IB DP Physics: Electric potential energy, potential and equipotentials (DP Physics) – Revision Notes

These notes condense the additional higher level (AHL) part of sub-topic D.2, Electric and magnetic fields, from the IB Physics guide (first assessment 2025). All of it is HL only. Read the study guide; to test yourself, use the practice questions. Check what you already know with the free 10-minute diagnostics.

Related: the course hub, the printable checklist and the syllabus guide. Constants in the examples are given as data: k = 8.99 × 10⁹ N m² C⁻², e = 1.60 × 10⁻¹⁹ C.

Definitions

• Electric potential energy, Ep: the work done by an external agent to assemble a system of charges from infinite separation.
• Electric potential, Ve: the work done per unit charge in bringing a small positive test charge from infinity to the point. Unit: volt, V = J C⁻¹.
• Scalar: Ve has size and sign only, no direction. Zero potential is defined at infinity.
• Potential gradient: the rate of change of Ve with distance. Field strength is minus this gradient.
• Equipotential surface: a surface on which Ve is the same at every point.
• Electronvolt, eV: the work done on charge e through 1 V. 1 eV = 1.60 × 10⁻¹⁹ J.

Formulas

Quantity	Formula	Notes
Ep of two point charges	Ep = k q₁q₂ / r	Keep signs; unlike charges give negative Ep
Potential of a point charge	Ve = kQ / r	Sign of Q kept; also outside a spherical charge
Field from potential	E = –ΔVe / Δr	Field points towards falling Ve
Uniform field	E = V / d	Constant gradient between plates
Work done	W = qΔVe	Joules; divide by e for eV
Conducting sphere surface	Ve = kQ / R	Inside: same constant value

Method boxes

Ep of a group of charges

1. List every pair once. Three charges have three pairs, four charges have six.
2. Work out k q q / r for each pair, with signs.
3. Add them. The result does not depend on the order of assembly.

Potential at a point from several charges

1. Find the distance from the point to each charge.
2. Work out kQ/r for each, with the sign of Q.
3. Add the numbers. No vectors, no angles.

Finding where Ve = 0 between two opposite charges

1. Let x be the distance from the first charge and (d – x) the distance from the second.
2. Set kQ₁/x + kQ₂/(d – x) = 0 with the signs included.
3. Solve for x. Look for a second solution outside the pair, nearer the smaller charge.

Speed after acceleration through a potential difference

1. Kinetic energy gained = qΔV (convert to joules).
2. Set this equal to ½mv² and solve.
3. If the answer is needed in eV, the energy is the charge in units of e multiplied by the voltage, so an electron through 200 V gains 200 eV and a doubly charged ion gains 400 eV.

Small worked reminders

Reminder 1. A +1.5 nC charge and a +3.0 nC charge are 0.090 m apart. Ep = 8.99 × 10⁹ × 1.5 × 10⁻⁹ × 3.0 × 10⁻⁹ / 0.090 = 4.5 × 10⁻⁷ J. Positive, since like charges repel.

Reminder 2. Ve at 0.25 m from –2.0 nC: 8.99 × 10⁹ × (–2.0 × 10⁻⁹) / 0.25 = –72 V.

Reminder 3. A uniform field region has Ve falling from 150 V to 90 V over 0.020 m. E = 60 / 0.020 = 3.0 × 10³ V m⁻¹, directed towards the lower potential.

Reminder 4. A proton moves from a point at 40 V to a point at 240 V. ΔVe = 200 V, so the agent does 200 eV = 3.2 × 10⁻¹⁷ J of work on it. If it were released from the higher point, it would gain 200 eV of kinetic energy.

Reminder 5. Three +1.0 nC charges lie in a line with 0.20 m between neighbours. Pairs: 0.20 m, 0.20 m and 0.40 m. Ep = 8.99 × 10⁹ × (1.0 × 10⁻⁹)² × (1/0.20 + 1/0.20 + 1/0.40) = 1.1 × 10⁻⁷ J.

Reminder 6. A conducting sphere of radius 0.050 m carries +2.0 nC. Surface Ve = 8.99 × 10⁹ × 2.0 × 10⁻⁹ / 0.050 = 3.6 × 10² V. At 0.15 m from the centre, Ve = 360 / 3 = 1.2 × 10² V. At 0.030 m from the centre (inside), Ve is still 3.6 × 10² V and E = 0.

Equipotentials: what to recognise

Source	Equipotentials	Field
Single point charge	Concentric circles (spheres), further apart as r grows	Radial, perpendicular to the circles
Two equal and opposite charges	Ve = 0 on the line (plane) halfway between; closed loops around each charge	Lines run from + to –
Four charges	Small circles near each charge; use symmetry and the net charge far away	Zero at the centre for symmetric arrangements
Charged conducting sphere, solid or hollow, outside	Concentric spheres as for a point charge at the centre	As for a point charge
Charged conducting sphere, solid or hollow, inside	One equipotential, equal to the surface value	Zero
Parallel plates	Equally spaced planes parallel to the plates, bending outward at the edges	Uniform in the middle

Rules to state in answers:

• No work is done in moving a charge along an equipotential, because ΔVe = 0.
• Field lines cross equipotentials at 90°.
• Close equipotentials (at equal steps in V) mean a strong field.
• Equipotentials never cross.

Must-know distinctions

• Ep and Ve: Ep belongs to a system and is in joules. Ve belongs to a point and is in joules per coulomb. Ep = qVe for a charge q placed at a point of potential Ve.
• Potential and field: Ve is a scalar and E is a vector. E is the gradient of Ve, not its value.
• 1/r and 1/r²: Ve and Ep fall as 1/r. E falls as 1/r².
• Zero potential and zero field: neither implies the other. Midway between two equal opposite charges Ve = 0 but E is not 0. Inside a charged conducting sphere E = 0 but Ve is not 0.
• Joule and electronvolt: both measure energy. Convert before using mass and speed.
• Work by the agent and work by the field: these have opposite signs. The agent raises Ep when it moves a charge against the field; the field lowers Ep when it accelerates a free charge.

Quick self-test

1. Find Ep for charges of +1.0 μC and +2.0 μC separated by 0.50 m.
2. Find Ve at 0.25 m from a point charge of +3.0 nC.
3. Ve falls uniformly from 300 V to 180 V over 0.040 m. Find the field strength.
4. How much work is needed to move +4.0 μC through a potential difference of 75 V?
5. An electron is accelerated through 1000 V. State its energy gain in J and in eV.
6. State the field and potential inside a charged hollow conducting sphere of surface potential 600 V.
7. At what distance from a +4.0 nC point charge is the potential 90 V?
8. A sphere has a surface potential of 600 V and radius R. What is the potential at 2R from its centre?
9. Explain why Ep is negative for a proton and an electron.
10. A charge is moved along an equipotential surface. State the work done and why.

Answers

1. Ep = 8.99 × 10⁹ × 1.0 × 10⁻⁶ × 2.0 × 10⁻⁶ / 0.50 = 3.6 × 10⁻² J.
2. Ve = 8.99 × 10⁹ × 3.0 × 10⁻⁹ / 0.25 = 1.1 × 10² V (108 V).
3. E = 120 / 0.040 = 3.0 × 10³ V m⁻¹, towards the 180 V end.
4. W = qΔV = 4.0 × 10⁻⁶ × 75 = 3.0 × 10⁻⁴ J.
5. Energy = 1.60 × 10⁻¹⁹ × 1000 = 1.60 × 10⁻¹⁶ J = 1000 eV.
6. The field is zero and the potential is 600 V everywhere inside.
7. r = kQ/Ve = 8.99 × 10⁹ × 4.0 × 10⁻⁹ / 90 = 0.40 m.
8. Ve = kQ/(2R) = half the surface value = 300 V.
9. They attract, so work must be done to separate them to infinity; the system has less energy than at infinite separation. Mathematically q₁q₂ is negative.
10. Zero, because ΔVe = 0 and W = qΔVe.

Where marks are usually lost

• Dropping the minus sign of a negative charge in Ep or Ve, so an attractive pair appears to store positive energy.
• Using the distance in centimetres or nanometres without converting to metres.
• Forgetting that the second charge in a pair is not counted twice: each pair is counted once.
• Adding potentials with vector diagrams, which wastes time and invites errors.
• Giving E = V/d for a point charge or any non-uniform field.
• Writing that zero field means zero potential inside a conducting sphere.
• Drawing equipotentials that cross each other, or field lines that meet them at a slant.
• Mixing eV with J in kinetic energy equations.
• Leaving out the direction of E, which is towards falling Ve.
• Treating Ve inside a conducting sphere as kQ/r with a small r. The interior value stays at the surface value kQ/R.
• Writing W = qΔVe with the wrong sign convention and not saying who does the work. State whether it is the external agent or the field.
• Giving a speed without checking that energy was converted to joules first.

Official syllabus

Source: Physics guide, first assessment 2025, International Baccalaureate Organization (Diploma Programme), sub-topic D.2 Electric and magnetic fields, additional higher level.