarrow 2$ to get the force on object 1

\vec{F}^{\,E}_{2,1} = k_e\,\frac{q_2 q_1}{r_{2,1}^{2}}\,\hat{r}_{2,1}

Step 3 — Compare, term by term

q_2 q_1 = q_1 q_2

(unchanged)

r_{2,1} = r_{1,2}

(unchanged)

\hat{r}_{2,1} = -\hat{r}_{1,2}

(the only thing that flips)

Step 4 — Substitute

\vec{F}^{\,E}_{2,1} = k_e\,\frac{q_1 q_2}{r_{1,2}^{2}}\,\big(-\hat{r}_{1,2}\big) = -\,\vec{F}^{\,E}_{1,2}

Step 5 — Newton's Third Law

\vec{F}^{\,E}_{2,1} = -\,\vec{F}^{\,E}_{1,2}\ \checkmark

Equal magnitude, opposite direction.

💡 Gravitation runs on the identical argument: swap the labels, only $\hat{r}$ flips, so $\vec{F}^{\,G}_{2,1} = -\vec{F}^{\,G}_{1,2}$ as well.

VQuiz Questions

Problem 1 · Compute the Force

Given: $q_1 = +3.0\ \mu\mathrm{C}$ and $q_2 = +2.0\ \mu\mathrm{C}$ separated by $r = 0.20\ \mathrm{m}$ — find the magnitude of the electric force.

✅ Correct!

F = k_e q_1 q_2 / r^2 = (8.9875\times10^{9})(3.0\times10^{-6})(2.0\times10^{-6})/(0.20)^2 = 1.35\ \mathrm{N}

❌ Close — check the exponent on $r$ . You divided by

r

instead of

r^2

;

(0.20)^2 = 0.04

, not

0.20

❌ Not quite. Work in SI:

1\ \mu\mathrm{C} = 10^{-6}\ \mathrm{C}

and

r

in metres, then

F = k_e q_1 q_2 / r^2

Show solution

Convert to SI and apply Coulomb's Law:

F = k_e\,\frac{q_1 q_2}{r^2} = (8.9875\times10^{9})\,\frac{(3.0\times10^{-6})(2.0\times10^{-6})}{(0.20)^2}

= (8.9875\times10^{9})\,\frac{6.0\times10^{-12}}{0.04} = 1.35\ \mathrm{N}

The numerator $k_e q_1 q_2 = 5.39\times10^{-2}$ ; dividing by $r^2 = 0.04$ gives $\mathbf{1.35\ \mathrm{N}}$ . The most common slip is dividing by $r = 0.20$ instead of $r^2 = 0.04$ , which gives $0.27\ \mathrm{N}$ .

Problem 2 · Which Way Does It Point?

Given: two charges $q_1 = -4\ \mu\mathrm{C}$ and $q_2 = -4\ \mu\mathrm{C}$ . Is the force between them attractive or repulsive?

✅ Correct!

q_1 q_2 = (-4)(-4) = +16 > 0

, so the charges repel — like charges always push apart.

❌ Watch the signs. The product of two negatives is positive:

(-)(-) = +

, so

q_1 q_2 > 0

and the force is repulsive.

❌ Not quite. Like charges (same sign) repel; only opposite signs attract. The electric force can push or pull.

Show solution

The direction of the Coulomb force is carried by the sign of the product $q_1 q_2$ :

q_1 q_2 = (-4\ \mu\mathrm{C})(-4\ \mu\mathrm{C}) = +16\ (\mu\mathrm{C})^2 > 0

A positive product means the force points along $+\hat{r}_{1,2}$ — away from the other charge. The two charges repel. Unlike gravity (always attractive), electricity does both, and here two like charges repel.

Problem 3 · Inverse-Square Scaling

Given: two fixed charges feel a force of magnitude $F$ at separation $r$ . The separation is increased to $3r$ (charges unchanged). The new force magnitude is:

✅ Correct!

F \propto 1/r^2

, so tripling

r

divides the force by

3^2 = 9

❌ Almost. The law is inverse-square, not

1/r

: the denominator is

r^2

, so the factor is

3^2 = 9

, giving

F/9

❌ Not quite. Moving the charges apart weakens the force, and it scales as

1/r^2

Show solution

Holding the charges fixed, the force depends only on the separation through $1/r^2$ :

\frac{F_{\text{new}}}{F} = \frac{1/(3r)^2}{1/r^2} = \frac{r^2}{9r^2} = \frac{1}{9}

So $F_{\text{new}} = \dfrac{F}{9}$ . Every doubling of distance quarters the force; tripling divides it by nine.

Problem 4 · Equal and Opposite (Video Example 8.1)

Given: a small charge $q_1 = +1\ \mu\mathrm{C}$ sits a distance $r$ from a much larger charge $q_2 = +9\ \mu\mathrm{C}$ . Compare the magnitude of the electric force each one feels.

✅ Correct! Both forces equal

k_e q_1 q_2 / r^2

— the same product

q_1 q_2

— so they're equal in size and opposite in direction. Newton's Third Law.

❌ Not the individual charges. Each force depends on the same product

q_1 q_2

, not on either charge alone — the tiny charge feels exactly the same magnitude.

❌ Not quite. Coulomb's Law is symmetric in the two charges, so the two magnitudes are always equal (Newton's Third Law).

Show solution

Write both forces from Coulomb's Law:

\big|\vec{F}^{\,E}_{1,2}\big| = k_e\,\frac{q_1 q_2}{r^2}, \qquad \big|\vec{F}^{\,E}_{2,1}\big| = k_e\,\frac{q_2 q_1}{r^2}

Since $q_2 q_1 = q_1 q_2$ and the separation $r$ is the same for both, the two magnitudes are identical — regardless of how much bigger $q_2$ is. Swapping the labels flips only the unit vector ( $\hat{r}_{2,1} = -\hat{r}_{1,2}$ ), so the directions are opposite:

\vec{F}^{\,E}_{2,1} = -\,\vec{F}^{\,E}_{1,2}

This is exactly Newton's Third Law, built into the form of the law.

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