Electric Potential & Potential Difference

Potential differences run the world across eleven orders of magnitude. Your own nervous system operates at about 70 millivolts: every neuron maintains that p.d. across its membrane, and every thought you are having right now is a wave of controlled collapses of it. Electric cars migrated from 400 V to 800 V platforms for a reason this article makes obvious — at fixed power, doubling the voltage halves the current, and thinner, cooler, lighter cables follow. The grid's transmission lines run at hundreds of kilovolts by the same per-coulomb logic (your AC chapter will finish that story). And the electron-volt has become the ruler of fundamental physics. An electron microscope accelerates electrons through up to 300 kV — worked example 2 with a bigger number — until their quantum wavelength resolves single atoms. The Large Hadron Collider's protons carry 6.8 TeV: six-point-eight trillion electron-volts, which is nothing more mysterious than one elementary charge's $Q\,\Delta V$ bill accumulated through an effective $6.8\times10^{12}$ volts of lift. The unit you learned in equation (5) scales, unchanged, from a AA cell to the energy frontier.

Picture the electrical world as terrain. Around every charge arrangement, space is a landscape of hills and valleys: positive charges build peaks, negative charges dig hollows. A positive test charge behaves exactly like a ball on this terrain — it rolls downhill, from high potential toward low, gaining kinetic energy as the ground drops. The potential is the altitude of the landscape; and just as a mountain's height exists whether or not anyone is climbing, the potential at a point exists whether or not a charge sits there. A battery, in this picture, is a ski lift. It hoists each coulomb up a fixed energy height — 9 joules of lift per coulomb for a 9 V battery, every trip, regardless of traffic. The wires are flat plateaus; the bulb is the downhill run where the altitude is cashed as light and heat; and the lift keeps the loop going. This is not a loose metaphor — lifting mass $m$ through height $h$ banks $mgh$, and pushing charge $Q$ up through potential difference $\Delta V$ banks $Q\,\Delta V$. Same bookkeeping, new landscape.

Two habits to steal from terrain-thinking immediately. First: only height differences matter. Sea level is a convention; what moves a ball is the slope between actual points. Declaring some point "0 V" — the earth, a battery's negative terminal — is bookkeeping convenience, and physics only ever responds to differences. That is the pigeon's whole survival strategy: its two feet span no difference. Second: the landscape's altitude is per coulomb, not total. A high hill with almost no charge available to roll down it (a rubbed balloon at several thousand volts) is a party trick; a modest 12 V hill with a huge reservoir behind it (a car battery) can weld a spanner. Volts price the trip; the charge that makes the trip decides the total energy.

The previous module, Electrostatic Potential Energy, priced a whole arrangement in joules. To turn that into a property of space, run one more step: take a location in someone's field, park a test charge $q$ there, and ask for the potential energy per unit of the test charge:

where: $V$ — the electric potential at the location (volts); $U$ — the potential energy of the test charge parked there (joules); $q$ — the test charge (coulombs). The division is the whole idea: $U$ doubles when $q$ doubles, so the ratio is blind to the traveller — it reports what the location itself offers. Positive $V$: a hill (positive charges resist coming here). Negative $V$: a hollow (positive charges are drawn in). Which specific arrangements build which landscapes — the $kQ/r$ hill of a point charge, most famously — is the next module's business; this one owns the currency. Between two points, the same division prices the trip. Move charge $Q$ from A to B and let $W$ be the energy converted on the way (work done on the charge, or delivered by it); then:

where: $\Delta V$ — the potential difference (what a voltmeter across A and B reads); $W$ — energy converted between A and B; $Q$ — the charge that made the trip. One volt is one joule per coulomb: $1\ \mathrm{V} = 1\ \mathrm{J/C}$. Read a battery label with it: "9 V" promises nine joules loaded onto every coulomb pumped terminal to terminal — whether the battery moves one coulomb per hour or one per second. The label fixes the per-coulomb price; the circuit decides the traffic. (How much current flows for a given price is exactly Ohm's law — the opening move of your Current Electricity chapter, which runs entirely on this equation.) In practice you run (2) backwards constantly: the voltage is known — printed on the supply, read off the meter — and the energy is what you want. Rearranged:

where: a charge $Q$ crossing a potential difference $\Delta V$ converts energy $W$ — into heat in a resistor, light in a filament, or kinetic energy in a vacuum. Two coulombs through a 6 V lamp: $2 \times 6 = 12$ J of light and heat, straight from the definition, no other physics needed. Equation (3) is the single most used equation in practical electricity, and it is also — fed one electron and one volt — the mint that coins physics' favourite small unit, the electron-volt, formalised below. One honest caveat guards all three equations: the "energy per charge" of a trip is only a well-defined price if it does not depend on the route. It does not — the electrostatic force is conservative, exactly the property that Conservative & Non-Conservative Forces certified for gravity and springs, and that the previous module used to make $U$ a function of arrangement alone. Any detour between A and B costs and refunds the same, so the trip has one price and each point has one honest altitude. (When you meet changing magnetic fields in the induction chapter, this guarantee is precisely what breaks — enjoy it while it lasts.)

Send one coulomb around a complete series circuit and follow its altitude. The battery lifts it by the full label value; each component drops it by that component's share; and arriving back at the starting terminal — the same point, hence the same potential — the coulomb's altitude is exactly where it began. Zero net change, guaranteed by definition. The lifts must equal the drops:

where: $\varepsilon$ — the battery's lift per coulomb; $V_1, V_2, \dots$ — the potential drops across the components in the loop. This is energy conservation retold per coulomb, and it is the seed of Kirchhoff's loop rule, which your circuits chapter will grow into a full method. If the books did not balance, a charge could harvest free energy by circling forever — nature declines.

Feed equation (3) the smallest charge nature sells — one elementary charge, $e = 1.6\times10^{-19}$ C — crossing one volt, and the energy is $1.6\times10^{-19}$ J. That number is so relentlessly useful that it gets its own name:

where: one electron-volt is the energy gained by one elementary charge crossing a potential difference of one volt. It is a unit of energy, not of voltage — the name trips half of every cohort. Its magic is mental arithmetic: an electron accelerated through 3000 V gains exactly 3000 eV, read straight off the voltage with no calculation; the 7.2 eV that the previous module computed for two protons at atomic spacing announces "chemistry-scale" at a glance. Ionisation energies, photon energies, particle beams — all quoted in eV, keV, MeV, GeV. Learn to hop between joules and eV in one multiplication; competitive papers assume it is reflex.

Board level — battery bookkeeping (ISC/IGCSE/IB pattern)

A battery does $60\ \mathrm{J}$ of work driving $5.0\ \mathrm{C}$ of charge around a circuit containing a bulb and a resistor in series. The bulb converts $35\ \mathrm{J}$ of that energy. Find (a) the battery's terminal potential difference, (b) the p.d. across the bulb, (c) the p.d. across the resistor.

Step 1. (a) The battery's price per coulomb, by equation (2): $\Delta V = \dfrac{W}{Q} = \dfrac{60}{5.0} = \boxed{12\ \mathrm{V}}$ — every coulomb receives 12 J.

Step 2. (b) The bulb takes 35 J from those same 5.0 coulombs: $V_{\text{bulb}} = \dfrac{35}{5.0} = \boxed{7.0\ \mathrm{V}}$.

Step 3. (c) Energy not yet spent must be spent in the resistor — the loop's books balance (equation 4): $V_{\text{res}} = 12 - 7.0 = \boxed{5.0\ \mathrm{V}}$. Cross-check by energy: $(60 - 35)/5.0 = 5.0$ V ✓.

Step 4. The marks-scheme sentence: "potential difference is the energy converted per unit charge, $V = W/Q$" — writing the definition in words is routinely a mark of its own on board papers.

Competitive level — the electron gun (JEE-Main/NEET/AP-2 pattern)

An electron starts from rest and is accelerated through a potential difference of $3000\ \mathrm{V}$ in a vacuum tube. Find its final kinetic energy in joules and electron-volts, and its final speed. Would a proton accelerated through the same p.d. move faster or slower, and by what factor? ($e = 1.6\times10^{-19}$ C, $m_e = 9.1\times10^{-31}$ kg, $m_p = 1.67\times10^{-27}$ kg.)

Step 1. Energy first, always: $W = e\,\Delta V = (1.6\times10^{-19})(3000) = \boxed{4.8\times10^{-16}\ \mathrm{J}}$. In the atomic currency, read it off the voltage: $\boxed{3000\ \mathrm{eV} = 3.0\ \mathrm{keV}}$ — zero arithmetic, that is the entire point of the unit.

Step 2. All of it becomes kinetic energy (vacuum, starts from rest): $\tfrac12 m_e v^2 = 4.8\times10^{-16}$ J.

Step 3. $v = \sqrt{\dfrac{2 \times 4.8\times10^{-16}}{9.1\times10^{-31}}} = \sqrt{1.05\times10^{15}} = \boxed{3.2\times10^{7}\ \mathrm{m/s}}$ — about 11% of light speed from a lab-bench voltage, because the electron's mass is so tiny. (Fast enough that relativity is knocking; at 30 kV it must be let in. State the check, earn the maturity mark.)

Step 4. The proton gains the same energy — equation (3) does not care about mass — but carries 1836 times the mass: $v \propto \sqrt{1/m}$ at fixed energy, so the proton is slower by $\sqrt{1836} \approx 43$. Numerically $v_p = 7.6\times10^5$ m/s. Same till, different getaway speeds. Timing note: 2 minutes. The kill-shots are (i) energy from $Q\Delta V$ before any mechanics, (ii) the $\sqrt{m_p/m_e}$ ratio done symbolically — never compute the proton's speed from scratch when a ratio is asked.

"Volts through" costs marks. P.d. is across; current is through. IGCSE and IB mark schemes look for the preposition explicitly; JEE distractors are built from the confusion.

Potential vs potential energy. $V$ belongs to a place (J/C); $U = qV$ belongs to a charge at that place (J). Doubling the charge doubles $U$ and leaves $V$ alone. Any question asking "what happens to the potential when the test charge is doubled" is testing exactly this — answer: nothing.

The electron's direction. Negative charge moves spontaneously toward higher potential (its energy $U = qV$ falls when $V$ rises, because $q < 0$). Free positive charge → lower $V$; free electron → higher $V$. Both lose potential energy; the field never pushes any free charge to higher $U$.

eV is energy, not voltage. "An electron at 500 eV" describes kinetic energy, not a 500 V location. Convert with one multiplication: $\mathrm{eV} \times 1.6\times10^{-19} = \mathrm{J}$. NEET options love planting the unconverted number.

Signs in $W = Q\Delta V$. Carry the sign of $Q$ and the sign of $\Delta V$; the sign of $W$ then reports who paid (positive: energy delivered to the charge's motion or surroundings along the drop; negative: an agent had to push). If a sign hunt confuses you mid-exam, fall back to energy logic — "does this charge want to make this trip?" — and set the sign by physics.

Constants and conventions. $e = 1.6\times10^{-19}$ C everywhere; boards state electron mass when needed ($9.1\times10^{-31}$ kg). CBSE/ISC expect the definition in words for one mark; IB expects $V = W/q$ quoted as in the data booklet; A-Level papers write V = W/Q at AS and expect fluent rearrangement.

Texts. NCERT Physics Part 1 (Class 12), Ch. 2 §2.1–2.3 — the CBSE/NEET canonical definitions · IGCSE 0625 syllabus §4.2.3 and IB Physics guide topic B.5, where $V = W/Q$ is the examinable statement · H.C. Verma, Concepts of Physics 2, Ch. 29 · Halliday, Resnick & Walker Ch. 24 for the AP-2/A-Level register.

Provenance. All relations are standard ($V = U/q$, $\Delta V = W/Q$, $W = Q\Delta V$, the loop balance, the eV). A verifier should recompute: battery example $60/5.0 = 12$ V, $35/5.0 = 7.0$ V, $12 - 7.0 = 5.0$ V with loop check $7 + 5 = 12$; electron gun $W = 1.6\times10^{-19} \times 3000 = 4.8\times10^{-16}$ J $= 3.0$ keV, $v = \sqrt{2W/m_e} = 3.2\times10^7$ m/s ($0.11c$), proton factor $\sqrt{1836} = 42.8$, $v_p = 7.6\times10^5$ m/s; lamp check $2\ \mathrm{C} \times 6\ \mathrm{V} = 12$ J.

Prerequisites. Electrostatic Potential Energy · The Electric Field · Conservative & Non-Conservative Forces. Leads to. Potential of Point Charges & Charge Systems (the landscapes themselves) · Equipotentials & the Field–Potential Link · Electric Current & Charge Flow (coming in E3 — the chapter this module bankrolls).