These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data for every question: k = 1/(4πε₀) = 9.0 × 10⁹ N·m²/C²; e = 1.6 × 10⁻¹⁹ C; g = 9.8 m/s². Charges are point charges unless a size is given, and source charges stay fixed. A scientific calculator is assumed. Round only at the end.
Question 1 (multiple choice · foundation)
A +4.0 μC test charge at point X feels an electric force of 0.10 N directed north. The test charge is removed and a −6.0 μC charge is placed at X. What force does it feel?
- (A) 0.15 N south
- (B) 0.15 N north
- (C) 0.067 N south
- (D) 0.10 N south
Answer and explanation
Answer: (A). The field at X is E = F/q = 0.10 N ÷ (4.0 × 10⁻⁶ C) = 2.5 × 10⁴ N/C, north. The new force has size |q|E = (6.0 × 10⁻⁶ C)(2.5 × 10⁴ N/C) = 0.15 N. The charge is negative, so the force is opposite to E: south.
- (B) has the right size but forgets that a negative charge is pushed opposite to the field.
- (C) scales the force by 4/6 instead of 6/4, as if a larger charge felt a smaller force.
- (D) assumes the force stays the same size. The field stays the same; the force is proportional to the charge placed there.
Question 2 (multiple choice · core)
The electric field 0.15 m from a small charged sphere is 720 N/C. What is the field 0.45 m from the sphere?
- (A) 80 N/C
- (B) 240 N/C
- (C) 2160 N/C
- (D) 6480 N/C
Answer and explanation
Answer: (A). E = k|q|/r², so E ∝ 1/r². The distance is 3 times larger, so the field is 3² = 9 times smaller: 720 ÷ 9 = 80 N/C.
- (B) divides by 3, treating the field as ∝ 1/r. That is not the point-charge law.
- (C) multiplies by 3, as if the field grew with distance.
- (D) multiplies by 9: right factor, wrong way round. Fields get weaker further from the charge.
Question 3 (multiple choice · core)
A solid metal sphere is given some excess positive charge. After a short time, no charges are moving. Which statement is correct?
- (A) The excess charge spreads evenly through the whole volume, and the field inside the metal is not zero.
- (B) The excess charge is on the outer surface, the field inside the metal is zero, and just outside the sphere the field is perpendicular to the surface.
- (C) The excess charge is on the outer surface, and inside the metal the field points outward from the centre.
- (D) The excess charge collects at the centre, so outside the sphere the field is the same as that of a point charge.
Answer and explanation
Answer: (B). In electrostatic equilibrium the free electrons have stopped moving, which can only happen if the field inside the metal is zero. Like charges repel and move as far apart as possible, to the outer surface. Any field component along the surface would move charge sideways, so the field at the surface is perpendicular to it.
- (A) describes a charged insulator, where charge cannot move freely. In a conductor the charge can move, and does.
- (C) puts the charge in the right place but leaves a field inside the metal. Such a field would keep moving electrons, so the sphere would not be in equilibrium.
- (D) confuses the result with its cause. Outside, the field does look like a point charge at the centre, but the charge itself is on the surface.
Question 4 (multiple choice · core)
A field-line diagram shows two small charged spheres, P and Q. Twelve field lines start on P. Four of them end on Q, and the other eight leave the edge of the diagram. No lines start on Q. Which conclusion does the diagram support?
- (A) P is positive, Q is negative, and the charge on P is 3 times the size of the charge on Q.
- (B) P is positive, Q is negative, and the two charges have the same size, because lines join them.
- (C) P is negative, Q is positive, and the charge on P is 3 times the size of the charge on Q.
- (D) P is positive, Q is negative, and the charge on P is 1.5 times the size of the charge on Q.
Answer and explanation
Answer: (A). Lines start on positive charges and end on negative charges, so P is positive and Q is negative. The number of lines touching a charge is proportional to its size: 12 at P and 4 at Q, so |q_P| = 3|q_Q|.
- (B) assumes every line from P must end on Q. Extra lines going off to infinity show P has more charge.
- (C) reverses the line direction. Field lines leave positive charges.
- (D) compares the 12 lines at P with the 8 lines that leave the diagram, instead of with the 4 lines that end on Q.
Question 5 (calculation · core)
A tiny charged oil droplet of mass 1.96 × 10⁻¹⁵ kg hovers at rest in a region where the electric field is uniform, 4.0 × 10⁴ N/C, pointing vertically downward.
(a) State whether the droplet’s charge is positive or negative. Explain. (b) Calculate the size of the charge on the droplet. (c) How many excess electrons does the droplet carry? (d) The field is suddenly doubled, with the same direction. Find the initial acceleration of the droplet.
Worked solution
- (a) The droplet is at rest, so the electric force balances the weight. The weight acts down, so the electric force must act up: opposite to the field. The charge is negative.
- (b) |q|E = mg, so |q| = mg/E = (1.96 × 10⁻¹⁵ kg)(9.8 m/s²) ÷ (4.0 × 10⁴ N/C) = 1.9208 × 10⁻¹⁴ N ÷ 4.0 × 10⁴ N/C = 4.80 × 10⁻¹⁹ C.
- (c) n = |q|/e = 4.80 × 10⁻¹⁹ ÷ 1.6 × 10⁻¹⁹ = 3 electrons.
- (d) The electric force doubles to 2mg upward. Net force = 2mg − mg = mg upward, so a = g = 9.8 m/s² upward.
Suggested mark points (4): 1 for “negative” with the reason that the force must be up while E points down; 1 for |q| = 4.80 × 10⁻¹⁹ C; 1 for 3 electrons; 1 for 9.8 m/s² upward. A whole number in (c) is a good check: charge comes in multiples of e.
Common error: answering (a) with “positive, because the force is up”. A positive charge would be pushed down, the same way as the field.
Question 6 (constructed response · core)
Point P is at the origin. A +5.0 nC charge is fixed at (0, 0.30 m) and a −5.0 nC charge is fixed at (0.40 m, 0).
(a) Draw a diagram of P showing the field from each charge as a labelled arrow, with the longer arrow for the stronger field. (b) Calculate the magnitude and direction of the net electric field at P. (c) Find the force on a proton placed at P. (d) A student says: “The fields are 500 N/C and 281 N/C, so the net field is 781 N/C.” Explain what is wrong.
Worked solution and suggested Marlbridge rubric
(a) The field from +5.0 nC points away from it: straight down (−y). The field from −5.0 nC points toward it: to the right (+x). The downward arrow should be longer.
(b) From +5.0 nC: E₁ = (9.0 × 10⁹)(5.0 × 10⁻⁹) ÷ (0.30)² = 500 N/C, in −y. From −5.0 nC: E₂ = (9.0 × 10⁹)(5.0 × 10⁻⁹) ÷ (0.40)² = 281.25 N/C, in +x. The two are at right angles: E = √(500² + 281.25²) = 574 N/C. Direction: tan θ = 500/281.25, so θ = 60.6° below the +x axis (down and to the right).
(c) F = eE = (1.6 × 10⁻¹⁹ C)(573.7 N/C) = 9.18 × 10⁻¹⁷ N, in the same direction as E (60.6° below +x), because the proton is positive.
(d) Fields are vectors. The student added the magnitudes as if both fields pointed the same way. Here they are perpendicular, so the components must be combined (Pythagoras), giving 574 N/C, not 781 N/C.
| Point | What earns it |
|---|---|
| 1 | Diagram: field from the positive charge points down (away), field from the negative charge points right (toward), with the downward arrow longer |
| 1 | Both magnitudes correct: 500 N/C and 281 N/C |
| 1 | Net magnitude 574 N/C from perpendicular components |
| 1 | Direction given as an angle (60.6° below +x, or 29.4° to the right of −y) |
| 1 | Force on the proton 9.18 × 10⁻¹⁷ N in the direction of E |
| 1 | Identifies that vector fields at right angles cannot be added as numbers |
Carry forward an error in the magnitudes from (b) into (c) once.
Question 7 (constructed response · stretch)
Two spheres each have radius 0.050 m and carry +3.0 nC. Sphere A is solid metal. Sphere B is solid plastic with its charge spread evenly through its volume. The spheres are far apart and nothing is moving.
(a) Calculate the field 0.15 m from the centre of each sphere. (b) Calculate the field just outside the surface of sphere A, and give its direction. (c) Compare the field at a point 0.025 m from the centre of each sphere. Justify your answer. (d) A student says: “In sphere A the charges repel each other, so they spread evenly through the whole metal, just like in sphere B.” Evaluate this claim.
Worked solution and suggested Marlbridge rubric
(a) Both spheres have spherically symmetric charge, so outside each one the field is that of a +3.0 nC point charge at the centre: E = (9.0 × 10⁹)(3.0 × 10⁻⁹) ÷ (0.15)² = 1200 N/C, radially outward, for both spheres.
(b) At r = 0.050 m: E = (9.0 × 10⁹)(3.0 × 10⁻⁹) ÷ (0.050)² = 1.08 × 10⁴ N/C, directed outward, perpendicular to the surface. (Check: 3 times closer, 9 times stronger: 9 × 1200 = 10 800.)
(c) Sphere A: the point is inside the metal, so E = 0. Free electrons would move if there were any field there, so in equilibrium the field is zero. Sphere B: the charge cannot move in plastic and stays spread through the volume, so some of it lies nearer the centre than the point. That enclosed charge makes a field at the point, so E inside B is not zero and points outward. So the field is larger at the point in B.
(d) The student is right that like charges repel, but wrong about the result. In the metal the charges are free to move, so they push each other as far apart as possible: onto the outer surface. In the plastic the charges cannot move, so they stay where they were put. The difference comes from conductor versus insulator, not from the repulsion.
| Point | What earns it |
|---|---|
| 1 | 1200 N/C at 0.15 m, stated for both spheres |
| 1 | Reason: spherical symmetry, so each acts like a point charge at its centre |
| 1 | 1.08 × 10⁴ N/C just outside A, perpendicular to the surface |
| 1 | Field inside A is zero, linked to free charges in equilibrium |
| 1 | Field inside B is nonzero, linked to charge that cannot move and stays in the interior |
| 1 | Rejects the claim: in a conductor the charges move to the outer surface; in an insulator they cannot move |
Answers to (c) may stay qualitative; no calculation of the field inside B is expected.
How did you do?
- Q1 or Q5 wrong: re-read “What an electric field is” and “Direction of the field and of the force” in the study guide. Check the sign of the charge first.
- Q2 wrong: revisit “Field of a point charge” and its inverse-square table.
- Q4 wrong: go back to “Field maps and field lines” and Figure 1(b).
- Q6 wrong: work through Worked examples 1 and 2 again, then redo the vector addition.
- Q3 or Q7 wrong: re-read “Charged conductors and insulators” and Worked example 3.
Then tick off the topic checklist.