How batteries work Chapter 0 Electrostatics

Chapter 0

Electrostatics for electrochemistry: charge, field, potential and current

A torch cell says 1.5 V on its side. What is a volt, and what is being pushed?

Where we are. Before the first cell, four words that every later screen uses: charge, the thing that pushes and is pushed; field, the push per unit charge at a place; potential, the energy per unit charge at a place; voltage, a difference of potential. Each gets one figure, one equation and one number.

Figure 0.1 Two point charges a distance r apart, with the force on each drawn as an arrow; the force is the same size on both, falls with the square of r, and reverses when one sign is flipped.
positive chargenegative chargeforcefield
Coulomb’s law. Two charges of one elementary charge each, in vacuum. Try this: set r to 1 nm, read the force, then set 2 nm and check that it is a quarter; then flip one sign.
SourceComputed from Coulomb’s law in SI form, F = qq′/(4πε0r2), and ε0 = 8.85419 × 10−12 C2 N−1 m−2, as printed in Bard, Faulkner and White 2022, section 14.3.1 (footnote 6) and 2.2.1; e = 1.602 × 10−19 C (CODATA). Arrow lengths scale with the square root of the force so that both ends of the slider stay visible.

Everything on this page moves because charges push on one another. Charge comes in two signs, and the names are a convention older than the electron: Benjamin Franklin called an excess of what we now call electronic charge negative and a deficiency of it positive.1

Between two charges q and q′ a distance r apart there is a force F = qq′/(4πε0r2), Coulomb’s law, where ε0 is a measured constant, the permittivity of free space.1 Read the formula: the force grows with either charge, falls with the square of the distance, and its sign follows the product of the charges, so like charges push apart and unlike charges pull together.

The unit of charge is the coulomb, and one electron carries 1.602 × 10−19 C of it. Put two electrons one nanometre apart, about the width of three water molecules, and the formula gives 0.23 nN pushing them apart: tiny in newtons, enormous for something that weighs 10−30 kg.

Charges push on each other with a force that falls as the square of the distance.

Figure 0.2 The electric field around one positive charge drawn as arrows that point outward and shorten with distance, a probe charge whose force is shown, and a graph of the potential against distance with the probe marked on it.
field (force per unit charge)potential φforce on the probe
Field and potential around a single charge. The map on the left and the graph on the right describe the same push in two languages. Try this: slide the probe from 1 nm to 4 nm and watch the force arrow shrink four times faster than the potential falls.
SourceField as the force on a unit charge and potential as φ = −∫E·dl from Bard, Faulkner and White 2022, section 2.2.1 (eqs. 2.2.1 and 2.2.2). The 1/r curve is that integral evaluated with Coulomb’s law for one elementary charge; potential zero taken at infinity.

Take the second charge away and ask what is left at its place. The answer is the electric field: the force that would act on a unit charge put there, a vector with a direction and a strength at every point in space.1 Around a positive charge it points outward and falls as 1/r2, because that is how the force falls.

Now bring a unit positive charge in from very far away, pushing against that field all the way. The work you do is the electric potential φ at the point you reach, and it does not depend on the path you took: φ = −∫E·dl.1

Only differences of φ between two points can be measured; the value at one point is a bookkeeping choice, and the usual choice sets φ = 0 infinitely far away.1 For a single elementary charge the integral gives φ = q/(4πε0r), which is 1.44 V one nanometre out and half that at two.

The two pictures are one: the field is the slope of the potential. Where φ falls steeply the push is strong; where φ is flat there is no push at all. That last case is about to matter.

Field is force per unit charge; potential is energy per unit charge; the field is the slope of the potential.

Figure 0.3 Two metal plates held at a potential difference V with a uniform field between them, a test charge that is pushed downhill if positive and uphill if negative, and a strip below showing the potential flat inside each plate and falling in a straight line across the gap.
fieldpotential φforce on the test charge
Voltage as a potential hill. Two conductors, a gap, and a charge that rolls. Try this: flip the test charge to −e, press Play, and watch it climb toward the higher potential; then double V and watch the field arrows.
SourceConductor at equilibrium (no field inside, one potential, excess charge on the surface) from Bard, Faulkner and White 2022, section 2.2.1; the linear potential and field E = V/l between parallel plates from their eq. 4.2.3; the energy qΔφ and the electron-volt from section 1.1.4 (footnote 9); 1 V = 1 J/C from section 1.1.2. Field strength computed; the strip is drawn to scale in volts.

A voltage is a difference of potential between two points, and its unit, the volt, is one joule per coulomb: the energy available to move each coulomb of charge from one point to the other.1 Move a charge q across a potential difference Δφ and its energy changes by qΔφ.1

For one electron across one volt that change is one electron-volt, and per mole of electrons it is 96.5 kJ, comparable to the energies of chemical bonds and reactions; which is why a few volts are enough to drive chemistry.1

Metals are conductors: they contain charges that are free to move. Leave a metal alone and those charges settle until there is no field left inside it, because any field would move them further; so a conductor at rest sits at a single potential throughout, and any excess charge it carries lies on its surface.1

The plates in the figure are two such flat regions of the potential; between them, where there is no conductor, the potential falls in a straight line and the field is simply V divided by the gap.1

A picture to keep: potential is height, field is slope, and a positive charge rolls downhill while a negative one is pushed uphill. The picture breaks in one place: charges also push on each other and rearrange, so the hill is not fixed scenery. Chapter 3 is about how a cell builds its own hill.

Worked example. How much charge does it take to raise a metal by one volt? Bard, Faulkner and White give the answer for a mercury drop of 0.5 mm radius in vacuum: about 5 × 10−14 C per volt.1 electrons per volt = 5 × 10−14 C ÷ 1.602 × 10−19 C ≈ 3 × 105 Three hundred thousand electrons, out of the 1020 or so in the drop, shift its potential by a full volt. Potentials are set by minute charge imbalances, which is why they are so easy to change and so hard to see.

A volt is a joule per coulomb; a conductor at rest is all at one potential.

Figure 0.4 A cell, a switch and a lamp joined by a wire, with electrons drawn moving from the negative terminal through the wire to the positive one, and a strip below showing the potential rising inside the cell and falling along the path, mostly across the lamp; with the switch open the whole drop sits across the switch.
electronfield along the wirepotential along the path
Why a current needs a voltage. The strip unrolls the whole loop, starting and ending at the − terminal. Try this: open the switch and watch where the drop goes; then close it and raise R until the lamp is barely lit.
SourceElectrons move through the wire from the more negative electrode to the more positive one (Bard, Faulkner and White 2022, 1.1.1); current as charge per second, i = dQ/dt (1.1.5); a potential difference is needed to drive current through a resistance (2.2.1, footnote 15) and metals take only a small share because they conduct far better than electrolytes (1.5, footnote 39); Ohm’s law E = IR from Winter and Brodd 2004, section 2.2; P = IV from Goodenough and Park 2013. The split of the drop, 3 % in each wire and the rest in the lamp, is illustrative; I and P are computed.

A cell keeps its two terminals at a fixed potential difference; chapter 1 shows how. Join them with a wire and electrons flow through the wire from the more negative terminal to the more positive one.1 The amount of charge passing a point each second is the current, i = dQ/dt, measured in amperes: one ampere is one coulomb per second.1

A wire carrying current is no longer at one potential. Between any two points on it there must be a potential difference to drive the current through the resistance between them.1 That is Ohm’s law, V = IR: the same current through twice the resistance needs twice the voltage.2

The resistance of a piece of conductor grows with its length and shrinks with its cross-section, R = ρl/A, where ρ is the resistivity of the material.1 Metals have a resistance too, but it is so much smaller than that of an electrolyte that on the scale of a cell the drop along the copper is invisible.1 In the figure, nearly the whole voltage is spent across the lamp.

Energy per coulomb times coulombs per second is energy per second: the power delivered is P = IV, in watts.3

Worked example. A 1.5 V cell drives a lamp of 3 Ω. I = V / R = 1.5 V / 3 Ω = 0.5 A, that is 0.5 C every second electrons per second = 0.5 C ÷ 1.602 × 10−19 C = 3.1 × 1018 P = I V = 0.5 A × 1.5 V = 0.75 W Each electron gives up 1.5 eV on its way round; three billion billion of them a second add up to three quarters of a watt of light and heat.

Current needs a potential difference to drive it through resistance; the power is current times voltage.

Figure 0.5 A liquid between two electrodes with a field across it; a cation is pushed toward the negative side and an anion toward the positive side, each reaching a steady drift speed when the drag of the liquid balances the electric force; a strip shows the potential sloping across the liquid.
our ion (cation)anionelectric forcedragfield
How an ion moves through a liquid. Our ion, the blue cation, will be followed by name through the rest of the page. Try this: double the ion radius and watch the drift halve; double the field and watch it double. Step through the launch to see the drag arrow grow until it matches the push.
SourceForce |z|eE, Stokes drag 6πηrv, terminal velocity and mobility u = |z|e/(6πηr) from Bard, Faulkner and White 2022, section 2.3.3 and Figure 2.3.4; cations driven down the sloping potential and anions up from their section 1.5.1; conductivity κ = FΣ|z|uC from eq. 2.3.10; electroneutrality from section 2.3.3. Speeds and forces are in relative units: no viscosity or radius is quoted because the sources give none for a battery electrolyte.

In a liquid the mobile charges are not electrons but ions, atoms or molecules that have lost or gained electrons; free electrons in a solution are so short-lived that they carry no current worth counting.1 Put an ion of charge number z in a field E and it feels the force |z|eE. As it speeds up the liquid drags on it, with a drag that grows with speed; when the drag balances the push the ion stops accelerating and drifts at a steady terminal velocity.1 The velocity per unit field is the ion’s mobility, u = |z|e/(6πηr): a smaller ion in a thinner liquid moves faster.1

When current flows through a liquid the potential slopes across it, exactly as it did along the wire, and the slope is the field: cations are driven down the slope and anions up it.1 Both carry current, and the liquid’s conductivity adds up every kind of ion, its charge, its mobility and its concentration: κ = FΣ|z|uC.1

One more rule holds everywhere in the liquid: on any scale you can see, positive and negative charge balance. An imbalance cannot build up, because it would create an enormous field that would move ions until it was gone.1

The constant F in that formula is the Faraday constant, the charge carried by one mole of electrons, 96 485.3 C.1,2 It is the bridge between the chemist’s moles and the physicist’s coulombs, and it returns in chapter 4 to count how much charge a gram of material can hold.

You can now explain what a volt is, why a field makes charges move and which way, why a current needs a potential difference to drive it, and why an ion in a liquid drifts at a steady speed rather than accelerating.

Now put two different metals into a liquid and see where a potential difference comes from.

Go deeper: Gauss’s law, and why the charge sits on the surface

Draw an imaginary closed surface anywhere. Gauss’s law says that the net charge inside it equals ε0 times the integral of the field over that surface, q = ε0∮E·dS.1 Inside a conductor at rest the field is zero everywhere, so the integral over any surface drawn inside the conductor is zero, so the charge inside is zero: whatever excess charge the conductor carries must lie on its boundary.1 In an electrolyte the same argument leaves the charge at the boundary too, but thermal motion smears it into a layer that can be tenths of a nanometre to thousands of nanometres thick, the space charge region; in a metal that layer is negligibly thin.1 Chapter 3 uses this to build the double layer at an electrode.

Bring an electrolyte next to the metal and the coulombic field of the metal’s surface charge is largely cancelled by the polarization of the liquid: the same mercury drop then needs about 10−6 C, some 6 × 1012 electrons, for a one-volt change, more than 107 times what it needed in vacuum.1

The mathematics of this chapter

Everything below is the equation behind a figure above, in the form Bard, Faulkner and White give it. Quantities: q, q′ charges (C); r distance (m); ε0 = 8.854 × 10−12 C2 N−1 m−2; e = 1.602 × 10−19 C; F = 96 485.3 C/mol.

F = qq′ / (4πε0r2)Coulomb’s law; positive means repulsion (figure 0.1)

E = F / q′ = q / (4πε0r2)the field of a point charge: force per unit charge, pointing away from a positive q

φ(r) = −∫∞r E·dl = q / (4πε0r)the potential: work per unit charge from infinity, path-independent (figure 0.2)

φ(B) − φ(A) = −∫AB E·dla potential difference is the field integrated along any path; in one dimension, E = −dφ/dx: the field is the slope of the potential

qenclosed = ε0 ∮ E·dSGauss’s law; with E = 0 inside a conductor at rest, the enclosed charge is zero and the excess charge sits on the surface

E = V / lthe uniform field between two parallel plates a distance l apart (figure 0.3)

ΔU = q Δφenergy change of a charge moved across Δφ; for q = e and Δφ = 1 V, ΔU = 1 eV = 1.602 × 10−19 J

F = NA e = 96 485.3 C/molthe Faraday constant: one electron-volt per electron is F × 1 V = 96.5 kJ per mole

i = dQ/dtcurrent, in amperes = coulombs per second (figure 0.4); at 1 A, 1/e = 6.24 × 1018 electrons pass each second

V = IR, R = ρl/A, G = 1/R = κA/lOhm’s law; resistance grows with length and falls with cross-section; ρ resistivity, κ conductivity

P = IVpower delivered, in watts

|z|eE = 6πηrv ⇒ u = |z|e / (6πηr), v = uEan ion of charge number z in a liquid of viscosity η: the electric force balances the Stokes drag at the terminal velocity v; u is the mobility (figure 0.5)

κ = F Σj |zj| uj Cj, tj = |zj|ujCj / Σk |zk|ukCkthe conductivity of an electrolyte adds up every ion’s charge, mobility and concentration; tj, the transference number, is the share of the current that ion j carries

Jj = −Dj ∇Cj − (zjF/RT) Dj Cj ∇φthe Nernst–Planck equation: an ion’s flux has a diffusion part (down the concentration gradient) and a migration part (along the field); with Dj = ujRT/(|zj|F) the two pictures agree. The modules on ion transport, still to come, are built on it.

References for this chapter

Numbers in the text refer to this list. All references of the course, with the chapters that use each one.

  1. 1 A. J. Bard, L. R. Faulkner, H. S. White, "Electrochemical Methods: Fundamentals and Applications", Wiley, 3rd edition, 2022.
  2. 2 M. Winter, R. J. Brodd, "What Are Batteries, Fuel Cells, and Supercapacitors?", Chemical Reviews 104, 2004, 4245-4269. doi:10.1021/cr020730k
  3. 3 J. B. Goodenough, K.-S. Park, "The Li-Ion Rechargeable Battery: A Perspective", Journal of the American Chemical Society 135, 2013, 1167-1176. doi:10.1021/ja3091438

Cite this page

Using this page in teaching, a thesis or an article? Cite it as follows; the date is the last update of the course.

Shyma Sajeevan, A. (2026). Electrostatics for electrochemistry: charge, field, potential and current. How batteries work. https://asajeevan.github.io/batteries/electrostatics/

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