Coulomb’s Law

You know that charges attract or repel each other, but by how much? If you double the charge, does the force double? If you move charges farther apart, how quickly does the force fade? Coulomb’s law answers all of these questions with one clean mathematical statement. It is the starting equation of electrostatics, and every concept you will encounter later, from electric fields to Gauss’s law, builds directly on it.

How Coulomb Discovered the Law

In 1785, the French physicist Charles Augustin de Coulomb set out to measure the exact force between two charged objects. He used a device called a torsion balance (a sensitive instrument that measures tiny forces by how much they twist a thin fibre). Coulomb placed small charged metallic spheres at the ends of a suspended rod and measured how the rod twisted when another charged sphere was brought near.

There was a practical challenge: Coulomb had no way to know the exact charge on any sphere. He needed to compare forces for different charge amounts, but he could not read the charge off an instrument. His clever workaround relied on symmetry. If a sphere carrying some unknown charge $q$ is touched to an identical uncharged sphere, the charge distributes itself equally between the two (since the spheres are identical, neither has a reason to take more). After contact, each sphere holds $q/2$. (This relies on two implicit assumptions: the additivity of charges, meaning two half-charges add back up to the original charge $q$, and the conservation of charge, meaning no charge is created or destroyed during the transfer.) Repeating this trick produces $q/4$, $q/8$, and so on. Coulomb could not measure absolute charges, but he could produce charges in precisely known ratios: half, quarter, eighth.

With this technique, Coulomb carried out a systematic set of measurements:

• Varying distance with fixed charges: He kept the charges constant and changed the separation, recording the force at each distance.
• Varying charges at fixed distance: He used his charge-halving trick to change the charges in known ratios, kept the separation fixed, and measured the force again.

By comparing results across all these trials, Coulomb found that the force:

• is directly proportional to the product of the magnitudes of the two charges
• is inversely proportional to the square of the distance between them
• acts along the straight line connecting the two charges

This relationship is what we now call Coulomb’s law.

Coulomb himself had a remarkable background. He began his career as a military engineer in the West Indies before returning to Paris in 1776 and retiring to a small estate to pursue scientific research. It is worth noting that the inverse-square law for electric charges had been anticipated by Priestley and also by Cavendish before Coulomb, though Cavendish never published his results. Coulomb was the first to establish it through systematic measurement in 1785. He later went on to discover the same inverse-square relationship for forces between unlike and like magnetic poles as well.

The Scalar Form of Coulomb’s Law

For two point charges $q_1$ and $q_2$ separated by a distance $r$ in vacuum, the magnitude of the electrostatic force between them is:

$F = k \frac{|q_1 \, q_2|}{r^2} \qquad \text{(1.1)}$

Here $k$ is a proportionality constant, and the absolute-value signs ensure we are talking about the magnitude (a positive number). The direction (attraction or repulsion) is handled separately.

The constant $k$ and the permittivity of free space

Here is a subtle but important point: Coulomb discovered his law without knowing the actual magnitude of any charge. It works the other way round. Once the law and the constant $k$ are fixed, they provide a way to define the unit of charge. The choice of $k$ determines how large one coulomb is.

The value of $k$ in SI units, determined by experiment, is:

$k \approx 9 \times 10^9 \; \frac{\text{Nm}^2}{\text{C}^2}$

For mathematical convenience in later formulas, $k$ is written as:

$k = \frac{1}{4\pi\varepsilon_0}$

so that Coulomb’s law becomes:

$F = \frac{1}{4\pi\varepsilon_0} \frac{|q_1 \, q_2|}{r^2} \qquad \text{(1.2)}$

The quantity $\varepsilon_0$ is called the permittivity of free space (a measure of how easily electric field lines can pass through empty space). Its SI value is:

$\varepsilon_0 = 8.854 \times 10^{-12} \; \text{C}^2 \text{N}^{-1}\text{m}^{-2}$

How large is one coulomb?

To get a feel for the unit, set $q_1 = q_2 = 1$ C and $r = 1$ m in Eq. (1.2):

$F = 9 \times 10^9 \text{ N}$

Two charges of 1 C each, placed 1 m apart in vacuum, would repel each other with a force of nine billion newtons. That is roughly the weight of a million tonnes. Clearly, the coulomb is an enormous unit for electrostatic purposes. In practice, you will almost always work with much smaller charges, typically measured in microcoulombs ($1 \; \mu\text{C} = 10^{-6}$ C) or millicoulombs ($1 \; \text{mC} = 10^{-3}$ C).

Scope and validity

Coulomb’s original experiment used macroscopic metallic spheres, but the inverse-square law has since been tested at subatomic scales (down to $r \sim 10^{-10}$ m) and found to hold. It is one of the most precisely verified laws in all of physics. However, Coulomb’s law in this simple form applies only to charges in vacuum. When matter fills the space between the charges, the situation becomes more complex because the charged particles inside the matter also participate, and that topic is taken up in the next chapter.

The Vector Form: Direction Matters

Force is a vector; it has both magnitude and direction. So a complete statement of Coulomb’s law must specify direction, not just size.

Setting up the geometry

Place $q_1$ and $q_2$ at positions $\mathbf{r}_1$ and $\mathbf{r}_2$ measured from some origin O. Define two vectors that connect them:

• $\mathbf{r}_{21} = \mathbf{r}_2 - \mathbf{r}_1$ points from $q_1$ toward $q_2$
• $\mathbf{r}_{12} = \mathbf{r}_1 - \mathbf{r}_2 = -\mathbf{r}_{21}$ points from $q_2$ toward $q_1$

Both vectors have the same length: $r_{21} = r_{12} = r$ (the distance between the charges). Their unit vectors (vectors of length 1 pointing in the same direction) are:

$\hat{\mathbf{r}}_{21} = \frac{\mathbf{r}_{21}}{r_{21}}, \qquad \hat{\mathbf{r}}_{12} = \frac{\mathbf{r}_{12}}{r_{12}}$

Since $\mathbf{r}_{12} = -\mathbf{r}_{21}$, the unit vectors point in opposite directions: $\hat{\mathbf{r}}_{12} = -\hat{\mathbf{r}}_{21}$.

Fig 1.3: (a) Geometry of two point charges with position and separation vectors; (b) Repulsive force for like charges and attractive force for unlike charges

The full vector equation

The force on $q_2$ due to $q_1$ is:

$\mathbf{F}_{21} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 \, q_2}{r_{21}^2} \hat{\mathbf{r}}_{21} \qquad \text{(1.3)}$

Notice an important difference from the scalar form: there are no absolute-value signs around $q_1 q_2$. The signs of the charges are left in, and they automatically tell you the direction:

• Same-sign charges ($q_1 q_2 > 0$): The product is positive, so $\mathbf{F}_{21}$ points along $\hat{\mathbf{r}}_{21}$, which means away from $q_1$. This is repulsion.
• Opposite-sign charges ($q_1 q_2 < 0$): The product is negative, so $\mathbf{F}_{21}$ points opposite to $\hat{\mathbf{r}}_{21}$, which means toward $q_1$. This is attraction.

One equation handles both cases. You never need separate formulas for attraction and repulsion.

Newton’s third law is built in

To find the force on $q_1$ due to $q_2$, swap the subscripts 1 and 2 in Eq. (1.3):

$\mathbf{F}_{12} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 \, q_2}{r_{12}^2} \hat{\mathbf{r}}_{12}$

Since $\hat{\mathbf{r}}_{12} = -\hat{\mathbf{r}}_{21}$ and $r_{12} = r_{21}$, this simplifies to:

$\mathbf{F}_{12} = -\mathbf{F}_{21}$

The force on $q_1$ is equal in magnitude to the force on $q_2$ but points in the opposite direction. This is precisely Newton’s third law: for every action there is an equal and opposite reaction. Coulomb’s law is fully consistent with this fundamental principle of mechanics.

Electric Force vs. Gravitational Force: A Dramatic Comparison

Both Coulomb’s electrostatic force and Newton’s gravitational force follow an inverse-square dependence on distance. But how do their strengths compare? The answer is astonishing.

The ratio for an electron and a proton

The electrostatic force between an electron and a proton separated by distance $r$ is:

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

The gravitational force between the same pair is:

$F_G = G \frac{m_p \, m_e}{r^2}$

Taking the ratio, the $r^2$ cancels out:

$\left|\frac{F_e}{F_G}\right| = \frac{e^2}{4\pi\varepsilon_0 \, G \, m_p \, m_e}$

Substituting the known values ($e = 1.6 \times 10^{-19}$ C, $G = 6.67 \times 10^{-11}$ Nm$^2$/kg$^2$, $m_p = 1.67 \times 10^{-27}$ kg, $m_e = 9.11 \times 10^{-31}$ kg):

$\left|\frac{F_e}{F_G}\right| \approx 2.4 \times 10^{39}$

The electric force is roughly $10^{39}$ times stronger than gravity for an electron-proton pair. That number is so large it is hard to grasp: a 1 followed by 39 zeros. Gravity is completely irrelevant at the atomic scale.

The ratio for two protons

A similar calculation for two protons gives:

$\left|\frac{F_e}{F_G}\right| = \frac{e^2}{4\pi\varepsilon_0 \, G \, m_p^2} \approx 1.3 \times 10^{36}$

Still overwhelmingly in favour of the electric force. Note, however, that the two forces have different signs here: gravity pulls the protons together while the electric force pushes them apart. Inside a nucleus (where protons sit about $10^{-15}$ m apart), the Coulomb repulsion between two protons is roughly 230 N, whereas the gravitational attraction is a mere $1.9 \times 10^{-34}$ N. Clearly, it is not gravity that holds the nucleus together; the nuclear force (a much shorter-range but even stronger force) does that job.

Worked Example 1.3: Force and Acceleration at Atomic Distances

Problem: An electron and a proton are separated by $1 \; \text{\AA} = 10^{-10}$ m. Find (a) the electrostatic force between them, and (b) the acceleration each particle experiences. ($m_p = 1.67 \times 10^{-27}$ kg, $m_e = 9.11 \times 10^{-31}$ kg)

Solution:

Step 1: Calculate the magnitude of the electrostatic force.

$|F| = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r^2}$

Substitute $k = 8.987 \times 10^9$ Nm$^2$/C$^2$, $e = 1.6 \times 10^{-19}$ C, $r = 10^{-10}$ m:

$|F| = 8.987 \times 10^9 \times \frac{(1.6 \times 10^{-19})^2}{(10^{-10})^2}$

Evaluate the numerator: $(1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38}$ C$^2$.

Evaluate the denominator: $(10^{-10})^2 = 10^{-20}$ m$^2$.

$|F| = 8.987 \times 10^9 \times \frac{2.56 \times 10^{-38}}{10^{-20}} = 8.987 \times 10^9 \times 2.56 \times 10^{-18}$

$|F| \approx 2.3 \times 10^{-8} \text{ N}$

This is a tiny force by everyday standards, but for particles of atomic mass, it produces enormous accelerations.

Step 2: Find the electron’s acceleration.

Apply Newton’s second law: $a = F/m$.

$a_e = \frac{2.3 \times 10^{-8}}{9.11 \times 10^{-31}} \approx 2.5 \times 10^{22} \text{ m/s}^2$

Compare this with $g \approx 10$ m/s$^2$. The electric acceleration is about $10^{21}$ times larger than gravitational acceleration. This confirms that gravity has no noticeable effect on electrons.

Step 3: Find the proton’s acceleration.

$a_p = \frac{2.3 \times 10^{-8}}{1.67 \times 10^{-27}} \approx 1.4 \times 10^{19} \text{ m/s}^2$

The proton accelerates less than the electron (because it is about 1836 times heavier), but its acceleration is still fantastically large compared to $g$. Both particles experience the same force (Newton’s third law), but the lighter electron responds with a much larger acceleration.

Worked Example 1.4: Charge Redistribution and the Inverse-Square Test

Problem: A charged sphere A is suspended by a nylon thread, and another charged sphere B (held by an insulating handle) is brought to a distance of 10 cm from A. The repulsion deflects A noticeably. Now, uncharged spheres C and D (identical in size to A and B respectively) touch A and B, halving the charge on each. After C and D are removed, B is moved to 5 cm from A. What happens to the force on A?

Solution:

Step 1: Write the original force.

Let the charge on A be $q$ and on B be $q'$. At separation $r$:

$F = \frac{1}{4\pi\varepsilon_0} \frac{q \, q'}{r^2}$

Step 2: Determine the new charges after contact.

When identical uncharged sphere C touches A, charge spreads equally between them (by symmetry). A now carries $q/2$. Similarly, D touching B leaves B with $q'/2$.

Step 3: Calculate the force at the new distance.

The separation is halved to $r/2$. The new force is:

$F' = \frac{1}{4\pi\varepsilon_0} \frac{(q/2)(q'/2)}{(r/2)^2}$

Simplify the numerator: $(q/2)(q'/2) = qq'/4$.

Simplify the denominator: $(r/2)^2 = r^2/4$.

$F' = \frac{1}{4\pi\varepsilon_0} \frac{qq'/4}{r^2/4} = \frac{1}{4\pi\varepsilon_0} \frac{qq'}{r^2} = F$

The factor of $1/4$ from reducing the charges and the factor of $4$ from reducing the distance cancel each other out exactly. The force on A remains unchanged. This is a neat illustration of how the inverse-square dependence and the charge dependence interact.