IB DP Physics: Electric potential energy, potential and equipotentials (DP Physics) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover the additional higher level part of sub-topic D.2, Electric and magnetic fields, from the IB Physics guide (first assessment 2025). Every question is HL only. Use k = 8.99 × 10⁹ N m² C⁻² and e = 1.60 × 10⁻¹⁹ C unless a question gives other data. If a topic is new, read the study guide first.

Questions

1. A charge of –1.2 nC and a charge of +2.5 nC are 6.0 cm apart in a vacuum. Calculate the electric potential energy of the pair and state what its sign shows. [2]

2. Define the electric potential at a point in an electric field, and state one property that shows it is a scalar quantity. [3]

3. A point charge of –8.0 nC is fixed in a vacuum. (a) Calculate the electric potential 0.50 m from the charge. [2] (b) Calculate the work done by an external agent in bringing a charge of +3.0 nC from infinity to this point, and comment on the sign. [2]

4. Two parallel plates are 15 mm apart with a potential difference of 900 V. The negative plate is at 0 V. (a) Calculate the electric field strength between the plates, away from the edges. [2] (b) Find the electric potential at a point 5.0 mm from the negative plate. [2]

5. An alpha particle (charge +2e, mass 6.64 × 10⁻²⁷ kg) is released from rest and accelerated through a potential difference of 1.2 kV. (a) Calculate the kinetic energy gained, in joules. [2] (b) State this energy in electronvolts. [1] (c) Calculate the final speed. [2]

6. An isolated solid conducting sphere of radius 0.12 m carries a charge of +25 nC. (a) Calculate the electric potential at its surface. [2] (b) Calculate the potential 0.30 m from the centre. [1] (c) State the potential and the electric field strength at the centre of the sphere. [2]

7. Two point charges of equal size and opposite sign are fixed a short distance apart. (a) Describe the equipotential halfway between them. [1] (b) State the angle between the electric field lines and the equipotential surfaces. Use W = qΔVe to explain why this must be so. [3]

8. The table gives the electric potential at distances r from a point charge of +2.0 nC in a vacuum.

r / m	0.10	0.20	0.30
Ve / V	180	90	60

(a) Use the values at 0.10 m and 0.30 m to estimate the field strength at r = 0.20 m. [2] (b) Calculate the field strength at r = 0.20 m from Ve = kQ/r and E = –ΔVe/Δr. [2] (c) Explain why the estimate in (a) differs from the value in (b). [2]

9. Three point charges lie in a straight line, 0.10 m apart: +2.0 nC, +2.0 nC and –1.0 nC, in that order. (a) Calculate the total electric potential energy of the system. [4] (b) Interpret the sign of your answer in terms of the work done in assembling the system. [2]

10. A charge of +4.0 nC and a charge of –4.0 nC are 0.60 m apart in a vacuum. Point P lies on the line joining them, 0.20 m from the positive charge. (a) Calculate the electric potential at P. [3] (b) Calculate the work done by an external agent in bringing a +3.0 nC charge from infinity to P. [2] (c) State the potential at every point on the plane halfway between the two charges, and explain why the electric field strength there is not zero. [2] (d) State the direction of the field at the midpoint of the line. [1]

11. Two protons are held 2.0 nm apart in a vacuum, then released from rest. The proton mass is 1.67 × 10⁻²⁷ kg. (a) Calculate the electric potential energy of the pair at release. [2] (b) Express this energy in electronvolts. [1] (c) Calculate the speed of each proton when they are very far apart. [3] (d) Explain why the two speeds are equal. [2]

12. An electron gains 3.2 × 10⁻¹⁵ J of kinetic energy in an accelerator. State this energy in electronvolts and the potential difference through which the electron moved from rest. [2]

Answers

1. Ep = kq₁q₂/r = 8.99 × 10⁹ × (–1.2 × 10⁻⁹) × (2.5 × 10⁻⁹) / 0.060 [1] = –4.5 × 10⁻⁷ J. The negative sign shows the charges attract, so work must be done to separate them [1]. [2] Examiner insight: A negative Ep with no comment on its meaning leaves the question half answered.

2. The electric potential at a point equals the work done for each coulomb of charge [1] in carrying a small positive test charge in from infinity to that point [1]. It is a scalar because it has size and sign but no direction, so potentials from several charges add as numbers [1]. [3] Examiner insight: A definition that leaves out “from infinity” or “per unit charge” is incomplete; include both, and say the test charge is positive.

3. (a) Ve = kQ/r = 8.99 × 10⁹ × (–8.0 × 10⁻⁹) / 0.50 [1] = –1.4 × 10² V (–144 V) [1]. [2] (b) W = qΔVe = 3.0 × 10⁻⁹ × (–144) [1] = –4.3 × 10⁻⁷ J. The negative sign shows that the field attracts the positive charge, so the agent’s work is negative [1]. [2] Examiner insight: Keep the sign of Q through both parts; a positive answer to (b) contradicts the attraction.

4. (a) E = V/d = 900 / 0.015 [1] = 6.0 × 10⁴ V m⁻¹ [1]. [2] (b) The gradient is constant, so Ve = E × 0.0050 [1] = 3.0 × 10² V [1]. [2] Examiner insight: Convert millimetres to metres before substituting, and say in (b) that the distance is measured from the 0 V plate.

5. (a) KE = qΔV = 2 × 1.60 × 10⁻¹⁹ × 1200 [1] = 3.8 × 10⁻¹⁶ J [1]. [2] (b) 2 × 1200 = 2400 eV (2.4 keV) [1]. [1] (c) ½mv² = 3.84 × 10⁻¹⁶, so v = √(2 × 3.84 × 10⁻¹⁶ / 6.64 × 10⁻²⁷) [1] = 3.4 × 10⁵ m s⁻¹ [1]. [2] Examiner insight: Forgetting the charge 2e changes both answers. Work in joules before using the mass.

6. (a) Ve = kQ/R = 8.99 × 10⁹ × 25 × 10⁻⁹ / 0.12 [1] = 1.9 × 10³ V [1]. [2] (b) Outside, the sphere acts as a point charge at its centre: Ve = 8.99 × 10⁹ × 25 × 10⁻⁹ / 0.30 = 7.5 × 10² V [1]. [1] (c) The potential is 1.9 × 10³ V, the same as the surface [1], and the field strength is zero [1]. [2] Examiner insight: Giving zero potential at the centre is a frequent slip. State potential and field separately.

7. (a) The plane halfway between them is at zero potential [1]. [1] (b) The field lines cross equipotential surfaces at 90° [1]. No work is done moving a charge along an equipotential because ΔVe = 0, so W = qΔVe = 0 [1]. A field component along the surface would do work on a charge moved along it, so the field must be perpendicular to the surface [1]. [3] Examiner insight: Linking W = qΔVe to the absence of a field component along the surface is stronger than quoting the 90° rule alone.

8. (a) Gradient = (60 – 180) / (0.30 – 0.10) = –600 V m⁻¹ [1], so E ≈ 600 V m⁻¹, directed away from the charge [1]. [2] (b) Ve = kQ/r gives –ΔVe/Δr = kQ/r² [1] = 8.99 × 10⁹ × 2.0 × 10⁻⁹ / 0.20² = 4.5 × 10² V m⁻¹ [1]. [2] (c) The graph of Ve against r is curved, so the gradient is not constant [1]. It is steeper near 0.10 m than at 0.20 m, so the average over the interval is larger than the value at the midpoint [1]. [2] Examiner insight: Name the curvature of the graph as the cause, not measurement error.

9. (a) There are three pairs [1]. Ep for the two +2.0 nC charges = 8.99 × 10⁹ × (2.0 × 10⁻⁹)² / 0.10 = +3.6 × 10⁻⁷ J [1]. Each +2.0 nC with –1.0 nC: –1.8 × 10⁻⁷ J at 0.10 m and –9.0 × 10⁻⁸ J at 0.20 m [1]. Total = 9.0 × 10⁻⁸ J [1]. [4] (b) The total is positive, so the external agent does net positive work of 9.0 × 10⁻⁸ J to assemble the system from infinity [1], because repulsion between the like pair outweighs attraction in the unlike pairs [1]. [2] Examiner insight: List each pair separately before adding. The 0.20 m pair is the one most often left out.

10. (a) The distance from the negative charge is 0.40 m [1]. Ve = 8.99 × 10⁹ × (4.0 × 10⁻⁹/0.20 – 4.0 × 10⁻⁹/0.40) [1] = 90 V [1]. [3] (b) W = qΔVe = 3.0 × 10⁻⁹ × 89.9 [1] = 2.7 × 10⁻⁷ J [1]. [2] (c) The potential is zero: every point on the plane is the same distance from both charges, so the two contributions cancel [1]. The field is not zero because potential is a scalar and cancels, while the two field vectors point the same way and add [1]. [2] (d) From the positive charge towards the negative charge [1]. [1] Examiner insight: Using a single distance for both charges at P is the commonest slip. Say that potential is a scalar when explaining why zero potential still allows a field.

11. (a) Ep = ke²/r = 8.99 × 10⁹ × (1.60 × 10⁻¹⁹)² / 2.0 × 10⁻⁹ [1] = 1.2 × 10⁻¹⁹ J [1]. [2] (b) 1.15 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 0.72 eV [1]. [1] (c) All of Ep becomes kinetic energy of two protons: 2 × ½mv² = Ep [1]. v = √(Ep/m) = √(1.15 × 10⁻¹⁹ / 1.67 × 10⁻²⁷) [1] = 8.3 × 10³ m s⁻¹ [1]. [3] (d) The forces on the protons are equal and opposite [1] and the masses are equal, so the accelerations and speeds are equal [1]. [2] Examiner insight: The energy is shared between two particles. Giving all of Ep to one proton overestimates the speed by a factor of √2.

12. Energy in eV = 3.2 × 10⁻¹⁵ / 1.60 × 10⁻¹⁹ = 2.0 × 10⁴ eV [1]. The electron has charge e, so the potential difference is 2.0 × 10⁴ V [1]. [2] Examiner insight: Show the division by e so the conversion to eV is visible.

Where marks are usually lost

• Dropping the sign of a charge in Ep or Ve, so an attractive system seems to have positive energy.
• Missing one pair when assembling three or more charges.
• Not converting cm, mm or nm to metres.
• Using r² in the potential or potential energy formula.
• Adding potentials as vectors.
• Claiming that zero potential and zero field go together.
• Giving the potential inside a conducting sphere as zero rather than the surface value.
• Giving all the energy to one particle in a two-body release.

Next steps

• Review the revision notes and the study guide.
• Return to the IB DP Physics course hub and tick off the printable checklist.
• Try all free 10-minute diagnostics at /diagnostics/.
• Book a free trial class at /trial/.

Official syllabus

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