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Edexcel International A Level Physics · Unit 4 · Further Mechanics, Fields and Particles
40 free, high-quality past-paper-style questions on Electric fields for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.
Which of the following units is equivalent to the unit of electric field strength, N C−1?
| Answer | Mark | Notes |
|---|---|---|
| For a uniform field E = V/d, so field strength can be measured in V m−1. Check: V m−1 = J C−1 m−1 = N m C−1 m−1 = N C−1. Options B and C are both the volt, the unit of potential. | ✓ 1 | Answer D |
Answer: D
Exam tip: N C−1 and V m−1 are interchangeable. Use V m−1 when the field comes from a potential gradient or V/d.
A particle with a charge of 3.0 μC is placed at a point where the electric field strength is 4.0 × 103 N C−1.
What is the size of the electric force on the particle?
| Answer | Mark | Notes |
|---|---|---|
| F = EQ = 4.0 × 103 × 3.0 × 10−6 = 1.2 × 10−2 N. | ✓ 1 | Answer B |
Answer: B
Common error: Misreading the prefix: μ means 10−6, m means 10−3. Write the charge in coulombs before substituting.
Two point charges exert an electrostatic force F on each other. One of the charges is trebled and the distance between the charges is doubled.
What is the new force between them?
| Answer | Mark | Notes |
|---|---|---|
| F ∝ Q1Q2/r2. New force = F × 3 ÷ 22 = 0.75F. | ✓ 1 | Answer B |
Answer: B
Exam tip: For ratio questions, write the factor for each change separately: × 3 for the charge, ÷ 4 for doubling the distance, then multiply.
A p.d. of 3.0 kV is applied between two parallel metal plates 1.5 cm apart.
What is the electric field strength between the plates?
| Answer | Mark | Notes |
|---|---|---|
| E = V/d = 3000 ÷ 0.015 = 2.0 × 105 V m−1. | ✓ 1 | Answer D |
Answer: D
Exam tip: Convert both values to base units first: 3.0 kV = 3000 V and 1.5 cm = 0.015 m.
A grain of dust carries a charge of −6.4 × 10−18 C. It is in a region where there is a uniform electric field of strength 2.5 × 104 N C−1, directed vertically upwards.
Define electric field strength.
Calculate the size of the electric force on the grain.
State the direction of the electric force on the grain and explain your answer.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the force per unit positive charge (acting on a small test charge) at that point | ✓ 1 | |
| Part (b) | ||
| F = EQ = 2.5 × 104 × 6.4 × 10−18 | ✓ 1 | |
| = 1.6 × 10−13 N | ✓ 1 | |
| Part (c) | ||
| vertically downwards, because the grain is negatively charged and the force on a negative charge is opposite to the field direction | ✓ 1 | |
Exam tip: The field direction is defined by the force on a positive charge. For a negative charge, reverse it.
Two horizontal metal plates are 2.5 cm apart. The upper plate is at a potential of +500 V and the lower plate is at 0 V.
Describe the electric field pattern in the region between the central parts of the plates.
Calculate the electric field strength between the plates.
State the potential at a point between the plates 1.0 cm above the lower plate.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| parallel, equally spaced field lines perpendicular to the plates | ✓ 1 | |
| directed from the upper (positive) plate to the lower plate / downwards | ✓ 1 | |
| Part (b) | ||
| E = V/d = 500 ÷ 0.025 | ✓ 1 | |
| = 2.0 × 104 V m−1 | ✓ 1 | |
| Part (c) | ||
| the potential rises uniformly at 2.0 × 104 V m−1 × 0.010 m: 2.0 × 102 V (200 V) | ✓ 1 | |
Model sketch
Exam tip: In a uniform field the potential changes by equal amounts over equal distances, so a point 1.0 cm into a 2.5 cm gap is 1.0/2.5 of the way through the p.d.
The figure shows the electric field lines around two point charges X and Y.
Which statement about X and Y is correct?
| Answer | Mark | Notes |
|---|---|---|
| The arrows point away from X and into Y, so X is positive and Y is negative. Twelve lines leave X but only six end on Y; the other six continue outwards. The number of lines is proportional to the size of the charge, so the charge on X is larger in magnitude (here twice as large). | ✓ 1 | Answer B |
Answer: B
Exam tip: A field-line diagram tells you two things: the arrows give the sign of each charge (out of positive, into negative), and the number of lines at each charge gives its relative size.
A small charged bead is placed at a fixed point P in an electric field. The charge Q on the bead is changed and the electric force F on it is measured each time. The graph shows the results.
What is the electric field strength at P?
| Answer | Mark | Notes |
|---|---|---|
| E = F/Q, which is the gradient of the graph: 12.0 × 10−3 N ÷ 5.0 × 10−6 C = 2.4 × 103 N C−1. | ✓ 1 | Answer D |
Answer: D
Common error: Ignoring the prefixes on the axes. Convert mN to N (× 10−3) and μC to C (× 10−6) before working out the gradient.
The figure shows equipotential lines, drawn at equal intervals of 50 V, around two point charges. Four points P, Q, R and S are marked.
At which point is the electric field strength greatest?
| Answer | Mark | Notes |
|---|---|---|
| Field strength is the potential gradient, ΔV/Δx. With equal 50 V steps, the field is strongest where the equipotentials are closest together. They are packed most tightly round the negative charge, so the field is greatest at R. | ✓ 1 | Answer C |
Answer: C
Exam tip: On an equipotential map drawn at equal steps of potential, judge field strength only from the spacing of the lines. The value of the potential at a point says nothing about the field there.
Two identical small conducting spheres carry equal charges. The graph shows how the electrostatic force F between them varies with the separation r of their centres.
Use the labelled points to show that the force obeys an inverse square law.
Calculate the charge on each sphere.
Predict the force between the spheres when their separation is 3.0 cm.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| e.g. from 4.0 cm to 12.0 cm, r increases by a factor of 3 and F falls from 9.0 mN to 1.0 mN, a factor of 9 = 32 | ✓ 1 | |
| Fr2 is the same for all three points: 9.0 × 10−3 × 0.0402 = 4.0 × 10−3 × 0.0602 = 1.0 × 10−3 × 0.1202 = 1.44 × 10−5 N m2 | ✓ 1 | |
| Part (b) | ||
| Q2 = Fr2 × 4πε0 = 1.44 × 10−5 ÷ 8.99 × 109 = 1.60 × 10−15 C2 | ✓ 1 | |
| Q = 4.0 × 10−8 C | ✓ 1 | |
| Part (c) | ||
| F = 1.44 × 10−5 ÷ 0.0302 = 1.6 × 10−2 N (16 mN) | ✓ 1 | |
Common error: Leaving r in centimetres. Convert to metres before squaring: 4.0 cm = 0.040 m, so r2 = 1.6 × 10−3 m2.
The figure shows the electric field lines around an isolated small sphere that carries a charge of magnitude 15 nC. Points P and Q are 0.10 m and 0.20 m from the centre of the sphere.
State the sign of the charge on the sphere and explain how the figure shows this.
Calculate the electric field strength at P.
Without using the equation again, state the field strength at Q. Explain your answer. (The figure shows only a flat slice through the field; in three dimensions the field lines point in towards the sphere from every direction.)
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| negative, because the field lines point towards the sphere: the field direction is the direction of the force on a positive charge, which is attracted to it | ✓ 1 | |
| Part (b) | ||
| E = 8.99 × 109 × 15 × 10−9 ÷ 0.102 | ✓ 1 | |
| = 1.3 × 104 N C−1 (1.35 × 104 N C−1) | ✓ 1 | |
| Part (c) | ||
| the field lines are spread over the surface of a sphere of area 4πr2; at twice the distance this area is four times as large, so the lines are a quarter as concentrated and the field is a quarter as strong (inverse square law) | ✓ 1 | |
| E = 1.35 × 104 ÷ 4 = 3.4 × 103 N C−1 (allow error carried forward) | ✓ 1 | |
Common error: Halving the field when the distance doubles. In a flat diagram the lines at twice the distance look only half as close together, but in three dimensions they spread over four times the area: the field of a point charge follows an inverse square law, so doubling the distance divides the field by four.
A small sphere carries a charge of −8.0 nC.
What is the size of the electric field strength at a distance of 15 cm from the centre of the sphere?
| Answer | Mark | Notes |
|---|---|---|
| E = Q/4πε0r2 = 8.99 × 109 × 8.0 × 10−9 ÷ 0.152 = 3.2 × 103 N C−1 (3196 N C−1), directed towards the sphere. | ✓ 1 | Answer C |
Answer: C
Common error: Using ε0 where 1/4πε0 is needed. Either use 8.99 × 109 N m2 C−2 directly, or keep the 4π in the denominator.
Which statement about equipotentials in an electric field is correct?
| Answer | Mark | Notes |
|---|---|---|
| Every point on an equipotential has the same potential, so the p.d. along it is zero and no work is done moving a charge along it. It follows that field lines cross equipotentials at right angles (B is wrong), that the field is strongest where equipotentials are closest (D is wrong), and around a point charge V ∝ 1/r makes equipotentials spread out with distance (C is wrong). | ✓ 1 | Answer A |
Answer: A
Exam tip: Field strength = −potential gradient. Closely spaced equipotentials mean a steep gradient and a strong field.
What is the electric potential at a distance of 0.25 m from a point charge of +2.0 nC?
| Answer | Mark | Notes |
|---|---|---|
| V = Q/4πε0r = 8.99 × 109 × 2.0 × 10−9 ÷ 0.25 = 72 V (71.9 V). | ✓ 1 | Answer B |
Answer: B
Common error: Squaring r in the potential equation. Potential falls as 1/r; only the force and the field strength fall as 1/r2.
An electron is accelerated from rest through a potential difference of 500 V.
What is its final speed?
| Answer | Mark | Notes |
|---|---|---|
| Energy gained = eV = 1.60 × 10−19 × 500 = 8.0 × 10−17 J = ½mv2, so v = √(2 × 8.0 × 10−17 ÷ 9.11 × 10−31) = 1.3 × 107 m s−1. | ✓ 1 | Answer C |
Answer: C
Exam tip: Kinetic energy gained = charge × p.d. It does not depend on the distance over which the p.d. acts.
In a simple model of a helium ion, He+, a single electron moves in a circular orbit of radius 2.6 × 10−11 m around a nucleus of charge +2e. The mass of the nucleus is 6.64 × 10−27 kg.
Calculate the electrostatic force on the electron.
This force provides the centripetal force on the electron. Calculate the speed of the electron.
Show that the gravitational force between the nucleus and the electron can be ignored.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| F = 8.99 × 109 × (2 × 1.60 × 10−19) × 1.60 × 10−19 ÷ (2.6 × 10−11)2 | ✓ 1 | |
| = 6.8 × 10−7 N | ✓ 1 | |
| Part (b) | ||
| F = mv2/r, so v = √(Fr/m) = √(6.81 × 10−7 × 2.6 × 10−11 ÷ 9.11 × 10−31) | ✓ 1 | |
| = 4.4 × 106 m s−1 (allow error carried forward) | ✓ 1 | |
| Part (c) | ||
| Fg = 6.67 × 10−11 × 6.64 × 10−27 × 9.11 × 10−31 ÷ (2.6 × 10−11)2 = 6.0 × 10−46 N, about 1039 times smaller than the electric force | ✓ 1 | |
Exam tip: Keep the charges as products of e: here Q1 = 2e and Q2 = e. Square r in metres.
The dome of a demonstration electrostatic generator is a metal sphere of radius 0.15 m. It carries a charge of +1.2 μC, which may be treated as a point charge at the centre of the sphere.
Calculate the electric field strength at the surface of the dome.
Calculate the electric potential at the surface of the dome.
The air next to the dome starts to conduct when the field strength exceeds 3.0 × 106 V m−1. Calculate the maximum charge the dome can hold.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| E = 8.99 × 109 × 1.2 × 10−6 ÷ 0.152 | ✓ 1 | |
| = 4.8 × 105 N C−1 | ✓ 1 | |
| Part (b) | ||
| V = 8.99 × 109 × 1.2 × 10−6 ÷ 0.15 | ✓ 1 | |
| = 7.2 × 104 V | ✓ 1 | |
| Part (c) | ||
| Q = 4πε0r2E = 3.0 × 106 × 0.152 ÷ 8.99 × 109 | ✓ 1 | |
| = 7.5 × 10−6 C (7.5 μC) | ✓ 1 | |
Common error: Using the same equation for field and potential. Field strength has r2 in the denominator; potential has r.
A small sphere A with a charge of +80 nC is fixed to a smooth horizontal insulating surface. A second small sphere B, of mass 2.0 g and charge +50 nC, is held at rest on the surface with its centre 0.10 m from the centre of A. Sphere B is then released. Ignore air resistance and friction.
Calculate the initial acceleration of B.
Calculate the electric potential due to A at the starting position of B.
Calculate the speed of B when it is a very long way from A.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| F = 8.99 × 109 × 80 × 10−9 × 50 × 10−9 ÷ 0.102 = 3.6 × 10−3 N | ✓ 1 | |
| a = F/m = 3.6 × 10−3 ÷ 2.0 × 10−3 = 1.8 m s−2 | ✓ 1 | |
| Part (b) | ||
| V = 8.99 × 109 × 80 × 10−9 ÷ 0.10 = 7.2 × 103 V (7190 V) | ✓ 1 | |
| Part (c) | ||
| electric potential energy of B at the start = QV = 50 × 10−9 × 7190 = 3.6 × 10−4 J | ✓ 1 | |
| far away the potential is zero, so all of this energy has become kinetic energy | ✓ 1 | |
| v = √(2 × 3.6 × 10−4 ÷ 2.0 × 10−3) = 0.60 m s−1 | ✓ 1 | |
Exam tip: The force changes as B moves away, so suvat cannot be used. Use energy: the potential energy QV at the start becomes kinetic energy.
A student investigates how the force between two charged spheres depends on their separation. A small conducting sphere on an insulating stand rests on a top-pan balance. An identical sphere is held vertically above it on an insulating rod. Both spheres are given equal positive charges, and the increase Δm in the balance reading is recorded for different centre-to-centre separations r.
Results: r = 4.0 cm, Δm = 0.311 g; 5.0 cm, 0.201 g; 6.0 cm, 0.138 g; 8.0 cm, 0.078 g; 10.0 cm, 0.050 g.
Explain why a graph of Δm against 1/r2 can be used to test Coulomb's law.
Calculate the electric force between the spheres when r = 4.0 cm.
Use your answer to (b) to determine the charge on each sphere.
The separation is measured with an uncertainty of ±1 mm. Calculate the percentage uncertainty in the value of 1/r2 when r = 4.0 cm.
With the spheres held at a fixed separation, the balance reading slowly decreases. Suggest a reason for this.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the electric force F = Δm × g, and Coulomb's law predicts F ∝ 1/r2 for fixed charges | ✓ 1 | |
| so if the law holds the graph is a straight line through the origin | ✓ 1 | |
| Part (b) | ||
| F = 0.311 × 10−3 × 9.81 = 3.05 × 10−3 N | ✓ 1 | |
| Part (c) | ||
| Q2 = Fr2 × 4πε0 = 3.05 × 10−3 × 0.0402 ÷ 8.99 × 109 = 5.4 × 10−16 C2 | ✓ 1 | |
| Q = 2.3 × 10−8 C (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| percentage uncertainty in r = 0.1 ÷ 4.0 × 100 = 2.5% | ✓ 1 | |
| percentage uncertainty in 1/r2 = 2 × 2.5 = 5.0% | ✓ 1 | |
| Part (e) | ||
| charge leaks from the spheres (into the air or through the supports), so the force between them decreases | ✓ 1 | |
Model sketch
Exam tip: When a quantity is raised to a power, multiply its percentage uncertainty by that power: 1/r2 has twice the percentage uncertainty of r.
In an experiment like Millikan's, charged oil drops are observed between two horizontal plates 6.0 mm apart. For each drop the p.d. between the plates is adjusted until the drop is stationary, and the mass of the drop is found by a separate method. Each drop carries excess electrons.
Drop 1: mass 2.4 × 10−15 kg, balancing p.d. 441 V. Drop 2: mass 4.0 × 10−15 kg, balancing p.d. 245 V.
Explain how a drop can be held stationary, and state which plate must be positive.
Calculate the electric field strength between the plates when drop 1 is stationary.
Calculate the charge on each drop.
State what these results suggest about electric charge.
The mass of drop 2 has an uncertainty of ±4% and the plate separation an uncertainty of ±0.1 mm; the uncertainty in the p.d. is negligible. Calculate the percentage uncertainty in the charge on drop 2 and deduce whether the drop certainly carries six excess electrons.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the electric force on the drop is upwards and equal in size to its weight, so the resultant force is zero | ✓ 1 | |
| the drop is negative, so the upper plate must be positive | ✓ 1 | |
| Part (b) | ||
| E = 441 ÷ 6.0 × 10−3 = 7.4 × 104 V m−1 (7.35 × 104 V m−1) | ✓ 1 | |
| Part (c) | ||
| Q = mgd/V: drop 1, 2.4 × 10−15 × 9.81 × 6.0 × 10−3 ÷ 441 = 3.2 × 10−19 C | ✓ 1 | |
| drop 2, 4.0 × 10−15 × 9.81 × 6.0 × 10−3 ÷ 245 = 9.6 × 10−19 C | ✓ 1 | |
| Part (d) | ||
| both charges are whole-number multiples of 1.6 × 10−19 C (2e and 6e), suggesting charge is quantised in units of e | ✓ 1 | |
| Part (e) | ||
| percentage uncertainty = 4 + (0.1 ÷ 6.0 × 100) = 4 + 1.7 = 5.7% | ✓ 1 | |
| so Q lies between about 9.1 × 10−19 C and 10.2 × 10−19 C (5.7e to 6.3e), which excludes 5e and 7e: the drop carries six electrons | ✓ 1 | |
Model sketch
Exam tip: For a product or quotient, add the percentage uncertainties of the measured quantities. Then turn the percentage back into a range of values before drawing a conclusion.
Three small charged spheres A, B and C are fixed in the positions shown in the figure. The angle at C is 90°.
What is the size of the resultant electrostatic force on C?
| Answer | Mark | Notes |
|---|---|---|
| Force from A: 8.99 × 109 × 6.0 × 10−9 × 5.0 × 10−9 ÷ 0.302 = 3.00 × 10−6 N (downwards, away from A). Force from B: 8.99 × 109 × 8.0 × 10−9 × 5.0 × 10−9 ÷ 0.402 = 2.25 × 10−6 N (to the left, away from B). The two forces are at right angles: resultant = √(3.002 + 2.252) × 10−6 = 3.7 × 10−6 N. | ✓ 1 | Answer C |
Answer: C
Exam tip: Forces from different charges are vectors. When they act at right angles, combine them with Pythagoras, never by simple addition.
The graph shows how the electric field strength E varies with distance r from the centre of a small charged sphere, which acts as a point charge.
What is the magnitude of the charge on the sphere?
| Answer | Mark | Notes |
|---|---|---|
| From the labelled point, E = 900 N C−1 at r = 20 cm = 0.20 m. Q = 4πε0r2E = 900 × 0.202 ÷ 8.99 × 109 = 4.0 × 10−9 C. | ✓ 1 | Answer B |
Answer: B
Exam tip: Read a value from a labelled point or a gridline intersection, then check the axis units: here r is in centimetres and must be converted to metres before it is squared.
The figure shows two parallel plates and the equipotentials between them. A particle with charge +3.0 μC is moved slowly in a straight line from point P to point Q.
How much work is done on the particle?
| Answer | Mark | Notes |
|---|---|---|
| The work done depends only on the potential difference between the end points: W = QΔV = 3.0 × 10−6 × (400 − 100) = 9.0 × 10−4 J. The sideways part of the movement is along equipotentials and needs no work. | ✓ 1 | Answer A |
Answer: A
Common error: Multiplying the force by the straight-line distance PQ. Only the displacement along the field counts, which is exactly what QΔV gives.
The electric potential V is measured at several distances r from a small charged sphere, which acts as a point charge. The graph shows V plotted against 1/r.
What is the charge on the sphere?
| Answer | Mark | Notes |
|---|---|---|
| V = Q/4πε0r, so the gradient of V against 1/r is Q/4πε0 = 900 ÷ 20 = 45 V m. Q = 45 ÷ 8.99 × 109 = 5.0 × 10−9 C. | ✓ 1 | Answer D |
Answer: D
Exam tip: When a graph is plotted to give a straight line, compare it with y = mx: here y = V, x = 1/r, so the gradient m = Q/4πε0.
Two small spheres with charges +4.0 nC and +9.0 nC are fixed 0.50 m apart. Points A, B, C and D lie on the line between them, as shown.
At which point is the resultant electric field strength zero?
| Answer | Mark | Notes |
|---|---|---|
| At the neutral point the two fields are equal and opposite: 4.0/x2 = 9.0/(0.50 − x)2, so (0.50 − x)/x = √(9.0/4.0) = 1.5 and x = 0.20 m from the +4.0 nC charge: point B. The neutral point is nearer the smaller charge. | ✓ 1 | Answer B |
Answer: B
Common error: Forgetting the square root. The field falls as 1/r2, so the distances are in the ratio √(9.0/4.0) = 1.5, not 9.0/4.0.
A small ball of mass 0.80 g carries a negative charge. It hangs from an insulating thread between two vertical parallel plates A and B, which are connected to a 4.5 kV supply. The figure shows the ball in equilibrium.
State which plate is positive. Give a reason for your answer.
Calculate the electric field strength between the plates.
By considering the forces on the ball, show that the electric force on it is about 1.2 × 10−3 N.
Calculate the charge on the ball.
The p.d. is increased to 9.0 kV. Calculate the new angle between the thread and the vertical.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| plate A, because the negatively charged ball is pulled towards it (the force on a negative charge is towards the positive plate) | ✓ 1 | |
| Part (b) | ||
| E = V/d = 4500 ÷ 0.080 = 5.6 × 104 V m−1 (56 250 V m−1) | ✓ 1 | |
| Part (c) | ||
| vertically T cos 9.0° = mg and horizontally T sin 9.0° = F, so F = mg tan 9.0° | ✓ 1 | |
| F = 8.0 × 10−4 × 9.81 × tan 9.0° = 1.24 × 10−3 N | ✓ 1 | |
| Part (d) | ||
| Q = F/E = 1.24 × 10−3 ÷ 5.625 × 104 = 2.2 × 10−8 C (allow error carried forward) | ✓ 1 | |
| Part (e) | ||
| the field, and so the electric force, doubles: tan θ = 2 × tan 9.0° = 0.317 | ✓ 1 | |
| θ = 18° (17.6°) | ✓ 1 | |
Exam tip: For a charged object hanging at rest, resolve the tension: its vertical component balances the weight and its horizontal component balances the electric force, so F = mg tan θ.
The graph shows how the electric potential V varies with distance r from the centre of a small charged sphere, which acts as a point charge. The tangent to the curve at r = 0.30 m has been drawn.
Use the labelled point (0.30, 360) to calculate the charge on the sphere.
Use the tangent to determine the electric field strength at r = 0.30 m.
Show that your answer to (b) agrees with the value given by E = Q/4πε0r2.
Calculate the work done in moving a charge of +2.0 nC from r = 0.60 m to r = 0.30 m.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Q = 4πε0rV = 360 × 0.30 ÷ 8.99 × 109 | ✓ 1 | |
| = 1.2 × 10−8 C | ✓ 1 | |
| Part (b) | ||
| gradient = (0 − 720) ÷ (0.60 − 0) = −1200 V m−1 | ✓ 1 | |
| E = −gradient = 1.2 × 103 V m−1 (allow 1.15–1.25 × 103 V m−1) | ✓ 1 | |
| Part (c) | ||
| E = 8.99 × 109 × 1.20 × 10−8 ÷ 0.302 = 1.2 × 103 N C−1, the same value | ✓ 1 | |
| Part (d) | ||
| ΔV = 360 − 180 = 180 V | ✓ 1 | |
| W = QΔV = 2.0 × 10−9 × 180 = 3.6 × 10−7 J | ✓ 1 | |
Exam tip: E = −ΔV/Δr: the field strength is the magnitude of the gradient of the V–r graph. Take the gradient from points far apart on the tangent, such as where it meets the axes.
A student investigates the force on a charged object in a uniform electric field. A small charged sphere on an insulating rod stands on a top-pan balance, midway between two horizontal metal plates 5.0 cm apart, as shown at the top of the figure. The lower plate is earthed and the potential V of the upper plate is increased in steps. The student converts each change in the balance reading into the extra downward force F on the sphere and plots the graph shown.
Explain why the graph is expected to be a straight line through the origin.
Draw a line of best fit and use it to determine the charge on the sphere.
The balance reading increases when the upper plate is made positive. Deduce the sign of the charge on the sphere.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the field between the plates is uniform, with E = V/d | ✓ 1 | |
| F = QE = QV/d, and Q and d are constant, so F ∝ V | ✓ 1 | |
| Part (b) | ||
| straight line of best fit through the origin with the points scattered evenly about it, e.g. through (5.0 kV, 2.50 mN) | ✓ 1 | |
| gradient = 2.50 × 10−3 ÷ 5000 = 5.0 × 10−7 N V−1 (allow 4.9–5.1 × 10−7) | ✓ 1 | |
| Q = gradient × d = 5.0 × 10−7 × 0.050 = 2.5 × 10−8 C (allow 2.45–2.55 × 10−8 C) | ✓ 1 | |
| Part (c) | ||
| the field points downwards, from the positive upper plate to the earthed plate, and the force on the sphere is also downwards, along the field, so the charge is positive | ✓ 1 | |
Exam tip: When a graph is expected to pass through the origin, use the gradient of the best-fit line, not one data point: a single point carries all of its own random error.
Two small spheres A and B carry charges of +9.0 nC and +1.0 nC. They are fixed 0.40 m apart. The graph shows how the electric potential V varies with distance x from A, along the line between the spheres. The tangent to the curve at x = 0.10 m has been drawn.
State the value of x at which the electric field strength is zero, and explain how the graph shows this.
Show, using the charges, that the field strength is zero at this point.
Use the tangent to determine the electric field strength at x = 0.10 m, and state its direction.
A proton is at rest at x = 0.30 m. Calculate the minimum kinetic energy it must be given to reach x = 0.10 m.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| x = 0.30 m | ✓ 1 | |
| the graph has zero gradient there (its minimum), and E = −ΔV/Δx | ✓ 1 | |
| Part (b) | ||
| EA = 8.99 × 109 × 9.0 × 10−9 ÷ 0.302 = 899 N C−1 and EB = 8.99 × 109 × 1.0 × 10−9 ÷ 0.102 = 899 N C−1: equal in size and opposite in direction | ✓ 1 | |
| Part (c) | ||
| gradient = (440 − 1240) ÷ (0.150 − 0.050) = −8.0 × 103 V m−1 | ✓ 1 | |
| E = 8.0 × 103 V m−1, directed away from A towards B (+x direction) | ✓ 1 | |
| Part (d) | ||
| potential difference = 839 − 360 = 479 V | ✓ 1 | |
| Ek = eΔV = 1.60 × 10−19 × 479 = 7.7 × 10−17 J | ✓ 1 | |
Exam tip: On a V–x graph the field is zero where the graph is flat, not where V is zero. Potentials from two positive charges add, so V never falls to zero between them.
A small bead with charge +8.0 nC can move along a smooth insulating track that points directly away from a fixed charged sphere. The graph shows how the electric force F on the bead varies with its distance r from the centre of the sphere. The area under the graph between r = 5.0 cm and r = 10.0 cm is shaded.
State what the shaded area represents.
Estimate the work done by the electric force as the bead moves from r = 5.0 cm to r = 10.0 cm.
Use the labelled point (5.0, 8.0) to calculate the charge on the sphere.
Calculate the potential due to the sphere at r = 5.0 cm and at r = 10.0 cm, and use these values to check your answer to (b).
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the work done on the bead by the electric force as it moves from 5.0 cm to 10.0 cm (equal to the loss of electric potential energy) | ✓ 1 | |
| Part (b) | ||
| about 20 grid squares under the curve (allow 19–21) | ✓ 1 | |
| each square represents 1.0 × 10−3 N × 0.010 m = 1.0 × 10−5 J | ✓ 1 | |
| work ≈ 2.0 × 10−4 J (allow 1.9–2.1 × 10−4 J) | ✓ 1 | |
| Part (c) | ||
| Q = Fr2/kq = 8.0 × 10−3 × 0.0502 ÷ (8.99 × 109 × 8.0 × 10−9) | ✓ 1 | |
| = 2.8 × 10−7 C | ✓ 1 | |
| Part (d) | ||
| V = kQ/r: 5.0 × 104 V at 5.0 cm and 2.5 × 104 V at 10.0 cm | ✓ 1 | |
| work = qΔV = 8.0 × 10−9 × 2.5 × 104 = 2.0 × 10−4 J, which agrees | ✓ 1 | |
Exam tip: The area under a force–distance graph is the work done, even when the force varies. Work out the energy one grid square represents before you count squares.
The electric potential V in a region varies with position x. At point P, the gradient of the graph of V against x is −3.0 × 103 V m−1.
What is the electric force on an electron at P?
| Answer | Mark | Notes |
|---|---|---|
| E = −ΔV/Δx = +3.0 × 103 V m−1, so the field points in the +x direction (the way the potential falls). Force = eE = 1.60 × 10−19 × 3.0 × 103 = 4.8 × 10−16 N. The electron is negative, so the force is opposite to the field: 4.8 × 10−16 N in the −x direction, towards higher potential. | ✓ 1 | Answer C |
Answer: C
Exam tip: Work out the field direction first (from high to low potential), then reverse it for a negative charge. Electrons are pushed towards higher potential.
A charged oil drop is held stationary between two horizontal parallel plates when the p.d. between them is 400 V. The p.d. is suddenly increased to 600 V.
What is the initial acceleration of the drop?
| Answer | Mark | Notes |
|---|---|---|
| When balanced, QE = mg. The electric force is proportional to V, so at 600 V it is 1.5mg upwards. Resultant force = 1.5mg − mg = 0.5mg, so a = 0.5 × 9.81 = 4.9 m s−2 upwards. | ✓ 1 | Answer A |
Answer: A
Exam tip: Write the balance condition first, then scale the electric force with the p.d. (E = V/d with d fixed). The weight does not change.
A student paints two long, straight, parallel electrodes 8.0 cm apart on a sheet of conducting paper. One electrode is connected to +6.0 V and the other to 0 V. Using a probe connected to a high-resistance voltmeter, she measures the potential V at points on a line perpendicular to the electrodes, at distance x from the 0 V electrode.
Results: x = 1.0 cm, V = 0.78 V; 2.0 cm, 1.49 V; 3.0 cm, 2.27 V; 4.0 cm, 3.02 V; 5.0 cm, 3.73 V; 6.0 cm, 4.52 V; 7.0 cm, 5.24 V.
Describe how she could use the probe and voltmeter to plot the 3.0 V equipotential on the paper.
Use the results to determine the electric field strength between the electrodes.
Explain why the equipotentials must be perpendicular to the electric field lines.
Near the ends of the electrodes the equipotentials are found to curve and spread apart. State what this shows about the field in that region.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| move the probe over the paper until the voltmeter reads 3.0 V and mark that point | ✓ 1 | |
| repeat for many points across the paper and join them with a line | ✓ 1 | |
| Part (b) | ||
| gradient of V against x = (5.24 − 0.78) ÷ (0.070 − 0.010) | ✓ 1 | |
| E = 74 V m−1, directed towards the 0 V electrode (accept 73–76 V m−1) | ✓ 1 | |
| Part (c) | ||
| there is no p.d. between points on an equipotential, so no work is done moving a charge along it | ✓ 1 | |
| so the electric force, and hence the field, can have no component along the equipotential | ✓ 1 | |
| Part (d) | ||
| the field is not uniform there: it is weaker where the equipotentials are further apart | ✓ 1 | |
Model sketch
Exam tip: Field strength is the magnitude of the potential gradient. Use points far apart on the line, convert cm to m, and give the direction: from high to low potential.
Protons, initially at rest, are accelerated through a p.d. of 2.0 kV. They then enter the region midway between two horizontal parallel plates, moving parallel to the plates. The plates are 4.0 cm long and 1.2 cm apart, and the p.d. between them is 300 V.
Calculate the speed of the protons as they enter the region between the plates.
Calculate the time each proton spends between the plates.
Calculate the vertical displacement of a proton as it leaves the plates, and state whether it hits a plate.
Explain why the path of a proton between the plates is a parabola.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| ½mv2 = eV = 1.60 × 10−19 × 2000 = 3.2 × 10−16 J | ✓ 1 | |
| v = √(2 × 3.2 × 10−16 ÷ 1.67 × 10−27) = 6.2 × 105 m s−1 | ✓ 1 | |
| Part (b) | ||
| t = 0.040 ÷ 6.19 × 105 = 6.5 × 10−8 s | ✓ 1 | |
| Part (c) | ||
| E = 300 ÷ 0.012 = 2.5 × 104 V m−1, a = eE/m = 2.4 × 1012 m s−2 | ✓ 1 | |
| y = ½at2 = ½ × 2.40 × 1012 × (6.46 × 10−8)2 = 5.0 × 10−3 m (allow error carried forward) | ✓ 1 | |
| this is less than 6.0 mm (half the gap), so the proton leaves without hitting a plate | ✓ 1 | |
| Part (d) | ||
| there is no force parallel to the plates, so the horizontal velocity is constant and horizontal distance ∝ t | ✓ 1 | |
| the constant vertical force eE gives a constant vertical acceleration, so vertical displacement ∝ t2 and therefore ∝ (horizontal distance)2 | ✓ 1 | |
Exam tip: Treat the two directions separately, as for a projectile: the time comes from the constant horizontal velocity, the deflection from the constant vertical acceleration.
Electric fields can be radial, such as the field around an isolated point charge, or uniform, such as the field between the central parts of two oppositely charged parallel plates.
Define the electric potential at a point.
Compare the electric field and the potential around an isolated positive point charge with those between two oppositely charged parallel plates. Your answer should refer to field lines, equipotentials, and how the field strength and potential vary with position.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the work done per unit charge in bringing a small positive test charge from infinity to that point | ✓ 1 | |
| Part (b) | ||
| Indicative content: | — | |
| point charge: field lines are radial, pointing outwards from the charge | ✓ 1 | |
| point charge: field strength E = Q/4πε0r2, so the field is non-uniform and weakens with distance (inverse square) | ✓ 1 | |
| point charge: potential V = Q/4πε0r; equipotentials are concentric spheres that get further apart for equal steps of potential | ✓ 1 | |
| plates: field lines are parallel, equally spaced and perpendicular to the plates, from the positive to the negative plate | ✓ 1 | |
| plates: field strength E = V/d is the same everywhere (uniform) away from the edges, and the potential falls linearly across the gap | ✓ 1 | |
| plates: equipotentials are equally spaced planes parallel to the plates; in both fields equipotentials are perpendicular to field lines and E is minus the potential gradient | ✓ 1 | |
Model sketch
Exam tip: In a comparison, make every point about both fields, or pair them up clearly. Six separate, correct statements are worth more than a long description of one field.
A proton enters the uniform field between two parallel plates, moving parallel to the plates. The figure shows its path; the scale at the exit edge of the plates is in millimetres.
The p.d. between the plates is halved, and a proton enters along the same line with twice the original speed. What is the vertical deflection of this proton as it leaves the plates?
| Answer | Mark | Notes |
|---|---|---|
| The deflection read from the scale is 4.0 mm. Inside the plates y = ½at2 with a = eV/md ∝ V and t = L/v, so y ∝ V/v2. New deflection = 4.0 × ½ ÷ 22 = 0.50 mm. | ✓ 1 | Answer A |
Answer: A
Exam tip: For a ratio question, write the deflection in terms of the quantities that change: y = eVL2/2mdv2. Doubling the speed halves the time in the field, and the deflection depends on the time squared.
The top graph in the figure shows how the electric potential V varies with position x along a straight line.
Which graph, A to D, shows how the electric field strength E in the +x direction varies with x?
| Answer | Mark | Notes |
|---|---|---|
| E = −ΔV/Δx. From 0 to 2.0 cm V is constant, so E = 0. From 2.0 to 4.0 cm, E = −(0 − 80) ÷ 0.020 = +4.0 × 103 V m−1. From 4.0 to 8.0 cm, E = −(−40 − 0) ÷ 0.040 = +1.0 × 103 V m−1. This is graph C: the field points the way the potential falls. | ✓ 1 | Answer C |
Answer: C
Exam tip: Work region by region: a flat section of a V–x graph means zero field, a steeper section means a stronger field, and a falling potential means a field in the +x direction.
Two point charges A (+6.0 nC) and B (−2.0 nC) are fixed 0.40 m apart. Point P is 0.30 m directly above B, as shown in the figure.
Calculate the electric field strength at P due to each charge separately.
Determine the magnitude and direction of the resultant electric field strength at P.
Calculate the size of the electric force on an electron placed at P, and state its direction.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| AP = √(0.402 + 0.302) = 0.50 m; EA = 8.99 × 109 × 6.0 × 10−9 ÷ 0.502 = 216 N C−1 (215.8 N C−1) | ✓ 1 | |
| EB = 8.99 × 109 × 2.0 × 10−9 ÷ 0.302 = 2.00 × 102 N C−1 (199.8 N C−1) | ✓ 1 | |
| Part (b) | ||
| EA acts along AP, away from A: components 215.8 × 0.8 = 172.6 N C−1 to the right and 215.8 × 0.6 = 129.5 N C−1 upwards; EB = 199.8 N C−1 acts downwards, towards B | ✓ 1 | |
| resultant components: 172.6 N C−1 to the right and 199.8 − 129.5 = 70.3 N C−1 downwards; magnitude √(172.62 + 70.32) = 1.9 × 102 N C−1 (186 N C−1) | ✓ 1 | |
| direction: tan−1(70.3 ÷ 172.6) = 22° below the horizontal, pointing to the right (away from A) | ✓ 1 | |
| Part (c) | ||
| F = eE = 1.60 × 10−19 × 186 = 3.0 × 10−17 N (allow error carried forward) | ✓ 1 | |
| opposite to the field: 22° above the horizontal, pointing to the left | ✓ 1 | |
Exam tip: Field strengths from different charges add as vectors. Draw each one at P with its direction (away from a positive charge, towards a negative one), then add horizontal and vertical components.
Two horizontal parallel plates are 5.0 cm apart. The lower plate is at 0 V and has a small hole in it; the upper plate is at −100 V. An electron passes up through the hole with a speed of 4.0 × 106 m s−1 at 30° to the plate, as shown in the figure. Gravitational effects are negligible.
Explain why the path of the electron between the plates is a parabola.
Calculate the acceleration of the electron.
Show that the electron rises about 6 mm above the lower plate, and so does not reach the upper plate.
Calculate the distance from the hole to the point where the electron lands on the lower plate.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the field is uniform and vertical, so the force on the electron is constant and vertical (downwards), and the horizontal component of velocity stays constant | ✓ 1 | |
| the vertical motion has constant acceleration, so vertical displacement depends on t2 while horizontal displacement is proportional to t, as for a projectile | ✓ 1 | |
| Part (b) | ||
| E = 100 ÷ 0.050 = 2.0 × 103 V m−1 | ✓ 1 | |
| a = eE/m = 1.60 × 10−19 × 2.0 × 103 ÷ 9.11 × 10−31 = 3.5 × 1014 m s−2 | ✓ 1 | |
| Part (c) | ||
| vertical component of velocity = 4.0 × 106 × sin 30° = 2.0 × 106 m s−1, and h = u2/2a = (2.0 × 106)2 ÷ (2 × 3.51 × 1014) | ✓ 1 | |
| h = 5.7 × 10−3 m, much less than the 5.0 cm gap | ✓ 1 | |
| Part (d) | ||
| time of flight t = 2 × 2.0 × 106 ÷ 3.51 × 1014 = 1.14 × 10−8 s | ✓ 1 | |
| distance = 4.0 × 106 × cos 30° × 1.14 × 10−8 = 3.9 × 10−2 m (allow error carried forward) | ✓ 1 | |
Exam tip: Resolve the initial velocity before anything else. Only the component along the field changes; the component parallel to the plates stays the same for the whole flight.
The figure shows the electric field lines and some equipotentials around two equal positive point charges. N is the point midway between the charges.
Explain why the electric potential at N is not zero, although the electric field strength at N is zero.
Using the figure, describe and explain how the electric field strength and the electric potential vary along the line joining the two charges, and along the line through N perpendicular to it.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| potential is a scalar, so the two positive potentials add; field strength is a vector, and the two equal fields at N point in opposite directions and cancel | ✓ 1 | |
| Part (b) | ||
| Indicative content: | — | |
| between the charges the field at each point points away from the nearer charge, and it is zero only at N | ✓ 1 | |
| along the joining line E increases towards each charge: the field lines and the equipotentials are closer together there | ✓ 1 | |
| along the joining line V is lowest at N and rises towards each charge; the minimum at N matches E = −ΔV/Δx = 0 | ✓ 1 | |
| along the perpendicular line V is highest at N and falls with distance from N, as more equipotentials are crossed | ✓ 1 | |
| along the perpendicular line E is zero at N, rises to a maximum and then decreases, pointing along the line away from N | ✓ 1 | |
| far from the charges the field lines become radial and the equipotentials nearly circular, as for a single charge of +2Q; everywhere the equipotentials cross the field lines at right angles | ✓ 1 | |
Exam tip: In a description question, take each line in turn and say what happens to both E and V. Link each statement to a feature of the diagram: line spacing for E, equipotentials crossed for V.
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