Electric Potential

We have met the electric field, which measures the force per unit charge at a point in space. In this lesson we meet its partner, the electric potential, which measures the potential energy per unit charge. The two are related in exactly the same way as $g$g and $\phi_{\text{grav}}$ϕgrav​ in the gravitational case — but with a crucial twist. Because charge can be positive or negative, electric potential can be positive or negative — different from gravitational potential, which is always negative (gravity is only attractive). OCR places particular emphasis on the analogy and contrast between electric and gravitational fields, and this lesson is the cleanest place to draw it.

Spec mapping (OCR H556 Module 6.2 — electric fields). This lesson covers the definition of electric potential as work done per unit positive charge, $V = W/Q$V=W/Q; the potential of a point charge $V = Q/(4\pi\varepsilon_0 r)$V=Q/(4πε0​r) with sign convention positive-for-positive and negative-for-negative; the potential energy of a charge pair $E_p = Q_1 Q_2/(4\pi\varepsilon_0 r)$Ep​=Q1​Q2​/(4πε0​r); equipotential surfaces and their perpendicularity to field lines; the relation $E = -\mathrm{d}V/\mathrm{d}r$E=−dV/dr between field and potential gradient; and the structural comparison with gravitational potential.

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Work Done Moving a Charge

Move a small positive test charge $q$q from point A to point B in an electric field. The field exerts a force on it, and so work is done either by the field (if the charge moves in the direction of the force) or against the field (if it moves in the opposite direction).

We define the potential difference between A and B as the work done per unit charge in moving the test charge from A to B:

$V_B - V_A = \frac{W_{A \to B}}{q}.$VB​−VA​=qWA→B​​.

The electric potential at a point is the work done per unit positive charge in bringing a small test charge from infinity to that point:

$V = \frac{W_\infty}{Q}.$V=QW∞​​.

Unit: volt (V) = joule per coulomb. This is the same volt you have used for e.m.f., potential difference and battery voltage — the concept runs right through GCSE electricity into A-Level electrostatics. A $1.5$1.5 V battery gives $1.5$1.5 J to every coulomb that passes through it.

Sign of the work

The definition uses a small positive test charge brought in from infinity. If the source charge is positive, the test charge experiences repulsion all the way in, so work is done against the field — $W_\infty > 0$W∞​>0 and $V > 0$V>0. If the source charge is negative, the test charge is attracted in, so work is done by the field — $W_\infty < 0$W∞​<0 and $V < 0$V<0. The sign of $V$V tracks the sign of the source charge, the opposite of the gravitational case.

W = qΔV — the work-energy bridge

For any charge $q$q (positive or negative) moved between two points of potential difference $\Delta V = V_B - V_A$ΔV=VB​−VA​, the work done by the electric field on the charge is

$W_{\text{field}} = -q\,\Delta V,$Wfield​=−qΔV,

and the work done on the charge by an external agent (assuming it ends up at rest) is

$W_{\text{ext}} = +q\,\Delta V.$Wext​=+qΔV.

Both signs of $q$q are allowed here, and the bookkeeping is important. A positive charge moved to higher potential gains potential energy; a negative charge moved to higher potential loses potential energy. The sign of $q$q matters.

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Potential Around a Point Charge

For a single point charge $Q$Q, the electric potential at distance $r$r is

$V = \frac{Q}{4\pi\varepsilon_0 r}.$V=4πε0​rQ​.

Note the $1/r$1/r dependence, not $1/r^2$1/r2. Potential falls off more slowly with distance than field strength does. A very common exam slip is to write $V = Q/(4\pi\varepsilon_0 r^2)$V=Q/(4πε0​r2) — that is the field formula, not the potential.

The potential is positive around a positive charge and negative around a negative charge. At infinity, $V \to 0$V→0 — infinity is the natural zero of electric potential.

Distance	$V$V if $Q = +1$Q=+1 nC	$V$V if $Q = -1$Q=−1 nC
$0.10$0.10 m	$+89.9$+89.9 V	$-89.9$−89.9 V
$0.50$0.50 m	$+18.0$+18.0 V	$-18.0$−18.0 V
$1.00$1.00 m	$+8.99$+8.99 V	$-8.99$−8.99 V
$10.0$10.0 m	$+0.899$+0.899 V	$-0.899$−0.899 V
$\infty$∞	$0$0 V	$0$0 V

Worked Example 1 — Potential Around a Point Charge

Calculate the electric potential $25$25 cm from a point charge of $+4.0$+4.0 nC.

$V = \frac{Q}{4\pi\varepsilon_0 r} = \frac{(4.0 \times 10^{-9})(8.99 \times 10^9)}{0.25} = \frac{35.96}{0.25} \approx 144\ \text{V}.$V=4πε0​rQ​=0.25(4.0×10−9)(8.99×109)​=0.2535.96​≈144 V.

Worked Example 2 — Adding Potentials (scalar superposition)

Two point charges $Q_1 = +5.0$Q1​=+5.0 nC and $Q_2 = -3.0$Q2​=−3.0 nC are placed $10$10 cm apart. Find the potential at the midpoint of the line joining them.

Reasoning. Electric potential is a scalar: just add the contributions algebraically, with signs. Each charge is $5.0$5.0 cm from the midpoint.

$V_1 = \frac{(5.0 \times 10^{-9})(8.99 \times 10^9)}{0.050} = +899\ \text{V}$V1​=0.050(5.0×10−9)(8.99×109)​=+899 V

$V_2 = \frac{(-3.0 \times 10^{-9})(8.99 \times 10^9)}{0.050} = -539\ \text{V}$V2​=0.050(−3.0×10−9)(8.99×109)​=−539 V

$V_{\text{mid}} = V_1 + V_2 \approx +360\ \text{V}.$Vmid​=V1​+V2​≈+360 V.

Note: at the midpoint, the field vectors from each charge both point from the positive toward the negative charge, so they add as vectors. But the potentials are scalars and we just sum the signed numbers. This is one of the great labour-saving features of working with potential.

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Potential Energy of a Charge Pair

The potential energy of a charge $Q_2$Q2​ sitting in the field of another charge $Q_1$Q1​ at separation $r$r is the work done bringing $Q_2$Q2​ from infinity to its current position. Since $V_1 = Q_1/(4\pi\varepsilon_0 r)$V1​=Q1​/(4πε0​r) is the potential at distance $r$r from $Q_1$Q1​,

$E_p = Q_2 V_1 = \frac{Q_1 Q_2}{4\pi\varepsilon_0 r}.$Ep​=Q2​V1​=4πε0​rQ1​Q2​​.

Signs matter — and you must substitute them literally:

• Like charges ($Q_1 Q_2 > 0$Q1​Q2​>0): positive potential energy. Work was done against the repulsion to assemble the pair; energy would be released if they flew apart.
• Unlike charges ($Q_1 Q_2 < 0$Q1​Q2​<0): negative potential energy. Like a bound state of gravitating masses, you would have to do positive work against the attraction to pull them apart to infinity.

This sign convention is the same as in mechanics: $E_p < 0$Ep​<0 for a bound state, $E_p = 0$Ep​=0 at infinity, $E_p > 0$Ep​>0 for a "stressed" configuration that wants to fly apart.

Worked Example 3 — Energy to Separate Two Charges

Two charges of $+2.0$+2.0 nC and $-3.0$−3.0 nC are $5.0$5.0 cm apart. How much work must be done by an external agent to separate them to infinity?

$E_p = \frac{Q_1 Q_2}{4\pi\varepsilon_0 r} = \frac{(+2.0 \times 10^{-9})(-3.0 \times 10^{-9})(8.99 \times 10^9)}{0.050}$Ep​=4πε0​rQ1​Q2​​=0.050(+2.0×10−9)(−3.0×10−9)(8.99×109)​

$E_p = \frac{-5.394 \times 10^{-8}}{0.050} \approx -1.08 \times 10^{-6}\ \text{J}.$Ep​=0.050−5.394×10−8​≈−1.08×10−6 J.

The potential energy of the bound pair is about $-1.08$−1.08 $\mu$μJ. To separate them to infinity, where $E_p = 0$Ep​=0, the external agent must do positive work of magnitude $1.08\ \mu$1.08 μJ. This is the binding energy of the pair.

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The Gravitational Analogy — OCR's Favourite Table

OCR H556 Module 6.2 explicitly asks candidates to compare electric fields with gravitational fields. Here is the comparison in full:

Quantity	Gravitational	Electric
Source	Mass $m$m (positive only)	Charge $Q$Q ($\pm$±)
Force on probe	$F = mg$F=mg	$F = qE$F=qE
Field strength (point source)	$g = GM/r^2$g=GM/r2	$E = Q/(4\pi\varepsilon_0 r^2)$E=Q/(4πε0​r2)
Potential (point source)	$\phi = -GM/r$ϕ=−GM/r	$V = Q/(4\pi\varepsilon_0 r)$V=Q/(4πε0​r)
Potential energy of pair	$E_p = -Gm_1 m_2/r$Ep​=−Gm1​m2​/r	$E_p = Q_1 Q_2/(4\pi\varepsilon_0 r)$Ep​=Q1​Q2​/(4πε0​r)
Sign of force	Always attractive	Attractive (unlike) or repulsive (like)
Sign of potential	Always negative	$\pm$± depending on sign of $Q$Q
Constant	$G \approx 6.67 \times 10^{-11}$G≈6.67×10−11	$1/(4\pi\varepsilon_0) \approx 8.99 \times 10^9$1/(4πε0​)≈8.99×109
Shielding	Impossible	Possible (conductors)
Zero of potential	At infinity ($\phi \to 0$ϕ→0)	At infinity ($V \to 0$V→0)

Key structural similarities:

1. Both have an inverse-square force law and an inverse-distance potential.
2. Both define field as force per unit "test quantity".
3. Both define potential as potential energy per unit test quantity.
4. Both take the zero of potential at infinity.

Key structural differences (mark-scheme gold):

1. Gravity is always attractive, so $\phi_{\text{grav}}$ϕgrav​ is always negative for an isolated mass. Electric potential can be positive or negative depending on the sign of the source.
2. The minus sign in $\phi = -GM/r$ϕ=−GM/r is definitional (because gravity is only attractive). The minus sign that may appear in $V = Q/(4\pi\varepsilon_0 r)$V=Q/(4πε0​r) comes from the sign of the source charge, not the law itself.
3. Electric forces are vastly stronger per unit source and can be shielded by a conductor.

Exam tip. If OCR ask you to compare the two fields, draw this table from memory. You will usually get a mark for each correct line, plus marks for the qualitative points about sign, strength and shielding.

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Field as the Gradient of Potential

A deep connection: the electric field strength is the negative gradient of the potential,

$E = -\frac{\mathrm{d}V}{\mathrm{d}r}.$E=−drdV​.

The negative sign says the field points from high potential to low potential — downhill on the potential "landscape". Verify for a point charge: $V = Q/(4\pi\varepsilon_0 r)$V=Q/(4πε0​r), so

$\frac{\mathrm{d}V}{\mathrm{d}r} = -\frac{Q}{4\pi\varepsilon_0 r^2}, \qquad E = -\frac{\mathrm{d}V}{\mathrm{d}r} = +\frac{Q}{4\pi\varepsilon_0 r^2}.$drdV​=−4πε0​r2Q​,E=−drdV​=+4πε0​r2Q​.

The same relationship holds for the uniform field between parallel plates: if the plates are separated by $d$d with potential difference $V$V, then

$E = \frac{V}{d}$E=dV​

(pointing from the + plate to the $-$− plate) is exactly the gradient of the linearly falling potential. So "field is gradient of potential" recovers all the field formulae you already know.

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Equipotentials

An equipotential is a surface on which the electric potential is the same everywhere. Two properties:

1. No work is done moving a charge along an equipotential, because $W = q\Delta V$W=qΔV and $\Delta V = 0$ΔV=0 on the surface.
2. Equipotentials are perpendicular to field lines at every point. If they weren't, the field would have a component along the equipotential, and moving a charge along it would require work — contradicting property 1.

For a point charge, equipotentials are spheres centred on the charge. For parallel plates, they are planes parallel to the plates. For a dipole (one + and one $-$−), the equipotentials are more interesting curves that bulge between the two charges.

flowchart LR
    A[Path along equipotential] -->|ΔV = 0| B[Work done = 0]
    C[Field component along path?] -->|yes| D[Work done ≠ 0 — contradiction]
    D --> E[Therefore field perpendicular to equipotential]
    B --> E

The mermaid above encapsulates the proof that equipotentials are perpendicular to field lines: it is a corollary of $W = q\Delta V$W=qΔV.

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Worked Example 4 — Electron Accelerated Through a Potential Difference

An electron is accelerated from rest through a potential difference of $2000$2000 V. Find its final speed.

Energy gained. A negative charge $-e$−e moved through $\Delta V = +2000$ΔV=+2000 V (from $-$− plate at low $V$V to $+$+ plate at high $V$V) has its potential energy decrease by

$\Delta E_p = q\,\Delta V = (-e)(+2000) = -3.2 \times 10^{-16}\ \text{J}.$ΔEp​=qΔV=(−e)(+2000)=−3.2×10−16 J.

Energy conservation: the kinetic energy gained equals the potential energy lost, so

$\tfrac{1}{2}m_e v^2 = +3.2 \times 10^{-16}\ \text{J}.$21​me​v2=+3.2×10−16 J.

Solve for v.

$v^2 = \frac{2 \times 3.2 \times 10^{-16}}{9.11 \times 10^{-31}} \approx 7.03 \times 10^{14}\ \text{m}^2\text{ s}^{-2}$v2=9.11×10−312×3.2×10−16​≈7.03×1014 m2 s−2

$v \approx 2.65 \times 10^7\ \text{m s}^{-1}.$v≈2.65×107 m s−1.

About 9 % of the speed of light — relativistic corrections are starting to bite but classical mechanics is still within about 0.5 %. The sign-of-charge bookkeeping is the conceptual content: a negative charge accelerated toward higher potential gains kinetic energy.

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