---
title: Coulomb's Law
module: Electric Fields
moduleNumber: 1
lessonNumber: 2
order: 102
summary: >
  Two charges at rest push or pull along the line joining them, and the whole of
  electrostatics is assembled by adding up such pairs. We measure that force — its
  inverse-square falloff, its linear dependence on each charge, the sign that says
  attract or repel — and write it as a vector so direction survives superposition.
  We work the magnitude and component forms on real numbers, check them against
  limiting cases and dimensions, and fix the point-charge approximation to source
  sizes small against every separation.
topics: [Electric Fields]
draft: false
sources:
  - book: Tipler & Mosca
    ref: "Ch. 21 — The Electric Field I; §21-3 Coulomb's Law"
---

## The two-charge model

Coulomb's law concerns the force between **two stationary point charges**. A point
charge has a specified charge $q$ and a position, while its physical size is small
compared with every separation used in the calculation. The term stationary refers
to the electrostatic configuration at the instant of analysis. Charged beads on
insulating stands, small charged spheres held by supports, and particles observed
over a sufficiently short interval can be modelled this way.

The model specifies three quantities:

- **Source charge** $q_1$: the charge whose electric interaction is being evaluated.
- **Test body** $q_2$: the body on which the force is reported.
- **Separation** $r$: the distance between their assigned positions.

The force magnitude in free space is

$$
F=\frac{1}{4\pi\varepsilon_0}\frac{|q_1q_2|}{r^2}
=k\frac{|q_1q_2|}{r^2},
\qquad
k=8.988\times10^9\ \mathrm{N\,m^2\,C^{-2}}.
$$

Here $\varepsilon_0$ is the permittivity of free space and
$k=1/(4\pi\varepsilon_0)$. The absolute value belongs only in the **magnitude**
formula. Charge signs determine the force direction and must be retained when a
vector or a signed component is required.

The historical torsion-balance arrangement used charged spheres much smaller than their
separation. A restoring torque measured their displacement, and the apparatus settled
before each reading. These conditions restrict the measurement to the point-charge,
electrostatic regime rather than arbitrary extended moving charge distributions.

$$
% caption: The torsion balance behind Coulomb's measurement. A charged sphere
% $q_2$ on a light arm hangs from a fibre; the force from a fixed sphere $q_1$ a
% distance $r$ away twists the fibre through a measured angle, and small spheres
% keep the point-charge approximation valid.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, line width=1.1pt] (-1.5,2.4) -- (1.5,2.4);
  \foreach \x in {-1.3,-0.9,-0.5,-0.1,0.3,0.7,1.1}{\draw[black] (\x,2.4) -- (\x-0.22,2.62);}
  \draw[black, thick] (0,2.4) -- (0,0.6);
  \node[black, anchor=west] at (0.12,1.6) {wire};
  \draw[black, thick] (0,0.6) -- (2.4,0.2);
  \filldraw[draw=black, fill=black!8] (2.4,0.2) circle (0.2);
  \node[black!70] at (2.4,0.2) {$q_2$};
  \filldraw[draw=black, fill=black!8] (4.7,0.2) circle (0.2);
  \node[black!70] at (4.7,0.2) {$q_1$};
  \draw[<->, black] (2.4,-0.5) -- (4.7,-0.5) node[midway, below] {$r$};
  \draw[->, acc, thick] (2.15,0.2) -- (1.25,0.2) node[midway, below] {$F$};
\end{tikzpicture}
$$

Each charge exerts a force on the other. Its magnitude depends on $q_1$, $q_2$, and
the separation $r$. A third charge adds a further force through superposition.

**Magnitude and inverse-square scaling.**

For fixed charges, Coulomb's law gives $F\propto r^{-2}$. A separation doubled from
$r$ to $2r$ reduces the magnitude to one quarter:

$$
\frac{F(2r)}{F(r)}
=\frac{k|q_1q_2|/(2r)^2}{k|q_1q_2|/r^2}
=\frac14.
$$

Tripling the separation gives $F(3r)=F(r)/9$. The exponent applies to the
**distance**, not to either charge. In experimental data,
a factor-of-four reduction may arise from doubling the distance, whereas halving
one charge gives only a factor-of-two reduction.

$$
% caption: Inverse-square falloff for one fixed charge pair. At separations $r$,
% $2r$, $3r$ the force is $F$, $F/4$, $F/9$; the curve flattens at large separation
% but keeps the same power law.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (5.3,0) node[right] {$r$};
  \draw[->, black] (0,0) -- (0,3.1) node[above] {$F$};
  \draw[acc, very thick] plot[domain=0.62:4.8, samples=120] ({\x},{2.7/(\x*\x)});
  \foreach \x/\lab in {1/{r},2/{2r},3/{3r}}{
    \draw[black, dashed] (\x,0) -- (\x,{2.7/(\x*\x)});
    \filldraw[acc] (\x,{2.7/(\x*\x)}) circle (0.045);
    \node[black, below] at (\x,0) {\lab};
  }
  \node[acc, anchor=west] at (3.2,1.55) {inverse square};
\end{tikzpicture}
$$

For fixed charges, the curve gives force as a function of prescribed separation. A
freely moving pair has changing separation and kinetic energy, so the force varies
along its motion. The static graph shows the geometry dependence alone.

Charge scaling is linear in the magnitude of either charge:

$$
F(\lambda q_1,q_2,r)=|\lambda|F(q_1,q_2,r).
$$

Thus a sphere carrying $4Q$ produces four times the force magnitude on a fixed test
charge at the same location as a sphere carrying $Q$. Reversing the sign of a charge
does not alter the magnitude; it reverses the direction of the force on the other
charge.

The force magnitude becomes singular as $r$ approaches zero in the mathematical
point-charge model. Physical spheres have finite size, quantum structure, and material
response; their centres cannot reach zero separation while retaining the point-source
assumptions. The divergence marks the end of the model range.

**Charge signs and force direction.**

Two charges of the same sign repel. Each charge is pushed along the line joining the
pair and away from the other charge. Two charges of opposite sign attract. Each
charge is pulled along that same line and toward the other charge. The direction
changes with the charge product $q_1q_2$:

$$
q_1q_2>0 \Longrightarrow \text{repulsion},
\qquad
q_1q_2<0 \Longrightarrow \text{attraction}.
$$

The two arrows in a force diagram must be assigned to different bodies. Drawing
opposite arrows does not mean that the forces cancel on one body. The force on
$q_2$ and the force on $q_1$ cancel only when the **two-charge system** is treated
as a whole; each individual charge still accelerates if no external support acts.

$$
% caption: The sign of the charge product sets the direction. Like charges
% ($q_1q_2>0$) repel, each force pointing away from the other; unlike charges
% ($q_1q_2<0$) attract, each force pointing inward. The magnitude $k|q_1q_2|/r^2$
% is the same in both rows.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % repel row
  \filldraw[draw=black, fill=black!8] (1.3,1.1) circle (0.24);
  \node[black!70] at (1.3,1.1) {+};
  \filldraw[draw=black, fill=black!8] (4.7,1.1) circle (0.24);
  \node[black!70] at (4.7,1.1) {+};
  \draw[->, acc, very thick] (1.02,1.1) -- (0.15,1.1);
  \draw[->, acc, very thick] (4.98,1.1) -- (5.85,1.1);
  \node[black, anchor=west] at (6.15,1.1) {repel};
  \draw[<->, black] (1.3,0.35) -- (4.7,0.35) node[midway, above] {$r$};
  % attract row
  \filldraw[draw=black, fill=black!8] (1.3,-0.9) circle (0.24);
  \node[black!70] at (1.3,-0.9) {+};
  \filldraw[draw=black, fill=black!8] (4.7,-0.9) circle (0.26);
  \node[black!70, font=\scriptsize] at (4.7,-0.9) {neg};
  \draw[->, acc, very thick] (1.54,-0.9) -- (2.5,-0.9);
  \draw[->, acc, very thick] (4.46,-0.9) -- (3.5,-0.9);
  \node[black, anchor=west] at (6.15,-0.9) {attract};
\end{tikzpicture}
$$

Unlike-charge arrows point inward. The magnitude remains
$k|q_1q_2|/r^2$; an attraction is not a negative magnitude. A signed component can
be negative in a chosen coordinate system, while the physical force magnitude is
always nonnegative.

Newton's third law follows from the symmetry of the pair force:

$$
\vec F_{1\to2}=-\vec F_{2\to1}.
$$

The notation reads “force exerted by 1 on 2.” Switching the subscripts changes both
the body receiving the force and the direction of the separation vector. A common
calculation error uses the correct magnitude but draws both arrows on the test
charge. Keeping the receiving-body subscript explicit prevents that error.

## Vector form and coordinate geometry

Place source charge $q_1$ at position $\vec r_1$ and receiving charge $q_2$ at
position $\vec r_2$. The displacement from source to receiver is

$$
\vec r_{12}=\vec r_2-\vec r_1,
\qquad
r_{12}=|\vec r_{12}|,
\qquad
\hat r_{12}=\frac{\vec r_{12}}{r_{12}}.
$$

The vector force on charge 2 due to charge 1 is

$$

\vec F_{1\to2}
=kq_1q_2\frac{\vec r_{12}}{r_{12}^3}
=k\frac{q_1q_2}{r_{12}^2}\hat r_{12}.

$$

The cubic power in the first expression is required because
$\vec r_{12}$ already carries one factor of distance. Taking its magnitude gives
$|\vec r_{12}|/r_{12}^3=1/r_{12}^2$, recovering the scalar inverse-square
law. It combines direction and sign without separate sign cases. If $q_1q_2>0$, the
coefficient is positive and the force follows $\hat r_{12}$. If
$q_1q_2<0$, the coefficient is negative and the force points opposite to
$\hat r_{12}$.

$$
% caption: The separation vector points from source to receiver. Starting at
% $q_1$ and ending at $q_2$, $\vec r_{12}=\vec r_2-\vec r_1$; its components are the
% coordinate differences, drawn here as the dashed horizontal and vertical legs.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (-0.3,0) -- (5.3,0) node[right] {$x$};
  \draw[->, black] (0,-0.3) -- (0,3.4) node[above] {$y$};
  \filldraw[draw=black, fill=black!8] (1.05,0.8) circle (0.19);
  \node[black!70, anchor=north east] at (0.88,0.66) {$q_1$};
  \filldraw[draw=black, fill=black!8] (4.1,2.6) circle (0.17);
  \node[black!70, anchor=south west] at (4.26,2.72) {$q_2$};
  \draw[->, acc, very thick] (1.24,0.91) -- (3.9,2.48) node[midway, above left] {$r_{12}$};
  \draw[black, dashed] (1.05,0.8) -- (4.1,0.8) -- (4.1,2.6);
  \node[black, anchor=north] at (2.55,0.74) {dx};
  \node[black, anchor=west] at (4.2,1.7) {dy};
\end{tikzpicture}
$$

In Cartesian components,

$$
\vec r_{12}=(x_2-x_1)\hat\imath
 +(y_2-y_1)\hat\jmath
 +(z_2-z_1)\hat k,
$$

and each force component follows from the same common denominator:

$$
F_x=kq_1q_2\frac{x_2-x_1}{r_{12}^3},
\quad
F_y=kq_1q_2\frac{y_2-y_1}{r_{12}^3},
\quad
F_z=kq_1q_2\frac{z_2-z_1}{r_{12}^3}.
$$

Component form retains the coordinate convention even in a two-charge problem. A
source located left and below the
receiving charge has $\Delta x>0$ and $\Delta y>0$. For equal sign charges, both
force components are positive; for unlike charges, both reverse sign.

The angle of a resultant component pair can be computed with the shared operator
defined for the notes:

$$
\theta=\atanTwo(F_y,F_x).
$$

The two-argument arctangent identifies the correct quadrant from the signs of both
components. A single ratio $F_y/F_x$ loses that quadrant information when $F_x$ is
negative or when the force is vertical.

> **Worked example (superposition of two forces).** A charge $q_0=+1.0\ \mathrm{nC}$
> sits at the origin. A second charge $q_A=+3.0\ \mathrm{nC}$ lies on the $x$ axis at
> $x=0.20\ \mathrm m$, and a third $q_B=+3.0\ \mathrm{nC}$ lies on the $y$ axis at
> $y=0.20\ \mathrm m$. Each pushes $q_0$ away, so $q_A$ drives it in the $-x$ direction
> and $q_B$ in the $-y$ direction. Both magnitudes are equal:
>
> $$
> F_A=F_B=k\frac{q_0 q}{r^2}
> =(8.988\times10^9)\frac{(1.0\times10^{-9})(3.0\times10^{-9})}{(0.20)^2}\ \mathrm N
> =6.74\times10^{-7}\ \mathrm N.
> $$
>
> Superposition adds the two forces as vectors, $\vec F=\vec F_A+\vec F_B$, with
> components $F_x=-6.74\times10^{-7}\ \mathrm N$ and $F_y=-6.74\times10^{-7}\ \mathrm N$:
>
> $$
> \lvert\vec F\rvert=\sqrt{F_x^2+F_y^2}=9.5\times10^{-7}\ \mathrm N,
> \qquad
> \theta=\atanTwo(F_y,F_x)=225^\circ.
> $$
>
> The resultant points into the third quadrant, $45^\circ$ below the $-x$ axis, along
> the bisector of the two source directions — as the symmetry of the equal charges
> requires.

$$
% caption: Superposition of two Coulomb forces. Equal positive charges $q_A$ (on
% the $x$ axis) and $q_B$ (on the $y$ axis) each repel $q_0$ at the origin; the two
% equal forces add to a resultant along the bisector, pointing into the third
% quadrant at $225^\circ$.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (-2.4,0) -- (2.6,0) node[right] {$x$};
  \draw[->, black] (0,-2.4) -- (0,2.6) node[above] {$y$};
  \filldraw[draw=black, fill=black!8] (0,0) circle (0.2);
  \node[black!70, anchor=south west] at (0.14,0.14) {$q_0$};
  \filldraw[draw=black, fill=black!8] (2.0,0) circle (0.18);
  \node[black!70, anchor=south] at (2.0,0.24) {$q_A$};
  \filldraw[draw=black, fill=black!8] (0,2.0) circle (0.18);
  \node[black!70, anchor=west] at (0.24,2.0) {$q_B$};
  \draw[->, thick] (-0.22,0.14) -- (-1.55,0.14) node[midway, above] {$F_A$};
  \draw[->, thick] (0.14,-0.22) -- (0.14,-1.55) node[midway, right] {$F_B$};
  \draw[->, acc, very thick] (-0.16,-0.16) -- (-1.4,-1.4);
  \node[acc] at (-1.02,-1.32) {$F$};
\end{tikzpicture}
$$

**Units and scale checks.**

The Coulomb constant has the units needed to convert a charge product divided by a
distance squared into newtons:

$$
[k]
=\mathrm{N\,m^2\,C^{-2}},
\qquad
\left[k\frac{q_1q_2}{r^2}\right]
=\frac{\mathrm{N\,m^2}}{\mathrm{C^2}}
\frac{\mathrm{C^2}}{\mathrm{m^2}}
=\mathrm N.
$$

Unit cancellation catches several common substitutions: nanocoulombs must be
converted to coulombs, centimetres to metres, and every separation must be squared.
A result in $\mathrm{N/C}$ is an electric-field unit rather than a force unit. A
result in $\mathrm{N\,m}$ contains an unremoved distance factor.

Order-of-magnitude arithmetic can be performed before a calculator is used. Two
charges of order $10^{-9}\ \mathrm C$ separated by order $10^{-1}\ \mathrm m$ give

$$
F\sim10^9\frac{10^{-9}10^{-9}}{(10^{-1})^2}\ \mathrm N
=10^{-7}\ \mathrm N.
$$

A final answer of $10^{11}\ \mathrm N$ for that scale signals a lost prefix or an
unsquared separation. The numerical coefficient then refines the estimate instead
of replacing physical checking.

> **Worked example (Coulomb force in hydrogen).** A hydrogen atom holds a proton of
> charge $+e$ and an electron of charge $-e$ about $r=5.3\times10^{-11}\ \mathrm m$
> apart. Coulomb's law gives the magnitude of the attractive force:
>
> $$
> F=k\frac{e^2}{r^2}
> =\frac{(8.988\times10^9)(1.602\times10^{-19})^2}{(5.3\times10^{-11})^2}\ \mathrm N
> =8.2\times10^{-8}\ \mathrm N.
> $$
>
> The signs fix the directions: the force on the electron points toward the proton,
> the force on the proton toward the electron, equal in magnitude. Equal force is not
> equal acceleration — with $a=F/m$, the far lighter electron accelerates much harder.

$$
% caption: The Coulomb force in a hydrogen atom. Proton ($+e$) and electron
% ($-e$) sit a distance $r\approx5.3\times10^{-11}\ \mathrm m$ apart; each is pulled
% toward the other with magnitude $8.2\times10^{-8}\ \mathrm N$.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \filldraw[draw=black, fill=black!8] (1.1,0) circle (0.24);
  \node[black!70] at (1.1,0) {+};
  \node[black, anchor=north] at (1.1,-0.34) {proton};
  \filldraw[draw=black, fill=black!8] (5.3,0) circle (0.26);
  \node[black!70, font=\scriptsize] at (5.3,0) {neg};
  \node[black, anchor=north] at (5.3,-0.34) {electron};
  \draw[->, acc, very thick] (1.36,0) -- (2.4,0) node[midway, above] {$F$};
  \draw[->, acc, very thick] (5.04,0) -- (4.0,0) node[midway, above] {$F$};
  \draw[<->, black] (1.1,-0.78) -- (5.3,-0.78) node[midway, below] {$r$};
\end{tikzpicture}
$$

The force is tiny on the scale of everyday pushes but enormous relative to the
electron's inertia. Atomic electrostatics needs care with exponents: the charge
product contributes roughly $10^{-38}$, the Coulomb constant $10^9$, and the squared
atomic separation in the denominator roughly $10^{-21}$.

> **Worked example (force between charged spheres).** Hold $q_1=+2.0\ \mathrm{nC}$
> and $q_2=-3.0\ \mathrm{nC}$ a distance $0.50\ \mathrm m$ apart in air. Treating each
> sphere as a point charge, the force magnitude is
>
> $$
> F=(8.988\times10^9)\,
> \frac{\lvert(2.0\times10^{-9})(-3.0\times10^{-9})\rvert}{(0.50)^2}\ \mathrm N
> =2.16\times10^{-7}\ \mathrm N.
> $$
>
> The charge product is negative before the magnitude bars apply, so the force is
> attractive. With $q_1$ at $x=0$ and $q_2$ at $x=0.50\ \mathrm m$, the force on $q_2$
> has negative $x$ component and the force on $q_1$ positive $x$ component. The signs
> come from the receiver's position and the attraction, not from a minus sign hung on
> the scalar magnitude.

A force of this size is easily overwhelmed by friction or mechanical forces in a
macroscopic apparatus. Conducting supports, humid air, and imperfect charge control
also change the intended configuration. Electrostatic measurements therefore use
light suspended objects, insulating mounts, and repeated comparison rather than
treating the formula as a guarantee that the force will be visually dramatic.

## The point-charge approximation

The separation in Coulomb's law is the distance between the positions assigned to
the charges. For a small charged bead of radius $a$ observed at distance $r$ from
its centre, a point model is reliable when $r$ is much larger than $a$ and the charge
distribution remains close to spherically symmetric. The dimensionless ratio
$a/r$ records the approximation:

$$
\frac ar\ll1.
$$

At a point far from a compact source, small details of its shape occupy a small
angular extent. Near its surface, different source elements are separated by
measurably different distances and a single centre-to-centre distance no longer
describes the geometry. Continuous charge distributions require integrating the
Coulomb contribution over the source.

$$
% caption: When a source counts as a point. The source has radius $a$; at the far
% point ($r_{\rm far}\gg a$) its size is unresolved and one centre distance
% suffices, while the near point resolves the extent and needs the distributed
% charge.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \fill[black!8] (0,0) circle (0.8);
  \draw[black, thick] (0,0) circle (0.8);
  \filldraw[draw=black, fill=black!8] (0,0) circle (0.06);
  \draw[<->, black] (0,0) -- (0.8,0) node[midway, above] {$a$};
  \filldraw[draw=black, fill=black!8] (1.6,0) circle (0.08);
  \node[black, anchor=south] at (1.6,0.14) {near};
  \filldraw[draw=black, fill=black!8] (5.1,0) circle (0.08);
  \node[black, anchor=south] at (5.1,0.14) {far};
  \draw[<->, black] (0,-1.1) -- (1.6,-1.1) node[midway, below] {$r_{\rm near}$};
  \draw[<->, black] (0,-1.7) -- (5.1,-1.7) node[midway, below] {$r_{\rm far}$};
\end{tikzpicture}
$$

The approximation is a statement about source extent, not about charge magnitude.
A large total charge may be treated as a point charge when its physical support is
small relative to the observation distance. A very small charge spread over a broad
sheet cannot be replaced by a point charge close to the sheet. A calculation must
state both the charge scale and the geometric scale.

Spherical symmetry is a special case. Outside a spherically symmetric charge
distribution, the force on an external point charge has the same inverse-square form
as if all the charge were concentrated at the centre. That result follows from
Gauss's law and is proved later. It does not apply to an arbitrary lopsided charged
object, even when its total charge is known.

**Conditions for an electrostatic calculation.**

Coulomb's law gives the electric force alone. A diagram with charges held at fixed
locations implicitly includes mechanical supports, tension, or other forces that
balance the electric force. If those supports are removed, the charges accelerate and
the separation changes; the force must then be evaluated along the resulting motion.

The free-space constant $k$ is used for vacuum and is an excellent approximation for
many air experiments. Matter between the charges can polarize and change the force;
dielectric response is treated with electric fields and capacitance. Rapidly varying
charge distributions also require electromagnetic propagation. Coulomb-force
calculations here require an electrostatic configuration whose charge arrangement is
fixed during the calculation.

**Calculation procedure and checks.**

Most two-charge problems reduce to a short sequence, provided magnitude and
direction are kept separate until the end.

1. Draw both charges, their signs, the assigned coordinate axes, and the
   centre-to-centre separation.
2. Convert all charge values to coulombs and all lengths to metres.
3. Compute the magnitude $k|q_1q_2|/r^2$.
4. Assign attraction or repulsion from the sign of $q_1q_2$.
5. For a vector answer, form $\vec r_{12}$ from source to receiver and apply the
   signed vector formula.
6. Check units, inverse-square scaling, and Newton's third-law partner force.

The following checks catch errors without repeating the full calculation.

| Question | Required result |
| --- | --- |
| What happens when $r$ doubles? | The magnitude becomes one quarter. |
| What happens when one charge changes sign? | The magnitude is unchanged and the direction reverses. |
| What force acts on the source charge? | Equal magnitude and opposite direction on the other body. |
| What happens at very large $r$? | The magnitude tends to zero. |
| Does a support-free configuration remain static? | No; the electric force produces acceleration. |

Coulomb's law determines one pair force. For several sources, form one such vector for
each source and sum the components. This applies Coulomb's law separately to every
source in the multi-charge configuration.

## Force ratios, sensitivity, and uncertainty

Coulomb-law ratios eliminate the constant from comparisons. Consider an initial configuration
$(q_1,q_2,r)$ and a second configuration $(q_1',q_2',r')$. Their force magnitudes
satisfy

$$
\frac{F'}{F}
=\frac{|q_1'||q_2'|}{|q_1q_2|}
\left(\frac r{r'}\right)^2.
$$

The ratio form avoids calculating two large or small force values when only a change
in force is required. If both charges are doubled and the separation is tripled,
then

$$
\frac{F'}F
=\frac{(2)(2)}{3^2}
=\frac49.
$$

Doubling both charges multiplies the force by $4$, while tripling the separation
divides it by $9$; the net ratio is $4/9$. This scaling check should precede
numerical substitution.

For small fractional changes, logarithmic differentiation gives the sensitivity
relation

$$
\frac{\d F}{F}
=\frac{\d q_1}{q_1}
+\frac{\d q_2}{q_2}
-2\frac{\d r}{r}.
$$

The minus sign states that an increase in separation reduces force. For uncertainty
estimates, magnitudes are used:

$$
\frac{\delta F}{F}
\approx
\frac{\delta q_1}{|q_1|}
+\frac{\delta q_2}{|q_2|}
+2\frac{\delta r}{r}
\qquad\text{(conservative bound).}
$$

Distance error receives double weight. A one-percent error in the measured
centre-to-centre separation contributes approximately two percent to the force
uncertainty before charge uncertainties are included. This is one reason an
electrostatic force experiment needs a clear geometric reference rather than a rough
visual estimate of the gap between objects.

For example, take $q_1=(5.00\pm0.10)\ \mathrm{nC}$,
$q_2=(4.00\pm0.12)\ \mathrm{nC}$, and
$r=(0.200\pm0.002)\ \mathrm m$. The fractional uncertainty contributions are
$2.0\%$, $3.0\%$, and $2(1.0\%)=2.0\%$, respectively. Adding absolute
contributions gives a conservative $7.0\%$ force bound. If the measurements have
independent random uncertainties, a root-sum-square estimate gives

$$
\left(\frac{\delta F}{F}\right)_{\rm rms}
\approx\sqrt{(0.020)^2+(0.030)^2+(0.020)^2}
=0.041,
$$

or about $4.1\%$. The two estimates answer different questions. The linear sum is a
safe maximum-style bound; the root-sum-square value estimates the spread expected
from independent random errors.

The linearized relation assumes small fractional errors. A distance uncertainty of
fifty percent should be handled by evaluating the force at the range limits:

$$
F_{\rm max}=k\frac{|q_1q_2|}{r_{\rm min}^2},
\qquad
F_{\rm min}=k\frac{|q_1q_2|}{r_{\rm max}^2}.
$$

The asymmetry of those bounds follows from the square. A symmetric uncertainty in
$r$ produces unequal upper and lower force changes when the uncertainty is large.

**What a measured separation means.**

For point charges, $r$ connects the assigned charge positions. For small uniformly
charged spherical objects, those positions are their centres. The visible gap between
surfaces is therefore generally not the separation used in Coulomb's law. If spheres
of radii $a_1$ and $a_2$ have a surface gap $g$, then

$$
r=a_1+g+a_2.
$$

Suppose two identical spheres each have radius $1.0\ \mathrm{cm}$ and a measured
surface gap of $10.0\ \mathrm{cm}$. The Coulomb separation is
$r=12.0\ \mathrm{cm}$, rather than $10.0\ \mathrm{cm}$. Substituting the gap in
place of the centre separation overestimates the force by

$$
\frac{F_{\rm gap}}{F_{\rm centre}}
=\left(\frac{0.120}{0.100}\right)^2
=1.44.
$$

The result is forty-four percent too large from a geometric substitution alone. The
error persists even when the charges and the Coulomb constant are known perfectly.

For extended objects with nonuniform charge, a single centre may not represent the
source. A nearby charged rod, plate, or irregular conductor contains source elements
at many distances and directions from the receiving charge. The appropriate force is
then built from differential charge elements, each with its own separation vector.
Replacing the object by its total charge at an arbitrary centre loses the geometry
that controls the inverse-square interaction.

There is one important special case. Outside a spherically symmetric charge
distribution, the field has radial symmetry, and the complete distribution acts as
if its total charge were concentrated at the centre. The result is exact outside the
distribution at every exterior point. This exact result depends on spherical
symmetry; a distorted conductor near another charge generally has a nonuniform
surface distribution and needs a more detailed treatment.

**One-dimensional signed force functions.**

An axis calculation benefits from a formula that carries direction algebraically.
For charge $q_1$ fixed at coordinate $x_1$, the $x$ component of the force on
charge $q_2$ at coordinate $x_2$ is

$$
F_x
=kq_1q_2\frac{x_2-x_1}{|x_2-x_1|^3}.
$$

This expression is the one-dimensional version of the vector law. The signed
displacement fixes the source-to-receiver direction, while the charge product changes
the sign for attraction. Separate left-right rules are unnecessary.

A positive source at the origin repels a positive receiving charge on the positive
$x$ axis. Here $x_2-x_1>0$ and $q_1q_2>0$, so $F_x>0$. The receiving charge is pushed to
the right. Put the same positive receiving charge on the negative $x$ axis:
$x_2-x_1<0$, hence $F_x<0$, and it is pushed to the left. Both results describe
repulsion away from the origin.

A negative receiving charge reverses the product $q_1q_2$ and both force
directions reverse. A negative charge placed right of a positive source has
$x_2-x_1>0$ but $q_1q_2<0$, giving $F_x<0$: it is attracted leftward toward the
source. The algebra agrees with the physical sign diagram while remaining usable
when positions and signs become less visually obvious.

The expression has a singularity at $x_2=x_1$. That coordinate is occupied by the
source charge, where a point charge cannot exert a finite force on itself. A force
plot therefore has separate intervals to the left and right of the source. The
vertical divergence is a property of the ideal point model and marks a coordinate
excluded from the physical configuration.

The plot is a signed component graph. Its negative branch does not indicate a
negative force magnitude. It indicates a force vector directed toward negative
$x$. Component sign, force magnitude, and charge sign are three distinct pieces of
information and should remain separate in written work.

**Measurement conditions and systematic effects.**

Coulomb's torsion-balance result requires charge values and geometry that remain
stable while the force is measured. Several systematic effects can change that
condition:

- **Charge leakage:** humid air, contaminated insulating supports, and imperfect
  contacts permit charge to escape over time.
- **Polarization:** a nearby conductor redistributes its mobile charge, altering the
  assumed source geometry even when its total charge remains fixed.
- **Finite size:** centres and surface gaps differ, and charge may occupy extended
  regions rather than one point.
- **Mechanical calibration:** the torsion fibre's restoring torque must be known to
  convert a twist angle into force.
- **Air ionization:** sufficiently large fields can transfer charge through air,
  changing the charges during the observation.

Reversing both charge signs provides a controlled comparison. The product $q_1q_2$
remains unchanged, so the force magnitude and attraction-or-repulsion category
remain unchanged. Reversing only one sign reverses the force direction while leaving
the magnitude unchanged. A measured magnitude that changes strongly under a
simultaneous sign reversal points to an uncontrolled geometry or charging effect.

The comparison works only when the magnitudes of the charges and the geometry are
reproduced. Rubbing two objects can change charge magnitude from trial to trial;
an electrometer or calibrated charge-transfer procedure is needed for a quantitative
test. The expected scaling follows from the law, while the apparatus determines
whether the assumed inputs are actually controlled.

## Torsion balance and the measured power law

Coulomb's apparatus converted a small electrostatic force into a measurable twist of
a suspended fibre. A light horizontal arm carried a charged sphere. A second charged
sphere was brought to a known separation from it. The electric force acted at a
known lever arm $L$, while the fibre supplied a restoring torque proportional to the
twist angle $\phi$ over its calibrated elastic range:

$$
\tau_{\rm fibre}=\kappa\phi,
\qquad
\tau_{\rm electric}=FL.
$$

Static equilibrium gives

$$
F=\frac{\kappa\phi}{L}.
$$

The force is therefore inferred from a mechanical measurement. The law does not
require the spheres to move appreciably. A small twist changes the geometry only
slightly, and the apparatus can be read after the torsional oscillation has decayed.
Large deflections need a geometrical correction because both the lever arm direction
and the charge separation change with the arm angle.

An inverse-square test holds the charges fixed as well as possible, measures the
force at several separations, and compares the ratios. For two separations $r_a$ and
$r_b$,

$$
\frac{F_a}{F_b}=\left(\frac{r_b}{r_a}\right)^2.
$$

If $r_b=2r_a$, the predicted twist-derived force at the closer position is four
times the force at the farther position. The comparison removes the unknown product
$|q_1q_2|$ and the calibration constant $\kappa/L$, provided neither changes between
measurements. This ratio method was central to distinguishing an inverse-square law
from a simple inverse-distance dependence.

A log-log representation expresses the same experimental test in linear form:

$$
\ln F=\ln\!\left(k|q_1q_2|\right)-2\ln r.
$$

Data plotted as $\ln F$ against $\ln r$ should lie near a straight line whose slope
is $-2$. The intercept depends on charge magnitude and calibration; the slope tests
the distance exponent. A consistent intercept with a slope far from $-2$ indicates a
failure of the assumed geometry, charge stability, or force model.

The measured twist contains effects beyond the desired electric force. A nearby wall
or conductor can polarize and alter the field around the spheres. A support that
leaks charge changes the charge product during the measurement. A sphere whose
radius is comparable with its separation samples a nonuniform force across its
surface. Good apparatus design places conducting objects far from the measurement
region, uses dry insulating supports, and works at separations large compared with
the sphere radii.

## A complete coordinate calculation

The vector law is most reliable when the geometry is collapsed to one displacement
vector before any number is substituted.

> **Worked example (vector Coulomb force in the plane).** A source $q_1=+5.0\ \mathrm{nC}$
> sits at $\vec r_1=(-0.30,0.10)\ \mathrm m$ and a receiving charge $q_2=-2.0\ \mathrm{nC}$
> at $\vec r_2=(0.10,0.40)\ \mathrm m$. The source-to-receiver displacement is
>
> $$
> \vec r_{12}=\vec r_2-\vec r_1=(0.40\,\hat\imath+0.30\,\hat\jmath)\ \mathrm m,
> \qquad
> r_{12}=\sqrt{0.40^2+0.30^2}\ \mathrm m=0.50\ \mathrm m.
> $$
>
> The charge product is $q_1q_2=-1.0\times10^{-17}\ \mathrm{C^2}$. The vector form gives
>
> $$
> \vec F_{1\to2}=k\,q_1q_2\,\frac{\vec r_{12}}{r_{12}^3}
> =(8.988\times10^9)(-1.0\times10^{-17})\frac{0.40\,\hat\imath+0.30\,\hat\jmath}{(0.50)^3}
> =(-2.88\,\hat\imath-2.16\,\hat\jmath)\times10^{-7}\ \mathrm N.
> $$
>
> The negative components point left and down, from $q_2$ toward the positive source
> $q_1$ — the attraction the sign of $q_1q_2$ demands. The magnitude
>
> $$
> \lvert\vec F_{1\to2}\rvert=\sqrt{2.88^2+2.16^2}\times10^{-7}\ \mathrm N
> =3.60\times10^{-7}\ \mathrm N
> $$
>
> matches the scalar check $k\lvert q_1q_2\rvert/r_{12}^2
> =(8.988\times10^9)(1.0\times10^{-17})/0.25=3.60\times10^{-7}\ \mathrm N$.

The direction can be checked before computing any components. The receiving charge
is negative and the source is positive, so attraction is required. The source lies
down and left of the receiver, so the force on the receiver must point down and
left. The calculated component signs agree with that geometric prediction.

The force on $q_1$ due to $q_2$ has the opposite vector:

$$
\vec F_{2\to1}
=(2.88\hat\imath+2.16\hat\jmath)\times10^{-7}\ \mathrm N.
$$

It would be incorrect to assign this vector to $q_2$ merely because it contains the
same numerical components with different signs. The receiver label in
$\vec F_{a\to b}$ is part of the answer.

## Boundary cases and model limits

The limits of the inverse-square law expose its physical range. As $r\to\infty$,

$$
\lim_{r\to\infty}k\frac{|q_1q_2|}{r^2}=0.
$$

Charges at very large separation still interact in the model, but the force may be
smaller than the sensitivity of the apparatus. The limit does not imply a sharp
cutoff distance. It states a continuous decrease with increasing separation.

As $r\to0$ in the point model, the magnitude diverges. The physical setup reaches a
different regime before that mathematical limit: finite spheres touch, charges may
redistribute across conducting surfaces, material deformation can occur, and quantum
description becomes relevant for atomic separations. The divergence records that a
zero-size source and zero centre separation cannot be combined with a classical
two-point-charge model.

The assumption of stationary sources also has a time scale. In electrostatics, charge
positions are treated as fixed while the force is evaluated. A source that moves
substantially during the observation changes the separation and therefore the force.
Changes in electromagnetic influence propagate at finite speed; the field treatment
introduced next provides the framework for time-dependent situations. Coulomb's law
remains the correct static pair limit within that broader theory.

The medium matters as well. The stated value of $k$ applies to vacuum and is a close
approximation in many air experiments. Polarizable matter between charges modifies
the interaction through its electrical response. A calculation should state whether
the source is in vacuum, air, or a material medium before applying a numerical
Coulomb constant.

**Independent checks before reporting a result.**

A correct numerical magnitude can still be attached to the wrong body or direction.
Independent checks use information not already consumed in arithmetic:

- **Sign check:** unlike charges require an inward force pair; equal signs require
  an outward force pair.
- **Coordinate check:** the force on the receiving charge must point toward the
  source for attraction and away from the source for repulsion.
- **Scale check:** doubling the chosen separation must reduce the reported
  magnitude by four.
- **Unit check:** the final unit is newtons for force, with no remaining coulombs or
  metres.
- **Pair check:** exchanging source and receiver reverses the force vector while
  preserving magnitude.
- **Model check:** centre-to-centre distance and point-source assumptions must match
  the physical objects described.

Check direction, centre-to-centre geometry, units, the receiving body, and nearby
conductors independently. Each check constrains a different part of the model.

A two-charge problem normally requires a vector statement with a
magnitude, a direction, and the body receiving the force. For example,

$$
\vec F_{1\to2}
=(-2.88\hat\imath-2.16\hat\jmath)\times10^{-7}\ \mathrm N,
\qquad
|\vec F_{1\to2}|=3.60\times10^{-7}\ \mathrm N.
$$

The component form fixes the coordinate direction; the magnitude gives the scalar
strength; the subscripts fix the physical body. Omitting any one of those pieces
leaves a force calculation incomplete.

In laboratory work, retain the sketch, charge signs, conversion factors, and
centre-to-centre measurement beside the final value. That record makes a later
comparison meaningful: a disagreement can be traced to a changed charge preparation,
a changed separation convention, an altered support geometry, or an arithmetic
conversion. A force value without its configuration cannot be independently checked.

## Comparison with Newtonian gravitation

Coulomb's law and Newton's gravitational law have the same inverse-square
geometrical form. For two masses,

$$
F_g=G\frac{m_1m_2}{r^2},
$$

whereas the electrostatic magnitude is

$$
F_e=k\frac{|q_1q_2|}{r^2}.
$$

Both forces act along the line joining the two idealized particles, and both weaken
with the square of the separation. Their physical content differs in the source
properties and in the allowed signs. Mass is positive in the ordinary Newtonian
model, so gravitational pair forces are attractive. Electric charge can be positive
or negative, giving either attraction or repulsion.

At the same separation, a proton and electron have an electric-to-
gravitational force magnitude is independent of distance:

$$
\frac{F_e}{F_g}
=\frac{ke^2}{Gm_pm_e}
\approx2.3\times10^{39}.
$$

The cancellation of $r^2$ makes the comparison especially clear. At atomic scales,
electric attraction between a proton and an electron exceeds their mutual
gravitational attraction by roughly thirty-nine orders of magnitude. Atomic and
molecular structure is therefore governed primarily by electromagnetic interactions,
while gravity is negligible for individual charged particles.

Macroscopic matter behaves differently because most objects contain nearly equal
positive and negative charge. The enormous individual electric contributions cancel
to a high degree outside an electrically neutral body. Mass contributions add with
the same sign, so gravity remains observable for planets, stars, and ordinary
neutral objects. A small net charge imbalance can still produce a readily measurable
electric force because the elementary electric interaction is so large.

$$
% caption: Sign rules for inverse-square pair forces. Two masses always attract;
% opposite charges attract; like charges repel. Each pair keeps the same $1/r^2$
% dependence.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % two masses
  \filldraw[draw=black, fill=black!8] (0.6,0.9) circle (0.2);
  \filldraw[draw=black, fill=black!8] (2.3,0.9) circle (0.2);
  \draw[->, black, thick] (0.86,0.9) -- (1.35,0.9);
  \draw[->, black, thick] (2.04,0.9) -- (1.55,0.9);
  \node[black, anchor=north] at (1.45,0.5) {two masses};
  % opposite charges
  \filldraw[draw=black, fill=black!8] (4.4,0.9) circle (0.2);
  \node[black!70] at (4.4,0.9) {+};
  \filldraw[draw=black, fill=black!8] (6.1,0.9) circle (0.22);
  \node[black!70, font=\scriptsize] at (6.1,0.9) {neg};
  \draw[->, thick] (4.66,0.9) -- (5.15,0.9);
  \draw[->, thick] (5.84,0.9) -- (5.35,0.9);
  \node[black, anchor=north] at (5.25,0.5) {unlike charges};
  % like charges
  \filldraw[draw=black, fill=black!8] (8.2,0.9) circle (0.2);
  \node[black!70] at (8.2,0.9) {+};
  \filldraw[draw=black, fill=black!8] (9.9,0.9) circle (0.2);
  \node[black!70] at (9.9,0.9) {+};
  \draw[->, thick] (7.94,0.9) -- (7.45,0.9);
  \draw[->, thick] (10.16,0.9) -- (10.65,0.9);
  \node[black, anchor=north] at (9.05,0.5) {like charges};
\end{tikzpicture}
$$

The parallel inverse-square form also exposes two common errors. A calculation
that treats an electric force as always attractive has imported the gravitational
sign rule by mistake. A calculation that writes a negative electric-force magnitude
has mixed component direction into a scalar quantity. The scalar law uses absolute
charge product; the vector law carries the sign and direction.

**Zero charge, equal charge, and limiting cases.**

Setting either charge equal to zero gives

$$
q_1=0\ \text{or}\ q_2=0
\quad\Longrightarrow\quad
\vec F=\vec 0.
$$

This result does not say that a neutral extended object is unaffected by a nearby
charged body. A neutral conductor can polarize, producing separated positive and
negative surface charge and a net force in a nonuniform environment. The point-charge
model represents one net charge at one position, so it cannot reproduce that induced
distribution. The distinction prevents an inappropriate use of $q=0$ in conductor
problems.

Equal charge magnitudes produce equal force magnitudes only when the separation is
the same. For example, a charge $Q$ at distance $r$ and a charge $2Q$ at distance
$2r$ produce force magnitudes in the ratio

$$
\frac{k(2Q)|q|/(2r)^2}{kQ|q|/r^2}
=\frac12.
$$

Doubling source charge only partly compensates for doubling separation. The ratio
tests charge and distance scaling in one expression.

The sign of the source-receiver vector is undefined at zero separation because the
two assigned points coincide. The unit vector
$\hat r_{12}$ therefore has no value at $r_{12}=0$, independently of the
divergence in the scalar magnitude. Both failures identify the same excluded point
of the model.

In the far-distance limit, the point-pair force tends smoothly to zero. A numerical
calculation may round a very small force to zero for an instrument or an application,
but the theoretical result remains a nonzero value at every finite separation when
both charges are nonzero. Stating the relevant measurement threshold makes that
approximation transparent.

## Recording a force result

A complete calculation records the inputs, the model assumptions, and the vector
result. The following compact format keeps those elements visible:

$$
\begin{gathered}
q_1=+2.0\ \mathrm{nC},\quad
q_2=-3.0\ \mathrm{nC},\quad
r=0.50\ \mathrm m,\\
F=k|q_1q_2|/r^2=2.16\times10^{-7}\ \mathrm N,\\
\vec F_{1\to2}
=-2.16\times10^{-7}\hat\imath\ \mathrm N
\quad\text{for the stated horizontal geometry.}
\end{gathered}
$$

The first line states the physical data and units. The second reports the magnitude
calculation. The third identifies the receiving body and coordinate
direction. A diagram should show the same direction as the component sign. This
format allows another reader to distinguish a unit conversion error from a direction
error without reconstructing the entire solution.

When measurements rather than exact values are supplied, attach a suitable
uncertainty:

$$
F=(2.16\pm0.15)\times10^{-7}\ \mathrm N
$$

for a seven-percent conservative uncertainty estimate. The uncertainty belongs to
the measured configuration and should not be reported with more significant digits
than the charge and distance data justify. The Coulomb constant itself is much more
precisely known than ordinary laboratory charge and spacing measurements, so it
rarely dominates the experimental uncertainty.

Apply the force law together with geometry, sign, unit, and model checks. Each check
identifies a distinct possible failure and yields a physically identified force
vector.

**Reading an inverse-square data set.**

An experimental test of Coulomb's law usually begins with force magnitudes measured
at several separations while the charge preparation is held fixed. Instead of fitting
the Coulomb constant directly, form the product

$$
C_{\rm data}=Fr^2.
$$

An ideal two-point-charge experiment gives

$$
C_{\rm data}=k|q_1q_2|,
$$

which is independent of $r$. The product provides a direct diagnostic: values that
remain constant within uncertainty support the inverse-square exponent, whereas a
systematic rise or fall with separation indicates a geometric or charge-control
problem.

For example, suppose a force magnitude $F_0$ is measured at
$r_0=0.20\ \mathrm m$. The inverse-square prediction at two larger separations is

$$
F(0.30\ \mathrm m)
=F_0\left(\frac{0.20}{0.30}\right)^2
=\frac49F_0,
$$

and

$$
F(0.40\ \mathrm m)
=F_0\left(\frac{0.20}{0.40}\right)^2
=\frac14F_0.
$$

The force ratios can be checked before any charge calibration is known. If the
measured values are $F_0$, $0.45F_0$, and $0.26F_0$, the deviations from
$1$, $0.444$, and $0.250$ should be compared with the experimental uncertainty
rather than treated as exact disagreement. Rounding, small separation errors, and
charge leakage can easily produce percent-level differences.

A log-log slope gives a second, independent exponent estimate. Model the force
data as

$$
F=Ar^{-n},
$$

where $A$ is a constant for fixed charges and $n$ is determined by the data. Taking
the logarithm yields

$$
\ln F=\ln A-n\ln r.
$$

For two measurements, the inferred exponent is

$$
n=
\frac{\ln(F_a/F_b)}{\ln(r_b/r_a)}.
$$

The denominator uses the ratio in the opposite order from the force ratio so that
the result is positive for a decreasing force. With $r_b=2r_a$ and
$F_b=F_a/4$, the result is

$$
n=
\frac{\ln 4}{\ln 2}
=2.
$$

Using many data points is preferable to relying on one pair. A straight-line fit to
all the log-log data estimates the slope and exposes curvature. Curvature can arise
when a background force was added to the electrostatic signal, when charges leaked
between readings, or when the separation approached the sphere radii. A constant
background force is particularly visible at large separation, where the true
electrostatic force is small and the offset becomes a large fraction of the reading.

Direction data should be recorded separately from magnitude data. A torsion balance
may reverse its twist when one charge sign is reversed, while the magnitude plot
still uses positive values of $F$. Mixing signed twist angles with unsigned force
magnitudes creates an artificial sign change in a log plot. The correct workflow
records the sign configuration, converts the mechanical deflection to a magnitude,
and compares the associated direction with the attraction-or-repulsion prediction.

The data test is meaningful only across a range where the same physical model
applies. Close points can violate the point-charge approximation; far points can be
dominated by leakage or instrumental offset. A clean inverse-square result comes
from a controlled interval, rather than from forcing every available measurement
onto one ideal curve.

**Prefix and significant-figure audit.**

Charge prefixes produce the largest numerical errors in introductory Coulomb-law
work. The required SI conversions are

$$
1\ \mathrm{mC}=10^{-3}\ \mathrm C,
\qquad
1\ \mathrm{\mu C}=10^{-6}\ \mathrm C,
\qquad
1\ \mathrm{nC}=10^{-9}\ \mathrm C,
\qquad
1\ \mathrm{pC}=10^{-12}\ \mathrm C.
$$

Both charges enter as a product. Replacing a stated $3.0\ \mathrm{nC}$ with
$3.0\times10^{-6}\ \mathrm C$ multiplies the calculated force by $10^3$ for that
one error. Replacing both nanocoulomb charges by microcoulomb values multiplies the
force by $10^6$. A numerical answer may still look plausible when written in
scientific notation, so every prefix should be converted before the product is
formed.

Distance prefixes are squared as well. A separation of $25\ \mathrm{cm}$ is
$0.25\ \mathrm m$, and its square is $0.0625\ \mathrm{m^2}$. Squaring the bare
number 25 while leaving the unit as metres inserts a factor of $10^4$ error. Writing
the converted distance with its unit on a separate line before applying the square
makes the operation visible.

The reported precision follows the least precise input. Charges quoted as
$2.0\ \mathrm{nC}$ and $3.0\ \mathrm{nC}$ and a separation quoted as
$0.50\ \mathrm m$ support a two-significant-figure force such as
$2.2\times10^{-7}\ \mathrm N$. Retaining all calculator digits in an intermediate
component calculation avoids rounding drift; rounding the final magnitude and angle
to the measurement precision communicates the actual quality of the input data.

A compact exponent estimate should accompany every prefix conversion. For two
nanocoulomb charges separated by a few tenths of a metre, the scale is
$10^9(10^{-9})(10^{-9})/(10^{-1})^2\sim10^{-7}\ \mathrm N$. For two microcoulomb
charges at the same separation, the scale rises to about $10^{-1}\ \mathrm N$.
Those estimates differ by six orders of magnitude because both charge factors
changed by $10^3$. Checking the exponent before calculating the leading decimal
coefficient prevents a unit prefix from passing unnoticed through a formally correct
algebraic expression.

State the force recipient whenever a result is reported. The force on charge $q_2$
from $q_1$ has the same magnitude and opposite direction to the force on $q_1$ from
$q_2$. A diagram, component signs, and a named receiving charge remove ambiguity
when attraction and repulsion are translated into Cartesian components.

A measured separation must be identified as centre-to-centre or as a
surface gap. Finite object size determines which distance belongs in the point-charge
approximation.

When the source objects are conducting spheres, the centre-to-centre distance may
still be inadequate if nearby charge redistribution is appreciable. Increase the
separation, model the conductor geometry, or state the approximation limit before
claiming a point-charge comparison.