IB DP Physics: Electric charge, Coulomb's law, electric and magnetic fields (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 standard level and higher level understandings of D.2, Electric and magnetic fields, in the IB Physics guide (first assessment 2025). The higher level potential content is not included. Take k as 8.99 × 10⁹ N m² C⁻², e as 1.60 × 10⁻¹⁹ C and g as 9.81 m s⁻².

Links: the study guide, the revision notes, the course hub and the printable checklist.

Questions

1. A small sphere carrying −12 nC hangs near a rod carrying +30 nC. State the direction of the force on the sphere, the direction of the force on the rod, and compare the sizes of the two forces. [3]

2. Two small spheres carry +12 nC and +30 nC. Their centres are 5.0 cm apart.

(a) Calculate the electric force on the 12 nC sphere and state its direction. [2] (b) Determine the separation at which the force would be four times as large. [2]

3. Three identical conducting spheres A, B and C are on insulating stands. A carries +9.0 µC, B is neutral and C carries −3.0 µC. A touches B and they are separated. Then B touches C and they are separated. Find the final charge on each sphere. [4]

4. In a Millikan-type experiment a negatively charged oil drop of mass 8.16 × 10⁻¹⁵ kg is held stationary between two horizontal plates 1.20 cm apart. The potential difference across the plates is 1.50 kV.

(a) State which plate is positive, with a reason. [2] (b) Calculate the charge on the drop. [3] (c) State the number of elementary charges on the drop. [1] (d) Explain how results from many drops provide evidence that charge is quantized. [2]

5. You have a negatively charged rod and a neutral metal sphere on an insulating stand. Describe how to give the sphere a positive charge without the rod touching it. [4]

6. An isolated point charge of −4.8 nC sits in air.

(a) Find the magnitude and direction of the field strength 0.25 m from it. [3] (b) Calculate the force on a proton placed at that point. [2]

7. Two parallel metal plates 2.5 cm apart have a potential difference of 1.8 kV.

(a) Calculate E in the central region of the gap. [2] (b) A particle of charge +3.2 × 10⁻¹⁹ C is in this region. Calculate the force on it and state its direction. [2] (c) Describe the field lines between and around the edges of the plates. [2]

8. Three point charges lie on a straight line: A (+3.0 nC) at x = 0, B (−2.0 nC) at x = 0.10 m and C (+5.0 nC) at x = 0.30 m. Calculate the resultant force on B, giving its direction. [6]

9. (a) Describe the electric field inside and outside an isolated charged conducting sphere. [3] (b) Explain how field lines show the way the field strength of a point charge changes with distance. [2] (c) Explain why two electric field lines never cross. [1]

10. (a) A long straight wire carries a current into the page. State the direction of the magnetic field at a point P to the right of the wire, in the plane of the page. [2] (b) Looking at the left end of an air-core solenoid, the current circulates clockwise. State which end is the north pole. [2] (c) State one way in which the field of a solenoid resembles that of a bar magnet. [1]

11. Two identical small spheres, each of mass 1.5 g, hang from insulating threads 0.50 m long attached to the same point. They carry equal charges and come to rest 4.0 cm apart.

(a) Identify the three forces on one sphere. [2] (b) Determine the electrostatic force between the spheres. [3] (c) Calculate the charge on each sphere. [3]

Answers

1. The sphere is pulled towards the rod, since the charges are unlike [1]. The rod is pulled towards the sphere [1]. The forces are equal in magnitude, even though the charges differ, as they form a Newton’s third law pair [1]. [3] Examiner insight: Comparing the forces needs a reason, and a statement that the larger charge gives a larger force is the common wrong answer here.

2. (a) F = (8.99 × 10⁹ × 12 × 10⁻⁹ × 30 × 10⁻⁹) / 0.050² [1]. F = 1.29 × 10⁻³ N, repulsive, directed away from the other sphere [1]. [2] (b) F is proportional to 1/r², so four times the force needs half the distance [1]. The separation is 2.5 cm [1]. [2] Examiner insight: Show the unit conversions in your substitution so that a slip is visible.

3. After A touches B: each has 9.0 / 2 = +4.5 µC [1]. B and C have a total of 4.5 + (−3.0) = +1.5 µC [1]. Shared equally, B and C each have +0.75 µC [1]. Final charges: A = +4.5 µC, B = +0.75 µC, C = +0.75 µC; the total is +6.0 µC, which equals the starting total [1]. [4] Examiner insight: Checking that the final total equals the initial total is a quick way to catch an error and shows conservation of charge explicitly.

4. (a) The drop is negative and the electric force must act upwards, so the field must point downwards [1]. Field lines run from positive to negative, so the top plate is positive [1]. [2] (b) E = V/d = 1500 / 0.0120 = 1.25 × 10⁵ V m⁻¹ [1]. qE = mg [1]. q = (8.16 × 10⁻¹⁵ × 9.81) / 1.25 × 10⁵ = 6.40 × 10⁻¹⁹ C [1]. [3] (c) 6.40 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 4 elementary charges [1]. [1] (d) Different drops carry different charges, but every measured value is a whole-number multiple of one smallest value [1]. No fractions of that unit appear, so charge comes in discrete units and is not continuous [1]. [2] Examiner insight: For an explain question about evidence, say what the data showed and what conclusion follows; stating only that charge is quantized does not answer how the results support it.

5. Bring the rod near the sphere without touching, so electrons in the sphere are repelled to the far side [1]. Connect the sphere to earth, so electrons flow from the sphere to earth [1]. Disconnect the earth connection while the rod is still in place [1]. Remove the rod, and the remaining positive charge spreads over the sphere [1]. [4] Examiner insight: The order of removing the earth and the rod is the point of the question, so state the order explicitly.

6. (a) E = kQ/r² = (8.99 × 10⁹ × 4.8 × 10⁻⁹) / 0.25² [1]. E = 6.90 × 10² N C⁻¹ [1], directed towards the charge, because the charge is negative [1]. [3] (b) F = eE = 1.60 × 10⁻¹⁹ × 6.90 × 10² [1]. F = 1.10 × 10⁻¹⁶ N, directed towards the charge [1]. [2] Examiner insight: A direction is part of the answer for a vector quantity, and a negative source charge reverses the direction you might assume.

7. (a) E = V/d = 1800 / 0.025 [1] = 7.2 × 10⁴ V m⁻¹ [1]. [2] (b) F = qE = 3.2 × 10⁻¹⁹ × 7.2 × 10⁴ [1] = 2.3 × 10⁻¹⁴ N, directed from the positive plate towards the negative plate [1]. [2] (c) Between the plates the lines are straight, parallel and evenly spaced, running from the positive plate to the negative plate [1]. Near the edges the lines curve outwards, so the field is not uniform there [1]. [2] Examiner insight: Convert cm to m before dividing; the answer in V m⁻¹ will otherwise be a hundred times too small.

8. Force from A on B: F = (8.99 × 10⁹ × 3.0 × 10⁻⁹ × 2.0 × 10⁻⁹) / 0.10² = 5.39 × 10⁻⁶ N [1], attractive, so towards A [1]. Force from C on B: the distance is 0.20 m, so F = (8.99 × 10⁹ × 5.0 × 10⁻⁹ × 2.0 × 10⁻⁹) / 0.20² = 2.25 × 10⁻⁶ N [1], attractive, so towards C [1]. The two forces are in opposite directions, so subtract [1]. Resultant = 3.1 × 10⁻⁶ N towards A (3.15 × 10⁻⁶ N before rounding) [1]. [6] Examiner insight: Draw both force arrows first; using 0.30 m for the force from C is a common slip.

9. (a) The charge sits on the surface of the sphere [1]. The field inside is zero, so there are no field lines inside [1]. Outside, the field lines are radial and meet the surface at right angles, as for a point charge at the centre [1]. [3] (b) Where the lines are closer, the field is stronger [1]. The radial lines spread apart as the distance from the charge increases, so the field gets weaker further away [1]. [2] (c) The field at a point has only one direction, and a crossing would give two directions at that point [1]. [1] Examiner insight: A description of the inside of a conductor should say zero field, not merely weak field.

10. (a) Using the right-hand grip rule with the thumb into the page, the field circles are clockwise [1]. At P, on the right of the wire, the field points down the page [1]. [2] (b) The fingers curl clockwise as seen from the left end, so the thumb points away from the viewer, to the right [1]. The field leaves the solenoid at the right end, so the right end is the north pole [1]. [2] (c) Outside the solenoid the field lines form loops running from the north end to the south end, like those of a bar magnet [1]. [1] Examiner insight: Mention conventional current and the grip rule you used; a bare direction with no reasoning leaves nothing to credit if it is wrong.

11. (a) The weight, acting downwards [1]. The tension in the thread, along the thread, and the horizontal electrostatic repulsion from the other sphere [1]. [2] (b) sin θ = 0.020 / 0.50, so θ = 2.29° [1]. In equilibrium, F = mg tan θ [1]. F = 1.5 × 10⁻³ × 9.81 × tan 2.29° = 5.9 × 10⁻⁴ N [1]. [3] (c) F = kq²/r² with equal charges, so q = √(F r² / k) [1]. q = √(5.89 × 10⁻⁴ × 0.040² / 8.99 × 10⁹) [1]. q = 1.0 × 10⁻⁸ C on each sphere [1]. [3] Examiner insight: Use the full separation of 0.040 m, not half of it, in Coulomb’s law.

Where marks are usually lost

• Leaving nC, µC or cm unconverted in a formula.
• Using the wrong distance, for example the radius of a sphere or half a separation, in Coulomb’s law.
• Giving a size without the direction, or the force on a negative charge along the field.
• Explaining Millikan’s results without stating the whole-number pattern.
• Describing induction with the steps in the wrong order, or missing the earth connection.
• Field lines lacking arrows, intersecting, or ending away from any charge.
• Using the direction of electron flow in a right-hand grip rule.

Next steps

• Review with the revision notes.
• Reread the study guide.
• Go to the IB DP Physics course hub and tick off the printable checklist.
• Try the free 10-minute diagnostics to find your weak spots.
• Or book a free trial class.

Official syllabus

These questions follow the IB Physics guide, first assessment 2025, published by the International Baccalaureate Organization, sub-topic D.2.