Coulomb's Law

Statement

The mutual electrostatic force between two point charges at rest is directly proportional to the product of the magnitudes of the charges, inversely proportional to the square of the distance between them, and directed along the line joining them.

$$F = k\dfrac{q_1q_2}{r^2} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2}$$

Here $k = \dfrac{1}{4\pi\varepsilon_0} = 9\times10^{9}\,\text{N m}^2\,\text{C}^{-2}$ and $\varepsilon_0 = 8.854\times10^{-12}\,\text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$ is the permittivity of free space.

Vector form — what a full-mark answer writes

Let $q_1$ and $q_2$ sit at position vectors $\vec{r}_1$ and $\vec{r}_2$. Write $\vec{r}_{21} = \vec{r}_2-\vec{r}_1$ for the vector from $1$ to $2$, and $\hat{r}_{21} = \dfrac{\vec{r}_{21}}{r_{21}}$ for the unit vector along it. Then the force on $q_2$ due to $q_1$ is

$$\vec{F}_{21} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r_{21}^{2}}\,\hat{r}_{21}$$

• The single equation covers both cases. If $q_1q_2>0$ (like charges) $\vec{F}_{21}$ is along $\hat{r}_{21}$, that is, away from $q_1$ — repulsion. If $q_1q_2<0$ (unlike charges) it is along $-\hat{r}_{21} = \hat{r}_{12}$ — attraction. So a pair of arrows drawn pointing towards each other tells you at once that $q_1q_2<0$.
• Interchanging $1$ and $2$ gives $\vec{F}_{12} = -\vec{F}_{21}$: Coulomb's law is consistent with Newton's third law.

Coulomb's experiment

• Coulomb used a torsion balance, a sensitive device for measuring small forces, later used by Cavendish for gravitation.
• The charges on his metallic spheres were unknown to begin with. He obtained known sub-multiples by touching a charged sphere of charge $q$ to an identical uncharged sphere: by symmetry each then carries $q/2$, and repeating gives $q/4$, $q/8$ and so on. (This assumes both additivity and conservation of charge.)
• He then varied the distance for a fixed pair of charges, and varied the charges at fixed distance, and compared the forces to arrive at the inverse-square product law.
• The law has been verified from the macroscopic scale down to subatomic distances of about $10^{-10}\,\text{m}$.

Definition of the coulomb that follows

Putting $q_1 = q_2 = 1\,\text{C}$ and $r = 1\,\text{m}$ into $F = 9\times10^{9}\dfrac{q_1q_2}{r^2}$ gives $F = 9\times10^{9}\,\text{N}$. So one coulomb is that charge which, placed at $1\,\text{m}$ from an equal charge in vacuum, experiences a repulsive force of $9\times10^{9}\,\text{N}$ — evidently far too big a unit for laboratory electrostatics.

Effect of a medium

In a medium of dielectric constant (relative permittivity) $K$,

$$F' = \dfrac{1}{4\pi\varepsilon_0 K}\dfrac{q_1q_2}{r^2} = \dfrac{F}{K}$$

Since $K>1$ for every insulator, inserting a plastic sheet between two charged balls reduces the force between them by the factor $K$. Two suspended identical charged balls therefore hang closer together when a dielectric sheet is placed between them.

Identical conductors brought into contact

Touching two identical conducting spheres carrying $q_1$ and $q_2$ leaves each with

$$q_1' = q_2' = \dfrac{q_1+q_2}{2}$$

• Example: $-Q$ and $+3Q$ give $\dfrac{-Q+3Q}{2} = +Q$ on each. Both are now positive, so the force after separation is repulsive.
• Example: $10\,\mu\text{C}$ and $-20\,\mu\text{C}$ give $-5\,\mu\text{C}$ each. If they are then placed at $r/2$, the new force is $F' = \dfrac{k(5)(5)\times4\times10^{-12}}{r^2}$ against the original $F = \dfrac{k(10)(20)\times10^{-12}}{r^2}$, so $\dfrac{F'}{F} = \dfrac{100}{200}$, giving $F' = \dfrac{F}{2}$.

Forces between multiple charges (NCERT 1.6, folded in here)

Coulomb's law alone cannot handle more than two charges. The extra experimental input is the principle of superposition:

The force on any one charge due to a number of other charges is the vector sum of the forces exerted on it by each of the others taken one at a time, and each of those pairwise forces is unaffected by the presence of the remaining charges.

For a system $q_1, q_2, \ldots, q_n$,

$$\vec{F}_1 = \vec{F}_{12}+\vec{F}_{13}+\cdots+\vec{F}_{1n} = \dfrac{q_1}{4\pi\varepsilon_0}\left[\dfrac{q_2}{r_{12}^2}\hat{r}_{12}+\dfrac{q_3}{r_{13}^2}\hat{r}_{13}+\cdots+\dfrac{q_n}{r_{1n}^2}\hat{r}_{1n}\right]$$

Each term is exactly the two-charge Coulomb force, written as though the other charges were absent. Add the terms by the parallelogram law, or resolve into components:

$$F_{\text{net}} = \sqrt{F_1^2+F_2^2+2F_1F_2\cos\alpha}$$

For two equal perpendicular contributions $F$ this gives $F_{\text{net}} = \sqrt{2}\,F$ at $45^\circ$ to each — the pattern behind the standard question with $-q$ on the $y$-axis, $+q$ on the $x$-axis and $q_0$ at the origin, whose answer is $\dfrac{\sqrt{2}\,kqq_0}{a^2}$.

Equilibrium of three collinear charges — the standard 3-mark derivation

Two identical charges $+q$ are fixed $2\,\text{m}$ apart at $A$ and $B$. A third charge $Q$ is placed on the line joining them so that the whole system is in equilibrium. Find its position and sign.

• Sign first. If $Q$ were positive, the net force on $Q$ itself could vanish, but each $+q$ would then feel repulsion from the other $+q$ and from $Q$ in the same direction, so the system as a whole could not be in equilibrium. Hence $Q$ must be negative.
• Position next. Let $AO = x$ and $OB = 2-x$. For $Q$ to be in equilibrium the two attractions must balance:

$$\dfrac{1}{4\pi\varepsilon_0}\dfrac{Qq}{x^2} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Qq}{(2-x)^2}$$

$$x^2 = (2-x)^2\ \Rightarrow\ x = 2-x\ \Rightarrow\ x = 1\,\text{m}$$

So $Q$ is negative and sits at the midpoint, $1\,\text{m}$ from either charge.

Standard results

• $F\propto\dfrac{1}{r^2}$: doubling the separation quarters the force; halving it quadruples the force.
• $F\propto q_1q_2$: doubling one charge doubles the force.
• A charge at a distance $r$ from a point on a square of side $d$: for the diagonal-type geometries CBSE sets, compute $r^2$ by Pythagoras before substituting. If one charge is at a corner and another at the midpoint of the opposite side, $r^2 = d^2+\dfrac{d^2}{4} = \dfrac{5d^2}{4}$, so that force is one fifth of the force at separation $d/2$.
• Coulomb force is much stronger than gravitation: $\dfrac{ke^2}{Gm_em_p}\approx2.4\times10^{39}$ for a proton and an electron.