EG1216 · Physics 2 — Electricity & Magnetism

Theme 1 · Electrostatics — charge & the electric field

Coulomb's law

A number for the push or pull between two charges — how it grows with charge and falls off with distance.

Source: Serway & Jewett, Physics for Scientists and Engineers, 7th ed., §23.3.

Before you start

What you need first

Topic 1 — electric charge — the two signs, and like repels, unlike attracts.
Charge in coulombs — and the prefixes $\mu\text{C}=10^{-6}$ C, $\text{nC}=10^{-9}$ C.
Adding arrows (vectors) in 1-D — left vs right (Physics 1).

What you'll be able to do

Find the force between two point charges with Coulomb's law.
Read off the direction (push or pull) from the signs.
Use the inverse-square rule: double the distance → quarter the force.
Add the forces from several charges (superposition).
Rearrange the law to find a distance or an unknown charge.

The idea

What decides how strong the force is?

From Topic 1 we know two charges push or pull. But how strong is that force? Charles-Augustin de Coulomb measured it and found it depends on just two things:

How big the two charges are — bigger charges give a bigger force.
How far apart they are — farther apart gives a much weaker force.

A point charge means a charge small enough that all of it sits at one point, so only the distance between the two points matters.

Two knobs set the force: the charges (top of the formula) and the distance (bottom). Nothing else.

The equation

Coulomb's law

$$ F = k\,\frac{|q_1\,q_2|}{r^{2}} $$

where:
Symbol	Meaning	SI unit
F	size of the force each charge feels from the other	N
q₁, q₂	the two point charges (use their sizes here)	C
r	distance between the two charges	m
k	Coulomb's constant — a fixed number for charges in air or vacuum, $=8.99\times10^{9}$	N·m²/C²

In words: the force is $k$ times the two charges multiplied together, divided by the distance squared. The bars $|\;|$ mean we use the sizes of the charges (drop the signs); the signs only decide the direction, which we handle next.

Is $k$ always the same? Yes — for charges in air or vacuum, always use $k=8.99\times10^{9}$. It changes only with the surroundings: inside a material (water, oil, plastic) the push is weaker. That weakening is the idea of a dielectric, in Theme 2.

What the formula is telling you

Top ($q_1 q_2$): double one charge and the force doubles. The charges drive the force straight up.

Bottom ($r^2$): the distance is squared. Move the charges twice as far apart and the force drops to $\left(\tfrac12\right)^2 = \tfrac14$ — a quarter, not a half.

This is the inverse-square rule. Distance weakens the force fast: $\times 2$ the distance → $\div 4$ the force; $\times 3$ → $\div 9$.

Twice the distance → a quarter of the force.

Which way does the force point?

The force always lies along the line joining the two charges. The signs tell you push or pull:

Same signs → apart (repel).
Opposite signs → together (attract).

The two charges always feel an equal force in opposite directions — that is Newton's third law. A big charge and a tiny charge still pull on each other with the same size of force.

Put the size into the formula; let the signs decide the arrow. Equal and opposite, always.

Forces act along the line, equal in size, opposite in direction.

📐 Worked example 1

Force between two charges

Two charges $q_1 = +2\ \mu\text{C}$ and $q_2 = +3\ \mu\text{C}$ sit $r = 0.50$ m apart in air. Find the force between them.

1Convert the charges to coulombs ($\mu\text{C}=10^{-6}$ C):

$$ q_1 = 2\times10^{-6}\ \text{C}, \qquad q_2 = 3\times10^{-6}\ \text{C} $$

2Put the sizes into Coulomb's law:

$$ F = (8.99\times10^{9})\,\frac{(2\times10^{-6})(3\times10^{-6})}{(0.50)^{2}} $$

3Work the numbers ($2\times3=6\times10^{-12}$, bottom $=0.25$):

$$ F = 8.99\times10^{9}\times\frac{6\times10^{-12}}{0.25} \approx 0.22\ \text{N} $$

Both charges are positive, so the $0.22\ \text{N}$ force is a push — each charge is shoved away from the other.

More than two charges? Add the forces

Coulomb's law is for a pair of charges. With several charges, use superposition: find the force from each other charge one at a time, then add them as arrows (with direction).

1Find the force from charge A using Coulomb's law.

2Find the force from charge B the same way.

3Add them, watching direction (here: left or right).

📐 Worked example 2

Net force from two other charges (a twist)

Three positive charges lie on a line: $q_1=+3\ \mu\text{C}$ at the left, $q_3=+1\ \mu\text{C}$ in the middle, $q_2=+2\ \mu\text{C}$ at the right. The spacing is $q_1\!-\!q_3 = 0.10$ m and $q_3\!-\!q_2 = 0.20$ m. Find the net force on the middle charge $q_3$.

$q_1$ pushes $q_3$ right; $q_2$ pushes it left.

1Force from $q_1$ on $q_3$ (both +, so it pushes $q_3$ to the right), $r=0.10$ m:

$$ F_{13} = (8.99\times10^{9})\,\frac{(3\times10^{-6})(1\times10^{-6})}{(0.10)^{2}} \approx 2.70\ \text{N} $$

2Force from $q_2$ on $q_3$ (pushes $q_3$ to the left), $r=0.20$ m:

$$ F_{23} = (8.99\times10^{9})\,\frac{(2\times10^{-6})(1\times10^{-6})}{(0.20)^{2}} \approx 0.45\ \text{N} $$

3Add as arrows (take right as positive):

$$ F_{\text{net}} = 2.70 - 0.45 = 2.25\ \text{N to the right} $$

The closer charge wins. Even though $q_2$ is bigger, it is twice as far, and $r^2$ makes distance count more than size here.

✏️ Try it yourself

Charges $q_1 = +6\ \mu\text{C}$ and $q_2 = -2\ \mu\text{C}$ are $r = 0.30$ m apart.

(a) Find the size of the force and say whether it is a push or a pull.
(b) Without recomputing, how far apart would they need to be for the force to drop to one-quarter of that value?

(a) Step 1. $F = (8.99\times10^{9})\dfrac{(6\times10^{-6})(2\times10^{-6})}{(0.30)^{2}}$ Step 2. $F \approx 1.20\ \text{N}$ — a pull (opposite signs attract). (b) A quarter of the force means double the distance: $r = 2\times0.30 = 0.60\ \text{m}$.

Helper — converting charge units

Charges are usually small, so they come with prefixes. Always convert to coulombs before substituting:

Written as	In coulombs
1 mC (millicoulomb)	$1\times10^{-3}$ C
1 μC (microcoulomb)	$1\times10^{-6}$ C
1 nC (nanocoulomb)	$1\times10^{-9}$ C
1 pC (picocoulomb)	$1\times10^{-12}$ C

The most common mistake in this topic is leaving a charge in $\mu\text{C}$. Convert to C first, every time.

Common mistakes

Mistake	Fix
Leaving charge in $\mu\text{C}$ or nC.	Convert to C first: $\mu\text{C}=10^{-6}$ C.
Forgetting to square $r$.	It is $r^2$ on the bottom — always.
Putting the $+/-$ signs into the formula.	Use sizes in the maths; the signs decide push or pull.
Adding forces as plain numbers.	Add them as arrows — mind left vs right.

Recap — the whole topic on one screen

Idea	What you own now
The law	$F = k\,\dfrac{\|q_1 q_2\|}{r^2}$, with $k = 8.99\times10^{9}$ N·m²/C².
Charges	Bigger charge → bigger force (force $\propto q_1 q_2$).
Distance	Inverse-square: $\times2$ distance → $\div4$ force.
Direction	Along the line; like → apart, unlike → together; equal & opposite.
Many charges	Superposition — add each pair's force as arrows.
Units	Convert charges to C first; $F$ comes out in N.

Next topic

The electric field & field lines

Coulomb's law needs two charges to talk about a force. Next we give a single charge a "reach" into the space around it — the electric field — so we can find the force on any charge we later place there.

→ Topic 3 · The electric field & field lines

Written as	In coulombs
1 mC (millicoulomb)	\(1\times10^{-3}\) C
1 μC (microcoulomb)	\(1\times10^{-6}\) C
1 nC (nanocoulomb)	\(1\times10^{-9}\) C
1 pC (picocoulomb)	\(1\times10^{-12}\) C

Mistake	Fix
Leaving charge in \(\mu\text{C}\) or nC.	Convert to C first: \(\mu\text{C}=10^{-6}\) C.
Forgetting to square \(r\).	It is \(r^2\) on the bottom — always.
Putting the \(+/-\) signs into the formula.	Use sizes in the maths; the signs decide push or pull.
Adding forces as plain numbers.	Add them as arrows — mind left vs right.

Idea	What you own now
The law	\(F = k\,\dfrac{\|q_1 q_2\|}{r^2}\), with \(k = 8.99\times10^{9}\) N·m²/C².
Charges	Bigger charge → bigger force (force \(\propto q_1 q_2\)).
Distance	Inverse-square: \(\times2\) distance → \(\div4\) force.
Direction	Along the line; like → apart, unlike → together; equal & opposite.
Many charges	Superposition — add each pair's force as arrows.
Units	Convert charges to C first; \(F\) comes out in N.