Electric Field and Potential

Beginner

Rub a balloon on your hair and it can lift the hair from a few centimetres away. Electric charges push and pull without touching, and the basic rule is short: like charges repel, opposite charges attract. The force also fades fast with distance: move two charges twice as far apart and it drops to a quarter.

Step 1: the field. Instead of tracking every charge, we say that each charge fills the space around it with an electric field. At any point, the field tells you which way, and how hard, a small positive "test" charge put there would be pushed. Around a + charge the field points outward; around a − charge it points inward.

Step 2: field lines. We draw the field as lines with arrows. They leave + charges and arrive at − charges. Where the lines crowd together the push is strong; where they spread out it is weak.

Step 3: how strong. Field strength is the force on each coulomb of charge, in newtons per coulomb (N/C). In a field of 100 N/C, a charge of half a coulomb feels $0.5\times 100=50\mathrm{N}$.

Step 4: voltage. Electric potential, measured in volts, works like height on a hill. A + charge "falls" from high voltage to low voltage and speeds up, the way a ball rolls downhill. A difference of one volt means each coulomb gains or loses one joule of energy.

In the animation, grey lines run from the + charge to the − charge and bunch up near each one. The small gold charge, released near the +, slides along the middle line, faster and faster.

Check yourself: in a field of 400 N/C, what force acts on a charge of 0.005 C? (Answer: 2 N.)

Intermediate

The field: force per unit charge

Charges push and pull each other across empty space, as Coulomb's Law describes. The field picture splits this into two steps: the source charges set up an electric field $\vec{E}$ everywhere around them, and a charge placed at a point feels the field right there.

To measure the field, place a small positive test charge $q$ at the point, measure the force on it and divide:
$E=\frac{F}{q}$

Doubling $q$ doubles the force, so the ratio belongs to the point, and the field is there even when nothing probes it. Its unit is the newton per coulomb, which equals the volt per metre. Turned around, $\vec{F}=q\vec{E}$: a positive charge is pushed along the field, a negative one against it. The animation's test charge, $q=+0.1\mathrm{nC}$, is ten times smaller than the fixed sources, $\pm 1\mathrm{nC}$.

Point charges and superposition

Dividing Coulomb's law by the test charge gives the field of a point charge $Q$ at a distance $r$:
$E=\frac{kQ}{r^2}$
with $k\approx 8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$. It points away from a positive $Q$ and towards a negative one.

With several charges, each produces its own field as if the others were absent, and the total is their vector sum. On the perpendicular bisector of the animation's pair, the two fields are equally strong, one pointing away from the + charge and one towards the − charge. Their components across the axis cancel and those along it add, so the total runs parallel to the axis, towards the − side. The blue grid arrows show these sums, their lengths compressed to fit the strong field near the charges.

Potential: energy per unit charge

The potential describes the same field with one number per point instead of an arrow. Moving a charge against the electric force takes work, stored as electric potential energy $U$, just as lifting a stone stores gravitational potential energy. Per unit charge this becomes the electric potential $V=U/q$, measured in volts: $1\mathrm{V}=1J/C$.

When a charge $q$ moves through a potential difference $\Delta V$, its potential energy changes by $\Delta U=q\Delta V$, and the field does work $W_{\mathrm{field}}=-\Delta U$ on it. A positive charge left to itself moves towards lower potential and speeds up, like a ball rolling downhill; an electron moves towards higher potential. An electron accelerated through 1 V gains one electron-volt, $1\mathrm{eV}=1.602\times 10^{-19}\mathrm{J}$.

With $V=0$ far away, a point charge has $V=kQ/r$: no square, and $Q$ keeps its sign, so the potentials of several charges add as signed numbers. The field points towards lower potential, and its strength is how fast $V$ falls per metre. Between parallel plates a distance $d$ apart the field is uniform, $E=\Delta V/d$. These are the working equations of this level:

• $E=\frac{F}{q}$
• $E=\frac{kQ}{r^2}$
• $V=\frac{kQ}{r}$
• $\Delta U=q\Delta V=-W_{\mathrm{field}}$
• $E=\frac{\Delta V}{d}\text{(uniform field)}$

Worked example: the dipole in the animation

The source charges $Q=\pm 1\mathrm{nC}$ are 180 mm apart, and the test charge $q=+0.1\mathrm{nC}$ starts at rest 20 mm from the + charge. At the midpoint each source is 0.090 m away and contributes
$\frac{kQ}{r^2}=\frac{8.99\times 10^9\cdot 10^{-9}}{0.090^2}\approx 1.11\times 10^{3}V/m$

Both contributions point towards the − charge, so $E\approx 2.22kV/m$ and $F=qE\approx 0.22\mu \mathrm{N}$. At the start the + charge, 0.020 m away, gives 22.5 kV/m and the − charge, 0.160 m away, adds 0.35 kV/m, so $E\approx 22.8kV/m$ and $F\approx 2.28\mu \mathrm{N}$, ten times more. These are the $E$ and $F$ readouts at the start and halfway, and the green force arrow is long at both ends of the trip and short in the middle.

At the start the potential is $V=kQ/0.020-kQ/0.160=449.5-56.2\approx 393\mathrm{V}$; by symmetry it is $-393\mathrm{V}$ at the end and zero at the midpoint. Over this drop of 786.6 V the field does work $W=10^{-10}\cdot 786.6\approx 7.87\times 10^{-8}\mathrm{J}$, the 78.7 nJ of kinetic energy on the expert layer's $K$ readout. Half of it is gained by the midpoint, so the speed there is $1/\sqrt{2}$, about 71 %, of the final speed; starting from rest, the charge takes 2.45 s of its 4.0 s trip to get there.

Worked example: an electron between two plates

In an electron gun, electrons are pulled from a negative plate towards a positive one. Take plates 1.0 cm apart at 500 V, with an electron starting at rest; the field between them is uniform:
$E=\frac{\Delta V}{d}=\frac{500}{0.010}=5.0\times 10^{4}V/m$

The force, $F=eE=1.602\times 10^{-19}\cdot 5.0\times 10^4\approx 8.0\times 10^{-15}\mathrm{N}$, points against the field, towards the positive plate; the electron's weight is about $10^{15}$ times smaller. Crossing the gap it gains 500 eV, or $8.0\times 10^{-17}\mathrm{J}$, and with its mass $m=9.11\times 10^{-31}\mathrm{kg}$, $\frac{1}{2}m{v}^2=e\Delta V$ gives
$v=\sqrt{\frac{2e\Delta V}{m}}=\sqrt{\frac{2\cdot 8.0\times 10^{-17}}{9.11\times 10^{-31}}}\approx 1.33\times 10^{7}m/s$

That is about 4.4 % of the speed of light. The acceleration is constant, so the average speed is $v/2$ and the crossing takes $t=2d/v\approx 1.5\mathrm{ns}$. Television tubes used about 25 kV, enough for nearly a third of the speed of light.

Reading the field map

A field line runs along $\vec{E}$ at every point, and its arrow shows where a positive charge would be pushed. Lines start on positive charges and end on negative ones, and in the animation as many reach the − charge as leave the + charge. They never cross, since the field has only one direction at each point, and where they crowd together, near the charges, the field is strong.

The dashed equipotentials of the expert layer cut every field line at right angles, since moving a charge along one takes no work; the straight one midway is $V=0$.

Common mistakes

• Field lines are not paths: a charge released off the axis cuts across the curved lines, since its inertia keeps it from turning with them; the gold charge follows the central line only because it is straight.
• The field is not the force: at the midpoint $E=2.2kV/m$ for any test charge, but the force is 0.22 µN on 0.1 nC and 2.2 µN on 1 nC.
• The potential falls as $1/r$, the field as $1/r^2$: at 10 cm from 4 nC, forgetting the square gives 360 V/m instead of 3600 V/m.
• Zero potential is not zero field: midway between the charges the potentials $+99.9$ and $-99.9\mathrm{V}$ cancel, while the fields add up to 2.22 kV/m.
• A negative charge is pushed against the field and gains speed towards higher potential: the electron between the plates gains 500 eV going from 0 V to +500 V.

Where you meet it

Electric fields steer electrons in electron microscopes and ions in mass spectrometers, clean smoke out of chimney gases and place toner in laser printers. Dry air breaks down at about 3 MV/m, so a spark jumps a 1 mm gap at about 3 kV. Nerve cells hold about 70 mV across a membrane 7 nm thick, some 10 MV/m.

Next, Capacitors shows how the field between two plates stores energy, Electric Current and Resistance how a field drives charge through a wire, and DC Circuits and Ohm's Law how a voltage keeps charge moving round a circuit. The same mathematics, for masses that only attract, governs Gravitation and Orbits.

Advanced

Superposition solves electrostatics

One rule settles every static field. Everything later in the lesson is either a shortcut to it or a way of picturing its answer. When the source charges stay still, two facts are enough. Coulomb's law gives the field of a single charge: it points along the line through the charge, and its size falls as the inverse square of the distance. The principle of superposition says that the fields of several charges simply add as vectors, each charge acting as if the others were not there.

Legend

$Q_1$	Positive source charge.	$\mathrm{C}$
$Q_2$	Negative source charge, here $-Q_1$.	$\mathrm{C}$
$P$	Point where the field is found.
$r_1$	Distance from $Q_1$ to $P$; equal to $r_2$ here.	$\mathrm{m}$
$r_2$	Distance from $Q_2$ to $P$; equal to $r_1$ here.	$\mathrm{m}$
${\vec{E}}_1$	Field of $Q_1$ alone at $P$, pointing away from $Q_1$.	$N/C$
${\vec{E}}_2$	Field of $Q_2$ alone at $P$, pointing towards $Q_2$.	$N/C$
$\vec{E}$	Total field at $P$: the vector sum ${\vec{E}}_1+{\vec{E}}_2$.	$N/C$

Figure 1 adds two fields at one point $P$. The positive charge $Q_1$ would push a positive test charge away, so its field ${\vec{E}}_1$ points from $Q_1$ through $P$ and beyond. The negative charge $Q_2$ would pull the test charge in, so its field ${\vec{E}}_2$ points from $P$ towards $Q_2$. Each arrow has the size $k|Q_i|/r_i^2$, where $r_i$ is the distance from charge $Q_i$ to $P$ and $k\approx 8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$ is the Coulomb constant.

Now look at where $P$ sits: on the perpendicular bisector of the pair. So $r_1=r_2$, and the two arrows are equally long. Their parts at right angles to the line of the charges are equal and opposite, so they cancel. What survives is the sum $\vec{E}$, the diagonal of the dashed parallelogram, and it runs parallel to the line from the positive charge to the negative one.

To write the same sum for any number of charges, put charge $Q_i$ at position ${\vec{r}}_i$ and ask for the field at position $\vec{r}$. The vector from the charge to the field point is $\vec{r}-{\vec{r}}_i$, of length $r_i$. Dividing it by $r_i^3$ leaves a unit vector divided by $r_i^2$: the inverse-square size, already pointing the right way. Adding over all the charges:
$\vec{E}(\vec{r})={\sum }_i\frac{k{Q}_i(\vec{r}-{\vec{r}}_i)}{|\vec{r}-{\vec{r}}_i{|}^3}$

Charge spread out with density $\rho$, in coulombs per cubic metre, is cut into small pieces of charge $\rho d^{3}s$ at points $\vec{s}$, and the sum becomes an integral:
$\vec{E}(\vec{r})=k\int \frac{\rho (\vec{s})(\vec{r}-\vec{s})}{|\vec{r}-\vec{s}{|}^3}{d}^{3}s$

In principle nothing more is needed. The animation draws its field lines from exactly this two-term sum. The potential, field lines and Gauss's law below reach the same answer with less work, or show its shape at a glance.

The potential: work that ignores the route

Adding vectors at every point is laborious. A single number at each point, the potential, would be far easier to add up. Such a number exists only if the work the field does between two points is the same along every route. This section shows that it is.

Legend

$Q$	Point charge that makes the field.	$\mathrm{C}$
$a$	Start of both routes.
$b$	End of both routes.
$r_a$	Distance from $Q$ to $a$.	$\mathrm{m}$
$r_b$	Distance from $Q$ to $b$.	$\mathrm{m}$
$d\vec{l}$	A small step along route 1: its length and direction.	$\mathrm{m}$
$dr$	Change in the distance from $Q$ over that step: the gap between the dashed arcs.	$\mathrm{m}$
$\vec{E}$	Field at the start of the step, pointing straight away from $Q$.	$N/C$

Figure 2 shows a point charge $Q$ and two routes from point $a$ to point $b$. Move a test charge along a route in small steps $d\vec{l}$. On each step the field does work $q\vec{E}\cdot d\vec{l}$ on a charge $q$, so the work per unit charge from $a$ to $b$ is the line integral ${\int }_a^b\vec{E}\cdot d\vec{l}$: the sum of $\vec{E}\cdot d\vec{l}$ over all the steps.

Look at the step drawn on route 1. The field $\vec{E}$ at its start points straight away from $Q$, so the dot product keeps only the part of the step along the radius. That part is $dr$, the gap between the two dashed arcs centred on $Q$: how much the step changes the distance from $Q$. Moving sideways, along an arc, costs no work at all. Hence
$\vec{E}\cdot d\vec{l}=Edr=\frac{kQ}{r^2}dr$
Adding the steps from distance $r_a$ out to distance $r_b$:
${\int }_a^b\vec{E}\cdot d\vec{l}={\int }_{r_a}^{r_b}\frac{kQ}{r^2}dr=kQ\left(\frac{1}{r_a}-\frac{1}{r_b}\right)$
The result depends on the end distances $r_a$ and $r_b$ and on nothing else. Route 2 is longer and swings far out above route 1, but between $r_a$ and $r_b$ it crosses the same set of circles around $Q$, so it collects exactly the same total.

Any static arrangement is a sum of point charges. Each term of that sum is independent of the route, so the whole field is too. Going out along one route and back along another therefore gives zero work round the closed loop:
$\oint \vec{E}\cdot d\vec{l}=0$
If this failed, a charge sent round the loop in the favourable direction would gain energy on every lap from charges that never change: a perpetual-motion machine.

Because the route does not matter, one number can be attached to each point: the electric potential $V$, in volts (joules per coulomb). Its change from $a$ to $b$ is defined as
$\Delta V=V_b-V_a=-{\int }_a^b\vec{E}\cdot d\vec{l}$
The minus sign makes $V$ fall in the direction the field points. A positive charge pushed along by the field then loses potential energy $qV$ and gains the same amount of kinetic energy. Choosing $V=0$ infinitely far away, and writing $s$ for the distance from $Q$ inside the integral, a point charge gives
$V(r)=-{\int }_{\infty }^r\frac{kQ}{s^2}ds=\frac{kQ}{r}$

Read the other way round, the field is minus the gradient of the potential. It points the way $V$ falls fastest, and its size is how many volts $V$ falls per metre in that direction. Energy conservation for a charge $q$ with kinetic energy $K$ then fits on one line:
$\vec{E}=-\nabla V,K+qV=\mathrm{const}$

An example from the animation. Its source charges are +1 nC and −1 nC, 180 mm apart. The gold test charge, $q=10^{-10}\mathrm{C}$, starts at rest 20 mm from the + charge, where $V=kQ(1/0.02-1/0.16)\approx +393\mathrm{V}$ with distances in metres. It ends at the mirror point, 20 mm from the − charge, where $V\approx -393\mathrm{V}$. So it arrives with
$K=q(V_a-V_b)=10^{-10}\cdot 786.6\approx 7.87\times 10^{-8}\mathrm{J}$
which is the 78.7 nJ of the $K$ readout. Guided along any other route between the same two points, by a frictionless track for example, it would arrive with the same energy.

Flux through a surface and Gauss's law

Gauss's law links the field on a closed surface to the charge inside it. Where there is symmetry it gives a field in one line, with no sum at all. Its idea starts with field lines. A fixed number of lines leaves a point charge, and every one of them crosses every sphere around it. The sphere's area grows as $4\pi r^2$, so the number of lines per square metre falls as $1/r^2$, exactly like the field: line density is a faithful picture of field strength. To turn "the number of lines through a surface" into a quantity that works for a surface of any shape, cut the surface into small patches.

Legend

$S$	Closed surface around the charge; its shape can be anything.
$Q$	Charge inside $S$: the $Q_{\mathrm{enc}}$ of Gauss's law.	$\mathrm{C}$
$dA$	Area of one small patch of $S$.	${\mathrm{m}}^2$
$d\vec{A}$	The patch as a vector: length $dA$, pointing out of $S$ at right angles to the patch.	${\mathrm{m}}^2$
$\vec{E}$	Field of $Q$ at the patch.	$N/C$
$\theta$	Angle between $\vec{E}$ and $d\vec{A}$.	$\mathrm{rad}$

Figure 3 shows a closed surface $S$ around the charge $Q$, with one small patch drawn enlarged. The patch has area $dA$. To record which way it faces, make it a vector $d\vec{A}$: of length $dA$, pointing out of $S$ at right angles to the patch. The flux through the patch, the amount of field crossing it, is
$d\Phi =\vec{E}\cdot d\vec{A}=EdA\cos \theta$
where $\theta$ is the angle between the field $\vec{E}$ and $d\vec{A}$. Only the part of the field that goes through the patch counts. A patch facing the charge squarely ($\theta =0$) catches the most lines; tilted by $\theta$, it catches fewer, by the factor $\cos \theta$; a patch the field only grazes ($\theta =90^{\circ }$) catches none.

Adding the flux of every patch of $S$ gives Gauss's law:
$\oint \vec{E}\cdot d\vec{A}=\frac{Q_{\mathrm{enc}}}{{\epsilon }_0}$
Here $Q_{\mathrm{enc}}$ is the total charge enclosed by $S$, and ${\epsilon }_0\approx 8.85\times 10^{-12}F/m$ is the permittivity of free space, related to the Coulomb constant by $k=1/(4\pi {\epsilon }_0)$.

Why does the shape of $S$ not matter? A solid angle is to a sphere what an angle is to a circle, and the whole sphere around a point is $4\pi$. Seen from $Q$, the patch at distance $r$ covers the small solid angle $d\Omega =dA\cos \theta /r^2$: its area as seen face-on from $Q$, divided by $r^2$. Its flux is therefore
$EdA\cos \theta =\frac{kQ}{r^2}dA\cos \theta =kQd\Omega$
The $r^2$ in the patch's apparent size cancels the $1/r^2$ of the field, so a small near patch and a large far one that cover the same solid angle carry the same flux.

The solid angles of a closed surface around $Q$ add up to $4\pi$, so the total flux is $4\pi kQ=Q/{\epsilon }_0$, however bumpy $S$ is. A charge outside $S$ sends each of its lines in through one patch and out through another, giving as much negative flux as positive. That is why only the enclosed charge counts.

For the animation's +1 nC charge, the flux through any closed surface around it is $10^{-9}/(8.85\times 10^{-12})\approx 113\mathrm{V}\mathrm{m}$. A surface around both charges has zero net flux: every line that leaves through it comes back in.

Two consequences follow. Shrinking the surface around a single point turns Gauss's law into a statement about that point, $\nabla \cdot \vec{E}=\rho /{\epsilon }_0$: field lines begin and end only where there is charge. And with symmetry, Gauss's law gives fields with almost no calculation. A uniformly charged sphere acts, outside itself, like a point charge at its centre. A large flat sheet carrying $\sigma$ coulombs per square metre sends its field straight out on both sides; a flat box of face area $A$ straddling the sheet then has flux $2EA$ and encloses $\sigma A$, so $E=\sigma /(2{\epsilon }_0)$ on each side.

Equipotentials and the V(x) graph

An equipotential is a surface on which $V$ has the same value everywhere. Moving a charge along it takes no work, and the spacing between neighbouring ones measures the field. A map of equipotentials therefore carries the same information as the field lines, often in a form that is easier to read.

In the expert layer of the animation, the dashed purple curves are equipotentials at ±400, ±150 and ±50 V, and the straight dashed line midway between the charges is $V=0$. They cross every field line at right angles. The reason: $-\nabla V$ points the way $V$ falls fastest, and that is straight across a surface on which $V$ does not change at all. The gold charge starts at 393 V, just outside the 400 V curve, which cuts the axis 19.7 mm from the + charge.

Close spacing means a strong field, because $V$ changes by the same amount over a shorter distance. The +50 V curve cuts the axis 21.3 mm from the midline, where $V=0$, so the average field over that gap is $50/0.0213\approx 2.35kV/m$. That is a little above the 2.22 kV/m at the midpoint itself, because the field grows towards the charge.

The $V(x)$ graph shows the same along the axis through the charges. Its slope is $-E_x$, minus the component of the field along the axis: about $-2.2$ V per millimetre at the midpoint, and ten times steeper where the trip starts. Outside the pair, $V(x)$ rises with $x$ on both sides, so there $E_x<0$: left of the + charge the field points away from it, right of the − charge it points towards it, and both directions are to the left. The gold dot slides down the curve as its potential energy $U=qV$ turns into kinetic energy $K$.

Laplace's equation: no resting place

No stable trap for a charge can be built from fixed charges. The reason is one property of a single equation for $V$ alone, found by combining the two results above.

Put $\vec{E}=-\nabla V$ into $\nabla \cdot \vec{E}=\rho /{\epsilon }_0$. This gives Poisson's equation, ${\nabla }^{2}V=-\rho /{\epsilon }_0$, and in empty space, where $\rho =0$, Laplace's equation, ${\nabla }^{2}V=0$. The Laplacian ${\nabla }^{2}V$ measures how much the average of $V$ over a small sphere around a point exceeds its value at the centre. In empty space it is zero, so $V$ at every point equals its average over any small sphere around it.

A value equal to the average of its surroundings can be neither above all of them nor below all of them. So in empty space $V$ has no maximum and no minimum, only saddles. A positive charge would need a minimum of $V$ to rest in stable equilibrium, so no arrangement of fixed charges can hold a charge stably (Earnshaw's theorem). For example, midway between two equal positive charges $E=0$, but a positive test charge there is pushed back only along the axis; sideways, it is pushed out. Atoms owe their stability to something else: quantum mechanics.

Conductors and the capacitor field

In a conductor some charges are free to move. That single fact fixes the field inside and just outside every conductor, and it explains the uniform field of a capacitor.

Free charges move as long as any field acts on them, so they stop only when the field inside the metal is zero. A conductor at rest is therefore a single equipotential. Any closed surface drawn inside the metal has zero flux, so by Gauss's law it encloses no charge: any excess charge sits on the surface. Just outside, the field is perpendicular to the surface, since a component along it would set the surface charges moving. A small flat box straddling the surface has flux only through its outer face, which gives the field's size, $E=\sigma /{\epsilon }_0$, with $\sigma$ the surface charge per square metre.

Two parallel plates carrying $+\sigma$ and $-\sigma$ each make the sheet field $\sigma /(2{\epsilon }_0)$. Between the plates the two fields point the same way and add; outside they point opposite ways and cancel. The result is the uniform field of a capacitor, $E=\sigma /{\epsilon }_0=\Delta V/d$ for plates a distance $d$ apart.

The dipole seen from far away

A dipole is a pair of equal and opposite charges, a common source in nature: many molecules are dipoles. Far from it the two potentials nearly cancel, and what is left obeys a simple law of its own.

Legend

$Q$	Size of each of the two charges, $+Q$ and $-Q$.	$\mathrm{C}$
$d$	Distance between the two charges.	$\mathrm{m}$
$\vec{p}$	Dipole moment: size $p=Qd$, pointing from $-Q$ to $+Q$.	$\mathrm{C}\cdot \mathrm{m}$
$P$	Point where the potential is found.
$r$	Distance from the centre of the dipole to $P$.	$\mathrm{m}$
$\theta$	Angle between $\vec{p}$ and the line from the centre to $P$.	$\mathrm{rad}$
$r_{+}$	Distance from $+Q$ to $P$.	$\mathrm{m}$
$r_{-}$	Distance from $-Q$ to $P$.	$\mathrm{m}$
$d\cos \theta$	How much farther $P$ is from $-Q$ than from $+Q$, when $P$ is far away: the part of $d$ pointing towards $P$.	$\mathrm{m}$

Figure 4 sets up the geometry. The charges $+Q$ and $-Q$ are a distance $d$ apart. The dipole moment $\vec{p}$ has size $p=Qd$ and points from $-Q$ to $+Q$. The point $P$ is at distance $r$ from the centre of the pair, in a direction at angle $\theta$ from $\vec{p}$, and at distances $r_{+}$ and $r_{-}$ from the two charges. Adding the two point-charge potentials:
$V=kQ\left(\frac{1}{r_{+}}-\frac{1}{r_{-}}\right)=kQ\frac{r_{-}-r_{+}}{r_{+}{r}_{-}}$

The result now depends on the small difference $r_{-}-r_{+}$. When $P$ is far away compared with $d$, the lines from the two charges to $P$ are nearly parallel. The dashed perpendicular from $+Q$ meets the line from $-Q$ to $P$; from that foot on, the remaining distance to $P$ is almost exactly $r_{+}$. So $r_{-}$ is longer than $r_{+}$ by the thick segment from $-Q$ to the foot, labelled $d\cos \theta$: the part of the separation $d$ that points towards $P$. In the denominator the two distances hardly differ, so $r_{+}{r}_{-}\approx r^2$. Hence
$V\approx \frac{kQd\cos \theta }{r^2}=\frac{kp\cos \theta }{r^2}$

Because the two contributions nearly cancel, the potential falls as $1/r^2$ instead of $1/r$, and the field, its gradient, falls as $1/r^3$. At a given distance, $V$ is largest along $\vec{p}$ ($\theta =0$), zero on the plane through the centre at right angles to $\vec{p}$ ($\theta =90^{\circ }$), and negative on the side of the negative charge.

A check with the animation's pair: $p=10^{-9}\cdot 0.18=1.8\times 10^{-10}\mathrm{C}\mathrm{m}$. At 1 m along the axis the formula gives 1.62 V, against the exact $kQ(1/0.91-1/1.09)\approx 1.63\mathrm{V}$. The relative error is about $d^2/(4r^2)$, here 0.8 %, so the approximation is already good at a distance of little more than five separations.

Worked problem: a proton and an electron

Energy conservation with the potentials found above gives speeds without solving any equation of motion. Comparing two particles then shows what depends on the mass and what does not.

Replace the gold charge by a proton, of charge $e$ and mass $m_p$, released from rest at the same point, 20 mm from the + charge. How fast is it at the midpoint, and at the end of the trip, 20 mm from the − charge? The potential falls by 393.3 V on the way to the midpoint, where $V=0$, and by 786.6 V on the way to the end. One electronvolt (eV), $1.602\times 10^{-19}\mathrm{J}$, is the energy a charge $e$ gains falling through one volt, so
$\frac{1}{2}{m}_p{v}_1^2=e\cdot 393.3\mathrm{V}=393\mathrm{eV},v_1\approx 2.74\times 10^{5}m/s$
$\frac{1}{2}{m}_p{v}_2^2=e\cdot 786.6\mathrm{V}=787\mathrm{eV},v_2\approx 3.88\times 10^{5}m/s$
Twice the energy gives $\sqrt{2}$ times the speed.

Now release an electron, of charge $-e$ and mass $m_e$, from rest at the far point. The force on it, $-e\vec{E}$, points against the field, so it is pushed back towards the + charge, up the potential. The arrangement is symmetric, and the electron's potential energy $-eV$ is the mirror image of the proton's, so it runs the same path in reverse and gains the same 787 eV, or $1.26\times 10^{-16}\mathrm{J}$. Its mass is 1836 times smaller, so its speed is $\sqrt{1836}\approx 42.8$ times larger: $1.66\times 10^{7}m/s$, 5.5 % of the speed of light.

The size of any relativistic correction is set by the kinetic energy compared with the rest energy, $K/(m_e{c}^2)=787/511000\approx 0.15\%$, and it lowers that speed by only about 0.1 %.

Energy gives speeds, not times. For times we need the equation of motion along the axis, $m\ddot{x}=qE(x)$. Divided by $m$, it depends on the particle only through $q/m$. So every particle with a charge of the same sign, released from rest at the same point, follows the same path, and if $q/m$ is larger by some factor, the whole motion runs faster by its square root: all times scale as $\sqrt{m/q}$.

The animation's charge is given a mass of about 49 mg so that its trip takes 4.0 s. The proton's $q/m$ is $4.7\times 10^{13}$ times larger, so it makes the trip in $4.0\mathrm{s}/\sqrt{4.7\times 10^{13}}\approx 0.58\mu \mathrm{s}$. The electron, 42.8 times quicker again, takes 13.6 ns.

Limits of validity

Every result above rests on assumptions. Knowing where they fail tells you when a larger theory is needed.

Sources at rest. Electrostatics assumes the charges stay put. News of a change at one charge reaches the other only after the light travel time, 0.6 ns across the animation's 180 mm, so the instantaneous Coulomb sum works only for charges that move slowly compared with light. When magnetic fields change, Faraday's law gives $\oint \vec{E}\cdot d\vec{l}\ne 0$: the work round a loop is no longer zero, and the field is no longer a pure gradient. It becomes $\vec{E}=-\nabla V-\partial \vec{A}/\partial t$, with $\vec{A}$ the magnetic vector potential.

Point charges. The inverse-square law is among the best-tested laws in physics: the exponent equals 2 to within about $10^{-15}$. The point charge, though, is an idealisation: its field energy, $\frac{1}{2}{\epsilon }_0{E}^2$ per cubic metre, adds up to infinity near a true point. At very short range quantum effects appear. Below about $4\times 10^{-13}\mathrm{m}$, the reduced Compton wavelength of the electron, a probe passes inside the cloud of quantum vacuum fluctuations that partly screens every charge. It then sees more of the charge, so the force grows slightly faster than $1/r^2$, and this measurably shifts hydrogen's energy levels. Inside the hydrogen atom the field is about $5\times 10^{11}V/m$; the electron there must be treated by quantum mechanics, though still in the Coulomb potential.

Matter and strong fields. In a material the field of a free charge is reduced by the relative permittivity, about 80 for water. Insulators also have limits: dry air breaks down and sparks at about 3 MV/m, and even empty space would start creating electron–positron pairs near $1.3\times 10^{18}V/m$.

Where this leads

In Capacitors, $E=\sigma /{\epsilon }_0$ and the energy $\frac{1}{2}{\epsilon }_0{E}^2$ per cubic metre describe the field between the plates. Electric Current and Resistance and DC Circuits and Ohm's Law follow charges driven through matter by a potential difference. Magnetic Force on Moving Charges adds $q\vec{v}\times \vec{B}$ to $q\vec{E}$, and Special Relativity shows that the two fields are parts of one: what one observer calls a pure electric field, another sees as partly magnetic. Electromagnetic Induction breaks $\oint \vec{E}\cdot d\vec{l}=0$, and Electromagnetic Waves shows the fields travelling on their own at $c=1/\sqrt{{\epsilon }_0{\mu }_0}$. The same $1/r$ potential describes gravity in Gravitation and Orbits and holds the electron in Atomic Models: the Bohr Atom, where hydrogen's binding energy is 13.6 eV.