· web viewthe graph below shows how the charge stored by a capacitor varies with time when it is...

Unit 4 mixed ABP FOR – 08-06-14 Hinchley Wood School

Q1. A wave of frequency 5 Hz travels at 8 km s–1 through a medium. What is the phase difference, in radians, between two points 2 km apart?

A 0

B

C π

D (Total 1 mark)

Q2.

A small loudspeaker emitting sound of constant frequency is positioned a short distance above a long glass tube containing water. When water is allowed to run slowly out of the tube, the intensity of the sound heard increases whenever the length l (shown above) takes certain values.

(a) Explain these observations by reference to the physical principles involved.

You may be awarded marks for the quality of written communication in your answer.


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(b) With the loudspeaker emitting sound of frequency 480 Hz, the effect described in part (a) is noticed first when l = 168 mm. It next occurs when l = 523 mm.

Use both values of l to calculate

(i) the wavelength of the sound waves in the air column,

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(ii) the speed of these sound waves.

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(Total 8 marks)

Q3. To find a value for the acceleration of free fall, g, a student measured the time of oscillation, T, of a simple pendulum whose length, l, is changed. The student used the results to plot a graph of T2 (y axis) against l (x axis) and found the slope of the line to be S. It follows that g is


A .

B 4π2S.

C .

D 2πS.(Total 1 mark)

Q4. A simple pendulum consists of a 25 g mass tied to the end of a light string 800 mm long. The mass is drawn to one side until it is 20 mm above its rest position, as shown in the diagram. When released it swings with simple harmonic motion.

(a) Calculate the period of the pendulum.

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(b) Show that the initial amplitude of the oscillations is approximately 0.18 m, and that the maximum speed of the mass during the first oscillation is about 0.63 m s–1.

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(c) Calculate the magnitude of the tension in the string when the mass passes through the lowest point of the first swing.

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(Total 8 marks)

Q5. A particle of mass m moves in a circle of radius r at a uniform speed with frequency f.What is the kinetic energy of the particle?

A

B

C 2π2 mf2r2

D 4π2 mf2r2

(Total 1 mark)

Q6. The Hubble space telescope was launched in 1990 into a circular orbit near to the Earth.It travels around the Earth once every 97 minutes.

(a) Calculate the angular speed of the Hubble telescope, stating an appropriate unit.

answer = .....................................(3)


(b) (i) Calculate the radius of the orbit of the Hubble telescope.

answer = ................................ m(3)

(ii) The mass of the Hubble telescope is 1.1 × 104 kg. Calculate the magnitude of the centripetal force that acts on it.

answer = ................................ N(2)

(Total 8 marks)

Q7. The diagram represents part of an experiment that is being used to estimate the speed of an air gun pellet.

The pellet which is moving parallel to the track, strikes the block, embedding itself. The trolley and the block then move along the track, rising a vertical height, h.


(a) Using energy considerations explain how the speed of the trolley and block immediately after it has been struck by the pellet, may be determined from measurements of h. Assume frictional forces are negligible.

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(b) The following data is collected from the experiment

mass of trolley and block 0.50 kgmass of pellet 0.0020 kgspeed of trolley and block immediately after impact 0.40 m s–1

Calculate

(i) the momentum of the trolley and block immediately after impact,

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(ii) the speed of the pellet just before impact.

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(c) (i) State what is meant by an inelastic collision.


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(ii) Use the data from part (b) to show that the collision between the pellet and block is inelastic.

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(Total 11 marks)

Q8. A body X moving with a velocity v makes an elastic collision with a stationary body Y of equal mass on a smooth horizontal surface.

Which line, A to D, in the table gives the velocities of the two bodies after the collision?

velocity of X velocity of Y

A

B

C v 0

D 0 v

(Total 1 mark)


Q9. Which one of the following statements about Newton’s law of gravitation is correct?

Newton’s law of gravitation explains

A the origin of gravitational forces.B why a falling satellite burns up when it enters the Earth’s atmosphere.C why projectiles maintain a uniform horizontal speed.D how various factors affect the gravitational force between two particles.

(Total 1 mark)

Q10.Two charges, each of + 0.8 nC, are 40 mm apart. Point P is 40 mm from each of the charges.

What is the electric potential at P?

A zero

B 180 V

C 360 V

D 4500 V(Total 1 mark)

Q11.Gravitational fields and electric fields have many features in common but also have several differences. For both radial and uniform gravitational and electric fields, compare and contrast their common features and their differences.

In your answer you should consider:

• the force acting between particles or charges

• gravitational field strength and electric field strength


• gravitational potential and electric potential.

The quality of your written communication will be assessed in your answer.(Total 6 marks)

Q12.The graph below shows how the charge stored by a capacitor varies with time when it is discharged through a fixed resistor.

(a) Determine the time constant, in ms, of the discharge circuit.

time constant ............................... ms(3)

(b) Explain why the rate of discharge will be greater if the fixed resistor has a smaller resistance.

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(Total 5 marks)

Q13. (a) In an experiment to illustrate electromagnetic induction, a permanent magnet is moved towards a coil, as shown in Figure 1, causing an emf to be induced across the coil.

Figure 1

Using Faraday’s law, explain why a larger emf would be induced in this experiment if a stronger magnet were moved at the same speed.

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(b) A conductor of length l is moved at constant speed v so that is passes perpendicularly through a uniform magnetic field of flux density B, as shown in Figure 2.

Figure 2


Show that the induced emf, , across the ends of the conductor is given by

= Blv.

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(c) A simple electrical generator can be made from a copper disc, which is rotated at right angles to a magnetic field, directed into the plane of the diagram (Figure 3). An emf is developed across the terminals P (connected to the axle) and Q (connected to a contact on the edge of the disc).

Figure 3

The radius of the disc is 64 mm and it is rotated at 16 revolutions per second in a uniform magnetic field of flux density 28 mT.

(i) Calculate the angular speed of the disc.

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(ii) Calculate the linear speed of mid-point M of a radius of the disc.

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(iii) Hence, or otherwise, calculate the emf induced across the terminals P and Q.

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(Total 11 marks)

Q14. The diagram shows a rigidly-clamped straight horizontal current-carrying wire held mid-way between the poles of a magnet on a top pan balance. The wire is perpendicular to the magnetic field direction.

The balance, which was zeroed before the switch was closed, reads 112 g after the switch is closed. If the current is reversed and doubled, what will be the new reading on the balance?

A –224 g

B –112 g

C zero

D 224 g(Total 1 mark)


M1. B[1]

M2. (a) reference to resonance (1)air set into vibration at frequency of loudspeaker (1)resonance when driving frequency = natural frequency of air column (1)more than one mode of vibration (1)stationary wave (in air column) (1) (or reference to nodes and antinodes)maximum amplitude vibration (or max energy transfer) at resonance (1)

[alternative answer to (a):first two marks as above, remaining four marks forwave reflected from surface (of water) (1)interference/superposition(between transmitted and reflected waves) (1)maximum intensity when path difference is nλ (1)maxima (or minima) observed when l changes by λ/2 (1)]

Max 4QWC 1

(b) (i) = 523 – 168 (1) (= 355 mm)

λ = 710 mm (1)

[if = 168, giving λ = 670 mm, (1) (1 max) (672 mm)]

(ii) c( = fλ) = 480 × 0.71 (1)= 341 m s–1 (1)(allow C.E. for incorrect λ from (i))[allow 480 × 0.67 = 320 m s–1 (1) (1max) (322 m s–1)]

4[8]


M3. A[1]

M4. (a) (use of T = 2π gives) T = 2π (1)

= 1.8 s (1)2

(b) mgh = ½ mv2 (1)

v = (1) (= 0.63 m s-1)

vmax = 2πfA = (1)

A = (1) (= 0.18m)

[or by Pythagoras A2 + 7802 = 8002

gives A = (1) ( = 180 mm)

(or equivalent solution by trigonometry (1) (1))

vmax = 2πfA or = (1)

= (1) (= 0.63 m s-1)4


(c) tension given by F, where F – mg = (1)

F = 25 × 10-3 = 0.26 N (1)2

[8]

M5. C[1]

M6. (a) ω = [or ]

= 1.1 × 10–3 (1.08 × 10–3) (1) [= 6.2 (6.19) × 10–2]

rad s–1 [accept s–1] (1) [degree s–1]3

(b) (i) or (1)

gives r3 = (1)

r = 6.99 × 106 (m) (1)3

(ii) F (= mω2r) = 1.1 × 104 × (1.08 × 10–3)2 × 6.99 × 106 (1)

= 9.0 × 104 (8.97 × 104) (N) (1)


[or (1)

= 9.0 × 104 (8.98 × 104) (N) (1)]2

[8]

M7. (a) kinetic energy changes to potential energy (1)potential energy calculated by measuring h (1)equate kinetic energy to potential energy to find speed (1) [or use h to find s (1) use g sinθ for a (1) use v2 = u2 + 2as (1)] [or use h to find s (1) time to travel s and calculate vav (1) v = 2vav (1)]

3

(b) (i) p(= mv) = 0.5(0) × 0.4(0) = 0.2(0) (1) N s(or kg m s–1) (1)

(ii) (use of mpvp = mtvt gives) 0.002(0) v = 0.2(0) (1)v = 100 m s–1 (1)

4

(c) (i) kinetic energy is not conserved (1)

(ii) initial kinetic energy = × 0.002 × 1002 = 10 (J) (1)

final kinetic energy = × 0.5 × 0.42 = 0.040 (J) (1)

hence change in kinetic energy (1)

(allow C.E. for value of v from (b))4

[11]


M8. D[1]

M9. D[1]

M10.C[1]

M11.The candidate’s writing should be legible and the spelling, punctuation and grammar should be sufficiently accurate for the meaning to be clear.

The candidate’s answer will be assessed holistically. The answer will be assigned to one of three levels according to the following criteria.

High Level (Good to excellent): 5 or 6 marks

The information conveyed by the answer is clearly organised, logical and coherent, using appropriate specialist vocabulary correctly. The form and style of writing is appropriate to answer the question.

The candidate gives a comprehensive account of the similarities and differences between gravitational and electric fields, referring to both radial and uniform fields. There are clear statements showing good to excellent understanding of the forces between masses/charges, gravitational and electric field strengths, and gravitational and electric potentials and of how aspects of them differ for gravitational and electric effects.

A High Level answer must refer to at least two valid similarities and at least two valid differences, and must also contain information about both radial and uniform fields.

Intermediate Level (Modest to adequate): 3 or 4 marks

The information conveyed by the answer may be less well organised and not fully coherent. There is less use of specialist vocabulary, or specialist vocabulary may be used


incorrectly. The form and style of writing is less appropriate.

The candidate’s comparisons are less complete but good understanding is shown of the factors affecting the respective forces and of the definitions of field strength and potential. There may be limited reference to radial and uniform fields. Similarities between gravitational and electric effects are better known than differences.

An Intermediate Level answer must refer to at least two valid similarities and at least one valid difference, and must also consider either radial or uniform fields, or both.

Low Level (Poor to limited): 1 or 2 marks

The information conveyed by the answer is poorly organised and may not be relevant or coherent. There is little correct use of specialist vocabulary. The form and style of writing may be only partly appropriate.

The candidate has a much weaker ability to convey the similarities and differences between gravitational and electric fields. There is likely to be little or no reference to radial and uniform fields. Factors affecting the respective forces are known, but understanding of field strength is likely to be weaker and understanding of potential may be poor or absent.

A Low Level answer must refer to at least one valid similarity but may not identify any valid differences.

The explanation expected in a competent answer should include a coherent selection of the following points.

(D) = valid difference.Good candidates are likely to mention other valid differences e.g. electrostatic forces depend on the medium whereas gravitational forces don’t, possibility of shielding an electric field but impossibility of shielding a gravitational field.

Forces

• In a radial field, both gravitational and electric involve an inverse square relationship.• In both cases the force is proportional to a product (masses / charges).• In both cases a spherical body may be considered to act as a point mass or charge

placed at the centre of the sphere.• In a uniform field the force is constant at all points.• Gravitational forces are always an attraction whilst electric forces may be attraction

or repulsion. (D)• Gravitational forces are usually much smaller than electric forces (unless very large

masses are involved). (D)

Field Strengths

• Both are defined as a force per unit mass or charge.• In both cases the field strength in a radial field is proportional to 1/r2

.• In both cases the field strength in a radial field is proportional to the magnitude of

the mass or charge that produces it.• In a uniform field the field has the same magnitude and same direction at all points.• A gravitational field is always directed towards the mass producing it whereas an

electric field is directed towards a negative charge but away from a positive charge. (D)


• A mass of 1 kg is small in terms of the gravitational field it produces but a charge of 1 C would produce a very strong electric field. (D)

Potentials

• Definitions of both involve work done in moving a mass or charge from infinity to a point.

• Both definitions involve the work done per unit mass or charge.• Both types of potential are proportional to 1/r in a radial field.• Both types of potential are proportional to the mass or charge producing them.• In a uniform field the potential varies linearly with distance.• The work done in moving a mass or charge across a potential difference is

calculated by multiplying the mass or charge by the potential difference.• Gravitational potential is always a negative quantity but electric potential is negative

for negative charges and positive for positive charges. (D)(max 6)

M12.(a) (Q = Q 0e−t /RC gives )1.0 = 4.0e−300 / RC

from which

[Alternative answer:time constant is time for charge to decrease to Q0 /e [or 0.37 Q0 ] 4.0/e = 1.47

reading from graph gives time constant = 216 ± 10 (ms) In alternative scheme, 4.0/e = 1.47 subsumes 1st mark. Also, accept T½ = 0.693 RC (or = ln 2 RC) for 1st mark.

3

(b) current is larger (for given V)(because resistance is lower)[or correct application of I = V / R] current is rate of flow of charge[or correct application of I =Δ Q / Δt]

larger rate of flow of charge (implies greater rate of discharge)[or causes larger rate of transfer of electrons from one plate back to the other]

[Alternative answer:time constant (or RC) is decreased (when R is decreased) explanation using Q = Q 0e−t / RC or time constant explained ]

Use either first or alternative scheme; do not mix and match.Time constant = RC is insufficient for time constant explained.

max 2[5]


M13. (a) greater flux (linkage) or more flux lines (at same distance)

[or stronger magnet produces flux lines closer together] (1)

greater rate of change of flux (linkage)

[or more flux lines cut per unit time] (1)

induced emf = [or =] rate of change of flux (linkage) (1)

[or using (1) ΔB is larger since magnet is stronger (1)

N, A and Δt are the same at the same speed is larger (1)]3

(b) area swept out ΔA = lvΔt (1)

ΔΦ (= BΔA) = BlvΔt (1)

gives result (1)3

(c) (i) ω (= 2πf) = 2π × 16 (1)

= 101 rad s–1 (1)

(ii) v (= rω) = 32 × 10–3 × 101 = 3.2(3) m s–1 (1)

(allow ecf for value of ω from (i))

(iii) (= Blv) = 28 × 10–3 × 64 × 10–3 × 3.23 (1)

= 5.7(9) × 10–3V (1)

(allow ecf for value of v from (ii))

[or accept solutions using = Bfπr2 to give 5.7(9) × 10–3 V]5


[11]

M14. A[1]


E1. This question was answered correctly by 58% of the candidates. Lack of understanding of radian measure when considering phase difference probably accounted for 27% of the candidates choosing distractor D (3π/2), rather than π/2.

E2. When dealing with free and forced vibrations, the specification requires ‘examples of these effects from more than one branch of Physics e.g. production of sound in a pipe instrument...’. Candidates’ responses to this question indicated that the great majority had studied these topics only in relation to mechanical vibrations, because references to them (or to resonance, even) were exceedingly rare in their answers. It was expected that candidates would appreciate that the loudspeaker sets the air into vibration at its driving frequency and that resonance occurs when this equals the natural frequency of the air column. Since the air column has more than one mode of vibration, further resonances occur at the same frequency for larger values of/. A high proportion of answers concentrated instead on how a stationary wave could be formed in the air column, by the superposition of the transmitted wave and its reflection from the water surface. Some credit could be obtained for this. Overall though, the mark for part (a) was rarely more than 2 out of 4.

The calculations in part (b) were easily solved by those who recognised that the distance between adjacent nodes in a stationary wave is λ/2. A common approach (only approximately correct) was to assume that λ/4 = 168 mm, and that 3 λ/4 = 523 mm; this gave two different values for λ that were then averaged. Half marks could be awarded for this method. Even less rewarding were the answers that began with λ = 523 – 168 = 355 mm. A compensatory single mark was given to those candidates who could achieve no more than quote the equation c = fλ in part (ii).

E3. This question was set in the context of a simple pendulum experiment, requiring candidates to show knowledge of how g could be found from the gradient of a graph of T2 against l. The facility was 69%. Distractor B, chosen by one in six, was the most popular incorrect response; this may suggest that these candidates had difficulty with algebraic re-arrangement.


E4. Answers to part (a) caused no serious difficulty and usually gained both marks by correct substitution of values into the well-known equation. Part (b) provided a greater challenge, but was usually met with partial success by the use of vmax = 2πfA. Many candidates attempted to produce the required two values by using this equation twice, once for vmax (by substituting 0.18 m) and then for A (by substituting 0.63 m s–1, which was also given in the question). This gained only two marks. It was necessary to break into the circular argument, either by energy conservation (giving vmax) or by use of Pythagoras (giving A), to access all four marks.

Most candidates were unable to marry oscillatory motion with the circular motion content of Unit 4 in order to solve part (c). In the vast majority of the work submitted this was treated as an equilibrium problem, with the tension equated to mg. A small minority of candidates, realising that centripetal force was involved, introduced mω2r rather than mv2/r. This approach was seldom successful, because of confusion between ω as the angular frequency of the SHM (which is constant) and ω as the angular velocity of the circular motion of the mass (which is not constant in this case).

E5. This question was the first of the re-banked questions from a previous examination. Three-quarters of the candidates were able to correctly combine v = 2πfr with Ek = ½ m v2 to arrive at the required algebraic result. Distractor D attracted one in eight responses, suggesting that the factor of ½ had been overlooked.

E6. This question as a whole was very rewarding for the candidates who were sufficiently familiar with the principles of gravitation to understand the mathematical conditions for a satellite in stable orbit, as required in part (b) (i). These candidates made good progress with all parts of the question, whereas many other candidates were only able to score well on parts (a) and (b) (ii). In part (a), the correct conversion of the orbital time of the Hubble satellite into seconds followed by correct use of ω = 2π/T, with a correct unit for angular speed, brought full marks for the majority of the candidates. Confusion of angular speed ω with linear speed v continues to be a problem, and giving the unit of ω as m s-1 inevitably caused the loss of one mark.


Part (b) (i) required candidates to appreciate that the radius of the orbit of a satellite can be found from the orbit equation GMm/r2 = mω2r. The angular speed ω had been determined in part (a), whilst the values for G and the Earth’s mass M could be taken from the Data and Formulae Booklet. Because the question had indicated that the Hubble telescope is in orbit close to the Earth, some candidates assumed that the radius of its orbit would be that of the Earth, 6.37 × 106 m.

Another common unsuccessful response was to attempt to determine the answer using the orbit relationship T2/r3 = constant, incorrectly treating the surface of the Earth as a satellite orbit and using T = 24 hours and r = 6.37 × 106 m.

Candidates who used F = mω2r, or F = GMm/r2, had very little difficulty in part (b) (ii), where both marks were still accessible to those who had worked out wrong values for ω and/or r in the earlier parts of the question. Attempts at this part using F = mv2/r were often incorrect because of inability to correctly work out the linear speed, v.

E7. This question was also done well although a significant minority of candidates could not explain in part (a) how the speed of the trolley and block might be determined from energy considerations. Part (b) was done well apart from the usual confusion over the unit for momentum.

The final part of the question produced more variable responses. Many candidates were able to explain correctly what is meant by an inelastic collision but were unable to carry out the necessary calculation to show that the collision of the pellet and the block was inelastic.

E8. An elastic collision between bodies of equal mass, one of which was stationary before the collision, was the subject of this question. Students who had witnessed such a collision on an air track, for example, should have had little difficulty in realising that the moving body stops whilst the second body moves off with all of the first body’s momentum. Fewer than expected (56%) gave the correct response. It is difficult to see why 36% of the students selected distractor B, in which neither momentum nor kinetic energy would be conserved.


E9. This question involving statements about Newton’s law of gravitation, had a facility of 85%. When pre-tested, this question had been found appreciably harder but was more discriminating than on this occasion.

E10.This question required students to understand that electric potential is a scalar, and that the potential at a point close to two charges is therefore the sum of the potentials due to each of them. This was understood by fewer than half of the candidates, and so the facility of the question was only 45%. Almost a third of responses were for distractor A (zero) and almost a fifth for distractor B (the potential due to one charge alone).

E11.This communications question on gravitational and electric fields gave plenty of scope for candidates to show what they had learned, and there were fewer totally unsatisfactory answers than have been presented in previous Unit 4 examinations. In this kind of question candidates should pay particular attention to the wording of the question and then make sure that their answer addresses its main requirements. On this occasion in a high level answer (5 or 6 marks) the examiners were looking for at least two valid similarities, at least two valid differences, and meaningful statements about radial and uniform fields. Relatively few answers satisfied these criteria, and so the majority of the candidates limited themselves to no more than 4 marks out of 6. Candidates who failed to identify at least one valid difference could not progress beyond 2 marks. One of the difficulties faced by candidates was that of recognising a valid difference. The differing constants of proportionality for gravitational and electric effects were not accepted as a valid difference, for example, since they are little more than conversion factors between different units.

The presentation of some of these answers was very impressive, whilst other answers were incoherent, disorganised and badly written. When deciding on the mark to award, examiners are inevitably influenced by the standard of presentation of the answer. It was often difficult for examiners to decide whether candidates were addressing what they thought were similarities, or differences, in their answer.

Free writing in communications questions gives candidates opportunities to reveal their misunderstandings and weaknesses. Amongst the answers there were many stating that forces are inversely proportional to separation (missing out the ‘squared’), whilst others referred to indirect proportion. Not uncommonly the product of the masses (or charges) was called the sum of them. Candidates quoted g = F / m together with E = V / d – whether intended as a similarity of a difference wasn’t clear – although these two equations are not really suitable to represent either. There were also many incorrect statements indicating that g = F /m and E = F / Q apply only to uniform fields. Some candidates thought it would be adequate to reproduce a comparison table of formulae, of the kind that might be used as a summary when teaching this topic. References to the definitions of gravitational and electric potential regularly quoted the incorrect direction of displacement of the mass or charge (‘from the point to infinity’ instead of ‘from infinity to


the point’).

E12.Part (a) required the evaluation of the time constant of an RC circuit from data on a graph of charge against time. This proved to be an easy test, and marks were high. The most economical solution followed from recognising that the charge falls to (1/e) of its initial value in a time equal to the time constant, or from appreciating that Q0 becomes Q0/2 in a time equal to ln 2 RC. More extensive answers that relied on a solution of Q = Q0e-t / RC were less common; in these it was essential for candidates to show their working correctly for full marks to be accessible. A few candidates knew that the time constant is equal to the time at which the capacitor would have discharged completely had the initial current been maintained. Therefore they drew a tangent to the curve at t = 0, continued the line to the time axis and then determined the required value by reading off the time.

Careless use of the language of physical quantities was sometimes an obstacle to progress in part (b). Loose terms such as “the current flows more quickly” (when the resistance is less) should be avoided: the candidate should have stated that the current is greater, or that more charge passes per second. The key to success in this part was to understand the meaning of a rate of change. Those who stated that the current is larger, and that current is the rate of flow of charge, readily scored both marks. Answers which stated that the time constant would be decreased were also accepted but it was then necessary to make reference to the implications, from Q = Q0e-t/RC, for the second mark to be awarded.

E14. This question involved a basic current balance and assumed familiarity with F = B I l. The simple idea here was that reversing the current and doubling it would produce a force in the opposite direction that would be twice the original. This eluded so many of the candidates that only 41% of them could select the correct response, whilst many of them chose distractors C or D.

· web viewthe graph below shows how the charge stored by a capacitor varies with time when it is...

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