Electric Potential and Energy

Electric potential provides an energy-based description of electric fields, just as gravitational potential does for gravitational fields. This topic (AQA 3.7.3) introduces the concept of potential, equipotentials, and energy changes when charges move through fields, culminating in Millikan's oil drop experiment.

Spec mapping (AQA 7408 §3.7.3 — Electric potential): This lesson covers the definition of electric potential V as work done per unit positive charge from infinity, the radial form V = Q/(4πε₀r) with its sign convention (positive near positive source charges, negative near negative source charges), equipotential surfaces and their relation to field lines (perpendicular intersections; no work along an equipotential), the scalar superposition of potentials due to multiple charges, the relationship E = −dV/dr, work done moving a charge W = qΔV, and Millikan's oil drop experiment as the quantitative demonstration of charge quantisation. Refer to the official AQA 7408 specification document for the exact wording and any limitations.

Synoptic links — where this resurfaces in 7408:

§3.7.2 Gravitational potential (Lesson 1 here) — same V = −GM/r vs V = +kQ/r mathematical structure; sign asymmetry is the headline difference and is a frequent compare-and-contrast prompt.

§3.7.4 Capacitance (Lessons 4–6 here) — the energy stored in a capacitor U = ½QV depends on the parallel-plate potential difference; the derivation hinges on the V = Q/C linear law.

§3.2.1 Quantum physics — Millikan's e measurement and the photoelectric effect (Einstein, 1905) jointly established the quantum nature of charge and the equivalence E = hf. Both used charged-drop / charged-plate apparatus; the historical synoptic link is examined.

§3.5 Electricity (§3.5.1.4 Potential difference and EMF) — V = W/Q is the very definition of potential difference; reading "potential" in the §3.7.3 sense and "potential difference" in the §3.5 sense are the same physics applied at different scales.

§3.4.1.7 Work, energy and power — W = QΔV is the electromagnetic instance of the general W = F·d framework; conservation of energy underpins every "work done" sub-question.

§3.7.5 Magnetic fields — particles accelerated through a known potential difference gain KE = qV; this links to mass spectrometers and Wien filters.

Electric Potential

Key Definition: The electric potential (V) at a point in an electric field is the work done per unit positive charge in bringing a small positive test charge from infinity to that point.

For a point charge Q:

V = kQ/r = Q/(4πε₀r)

where V is the electric potential (V or J C⁻¹), k = 8.99 × 10⁹ N m² C⁻², Q is the source charge (C), and r is the distance from the charge (m).

Key differences from gravitational potential:

Electric potential can be positive (near a positive charge) or negative (near a negative charge).
Gravitational potential is always negative.
The sign of V matches the sign of the source charge Q.

Worked Example 1 — Electric Potential at a Point

Question: Calculate the electric potential at a distance of 0.20 m from a point charge of +4.0 μC.

Solution:

V = kQ/r = (8.99 × 10⁹ × 4.0 × 10⁻⁶) / 0.20

V = 3.596 × 10⁴ / 0.20

V = 1.80 × 10⁵ V = 180 kV

Since the charge is positive, the potential is positive. A positive test charge would need to have work done on it to bring it from infinity to this point (against the repulsive force).

Equipotential Lines and Surfaces

Equipotential surfaces connect points of equal electric potential.

Described diagram — Equipotentials around a positive charge: Concentric circles centred on the charge, with potential values decreasing (but remaining positive) further from the charge. Field lines radiate outward and are perpendicular to every equipotential circle.

Described diagram — Equipotentials between parallel plates: Equally spaced straight lines parallel to the plates. The potential decreases uniformly from the positive plate to the negative plate. Field lines are perpendicular to these equipotential lines, running from positive to negative plate.

Properties:

No work is done when a charge moves along an equipotential surface (ΔV = 0, so W = QΔV = 0).
Equipotentials are always perpendicular to field lines.
For a uniform field (parallel plates), equipotentials are equally spaced.
For a radial field (point charge), equipotentials become more widely spaced at greater distances.

Potential Due to Multiple Charges

Since electric potential is a scalar, the total potential at a point due to multiple charges is simply the algebraic sum:

V_total = V₁ + V₂ + V₃ + ... = Σ kQᵢ/rᵢ

This is much simpler than adding fields (which requires vector addition).

Worked Example 2 — Potential Between Two Charges

Question: A charge of +6.0 μC is placed at the origin and a charge of −3.0 μC is placed 0.40 m away. Calculate the electric potential at the midpoint between the charges.

Solution:

At the midpoint, the distance from each charge is 0.20 m.

V₁ = kQ₁/r₁ = (8.99 × 10⁹ × 6.0 × 10⁻⁶) / 0.20 = 2.70 × 10⁵ V

V₂ = kQ₂/r₂ = (8.99 × 10⁹ × (−3.0 × 10⁻⁶)) / 0.20 = −1.35 × 10⁵ V

V_total = V₁ + V₂ = 2.70 × 10⁵ + (−1.35 × 10⁵) = 1.35 × 10⁵ V = 135 kV

Note: Although the potential at the midpoint is not zero, the field at the midpoint is not zero either (both fields point in the same direction — away from the positive charge and towards the negative charge).

Relationship Between E and V

The electric field strength is the negative of the potential gradient:

E = −dV/dr

In a uniform field (parallel plates):

E = V/d (magnitude)

On a V-r graph for a point charge, the gradient at any point gives −E at that distance.

Electric Potential Energy

The electric potential energy of a charge q at a point where the potential is V:

E_p = qV

For two point charges Q₁ and Q₂ separated by distance r:

E_p = kQ₁Q₂/r = Q₁Q₂/(4πε₀r)

If the charges are like (both positive or both negative), E_p is positive — energy must be supplied to bring them together. If unlike, E_p is negative — energy is released when they come together.

Work Done Moving a Charge

W = qΔV = q(V₂ − V₁)

Worked Example 3 — Energy to Assemble a System of Charges

Question: Three charges of +2.0 μC each are placed at the corners of an equilateral triangle with side length 0.10 m. Calculate the total electric potential energy of the system.

Solution:

There are three pairs of charges, each separated by 0.10 m.

Energy of one pair: E = kQ₁Q₂/r = (8.99 × 10⁹ × 2.0 × 10⁻⁶ × 2.0 × 10⁻⁶) / 0.10 E = (8.99 × 10⁹ × 4.0 × 10⁻¹²) / 0.10 E = 3.596 × 10⁻² / 0.10 E = 0.360 J

Total energy for 3 pairs: E_total = 3 × 0.360 = 1.08 J

This is the work that would need to be done (against the repulsive forces) to assemble this configuration starting from all charges at infinity.

Millikan's Oil Drop Experiment

Millikan's oil drop experiment (1909–1913) provided the first accurate measurement of the elementary charge e and demonstrated that electric charge is quantised.

Experimental Setup

Described diagram — Millikan's apparatus: Two horizontal parallel metal plates are separated by a known distance d. A potential difference V is applied across the plates, creating a uniform electric field E = V/d between them. Oil drops are sprayed through a small hole in the top plate using an atomiser. Some drops become charged by friction or by exposure to X-rays. A microscope with a calibrated scale is used to observe individual drops.

Method

An oil drop is selected and observed through the microscope.
With the electric field switched off, the drop falls under gravity, quickly reaching terminal velocity. By measuring the terminal velocity and using Stokes' law (F_drag = 6πηrv, where η is the viscosity of air and r is the drop's radius), the radius and hence the mass of the drop can be determined.
The electric field is then switched on and adjusted until the drop is held stationary (hovering).

Analysis — Hovering Drop

When the drop hovers, the upward electric force balances the downward gravitational force (ignoring buoyancy for simplicity):

QE = mg

Q(V/d) = mg

Q = mgd/V

Worked Example 4 — Millikan Calculation

Question: An oil drop of mass 9.79 × 10⁻¹⁵ kg is held stationary between two parallel plates separated by 6.0 mm when a potential difference of 360 V is applied. Calculate the charge on the drop and determine how many excess electrons it carries.

Solution:

Step 1: Calculate the electric field. E = V/d = 360 / (6.0 × 10⁻³) = 6.0 × 10⁴ V m⁻¹

Step 2: At equilibrium, QE = mg. Q = mg/E = (9.79 × 10⁻¹⁵ × 9.81) / (6.0 × 10⁴) Q = 9.60 × 10⁻¹⁴ / 6.0 × 10⁴ Q = 1.60 × 10⁻¹⁸ C

Step 3: Find the number of excess electrons. n = Q/e = 1.60 × 10⁻¹⁸ / 1.60 × 10⁻¹⁹ = 10 electrons

Key Results

Millikan found that the charge on every drop was always a whole-number multiple of 1.60 × 10⁻¹⁹ C. This proved:

Charge is quantised — it exists only in discrete multiples of e.
The elementary charge is e = 1.60 × 10⁻¹⁹ C.

Exam Tip: In Millikan's experiment, you should be aware that the weight of the drop also includes an upthrust correction (Archimedes' principle). In a more precise treatment, the net downward force is (m_drop − m_air displaced)g = (ρ_oil − ρ_air) × V_drop × g. At A-Level, this correction is usually stated or neglected, but you should know it exists.

Worked Example 5 — Potential and Field at the Same Point

Question: A point charge of +8.0 × 10⁻⁹ C is placed at the origin. At a distance of 0.25 m from the charge, calculate (a) the electric potential V; (b) the electric field strength E; (c) the work that would need to be done by an external agent to move a small test charge q = −1.5 × 10⁻⁹ C from infinity to this point.

Solution:

(a) V = kQ/r = (8.99 × 10⁹ × 8.0 × 10⁻⁹) / 0.25 = 71.92 / 0.25 = 288 V (positive, because the source is positive).

(b) E = kQ/r² = (8.99 × 10⁹ × 8.0 × 10⁻⁹) / (0.25)² = 71.92 / 0.0625 = 1.15 × 10³ N C⁻¹ (radially outward, away from the positive source).

(c) The work done by an external agent moving the test charge q from infinity (where V = 0) to the point (where V = +288 V): W = qΔV = q(V_final − V_initial) = (−1.5 × 10⁻⁹)(288 − 0) = −4.32 × 10⁻⁷ J

The negative sign indicates the external agent does negative work — equivalently, the field does positive work on the test charge. This makes physical sense: a negative test charge is attracted to a positive source, so it would accelerate inward on its own; the external agent must hold it back (apply a force opposing the motion), and the work done by that holding force is negative.

Common Misconception: Students sometimes assume that "work done to move from infinity" is always positive. The sign depends on whether the displacement is in the same direction as the external force (positive work) or opposite (negative work). For a test charge moving spontaneously towards a source it is attracted to, the external agent does negative work.

Worked Example 6 — Energy Released by Two Charges Coming Together

Question: A charge of +3.0 μC and a charge of −2.0 μC, initially infinitely far apart, are brought together until they are 5.0 cm apart. Calculate the change in electric potential energy of the system and state whether energy is absorbed or released.

Solution:

Initial PE (at infinity): E_p,initial = 0 (by convention).

Electric Potential and Energy

Electric Potential and Energy