Animated Science – A Science & Food Blog · Web view2021/03/03 · A positive ion with a mass of 3.4 × 10–26 kg and a charge of 3.2 × 10–19 C is initially at rest at a point

Colonel Frank Seely School

3.7.3.2 Electric Field Strength

Q1.A satellite orbiting the Earth moves to an orbit which is closer to the Earth.

Which line, A to D, in the table shows correctly what happens to the speed of the satellite and to the time it takes for one orbit of the Earth?

Speed of satellite Time For One Orbit Of Earth

A decreases decreases

B decreases increases

C increases decreases

D increases increases

(Total 1 mark)

Q2.A positive ion has a charge−to−mass ratio of 2.40 × 107 C kg–1. It is held stationary in a vertical electric field.Which line, A to D, in the table shows correctly both the strength and the direction of the electric field?

Electric field strength / V m–1

Direction

A 4.09 × 10–7 upwards

B 4.09 × 10–7 downwards

C 2.45 × 106 upwards

D 2.45 × 106 downwards

(Total 1 mark)

Q3.In the equation X = , X represents a physical variable in an electric or a gravitational field, a is a constant, b is either mass or charge and n is a number.

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Which line, A to D, in the table provides a consistent representation of X, a and b according to the value of n?

The symbols E, g, V and r have their usual meanings.

n X a b

A 1 E charge

B 1 V mass

C 2 g G mass

D 2 V G charge

(Total 1 mark)

Q4.Which one of the following statements is correct?

An electron follows a circular path when it is moving at right angles to

A a uniform magnetic field.

B a uniform electric field.

C uniform electric and magnetic fields which are perpendicular.

D uniform electric and magnetic fields which are in opposite directions.(Total 1 mark)

Q5.The figure below shows a system that separates two minerals from the ore containing them using an electric field.

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The crushed particles of the two different minerals gain opposite charges due to friction as they travel along the conveyor belt and through the hopper. When they leave the hopper they fall 4.5 metres between two parallel plates that are separated by 0.35 m.

(a) Assume that a particle has zero velocity when it leaves the hopper and enters the region between the plates.

Calculate the time taken for this particle to fall between the plates.

time taken = _____________s(2)

(b) A potential difference (pd) of 65 kV is applied between the plates.

Show that when a particle of specific charge 1.2 × 10–6 C kg–1 is between the plates its horizontal acceleration is about 0.2 m s–2.

(3)

(c) Calculate the total horizontal deflection of the particle that occurs when falling between the plates.

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horizontal deflection = _____________m(1)

(d) Explain why the time to fall vertically between the plates is independent of the mass of a particle.

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(e) State and explain two reasons, why the horizontal acceleration of a particle is different for each particle.

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(Total 12 marks)

Q6.The diagram shows a small negative charge at a point in an electric field, which is represented by the arrowed field lines.

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Which of the following statements, about what happens when the charge is displaced, is correct?

When the negative charge is displaced

A to the left the magnitude of the electric force on it

decreases.

B to the right its potential energy increases.

C along the line PQ towards Q its potential energy

decreases.

D along the line PQ towards P the magnitude of the

electric force on it is unchanged.

(Total 1 mark)

Q7.Two parallel metal plates are separated by a distance d and have a potential difference V across them. Which expression gives the magnitude of the electrostatic force acting on a charge Q placed midway between the plates?

A

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B

C

D

(Total 1 mark)

Q8.The diagram shows the path of an α particle deflected by the nucleus of an atom. Point P on the path is the point of closest approach of the α particle to the nucleus.

Which of the following statements about the α particle on this path is correct?

A Its acceleration is zero at P.

B Its kinetic energy is greatest at P.

C Its potential energy is least at P.

D Its speed is least at P.

(Total 1 mark)

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Q9.The electric potential at a distance r from a positive point charge is 45 V. The potential increases to 50 V when the distance from the point charge decreases by 1.5 m. What is the value of r?

A 1.3m

B 1.5m

C 7.9m

D 15m

(Total 1 mark)

Q10.The diagram shows two particles at distance d apart. One particle has charge +Q and the other –2Q. The two particles exert an electrostatic force of attraction, F, on each other. Each particle is then given an additional charge +Q and their separation is increased to distance 2d.

Which of the following gives the force that now acts between the two particles?

A an attractive force of

B a repulsive force of

C an attractive force of

D a repulsive force of

(Total 1 mark)

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Q11.Which of the following statements about a parallel plate capacitor is incorrect?

A The capacitance of the capacitor is the amount of charge

stored by the capacitor when the pd across the plates is 1 V.

B A uniform electric field exists between the plates of the

capacitor.

C The charge stored on the capacitor is inversely proportional

to the pd across the plates.

D The energy stored when the capacitor is fully charged is

proportional to the square of the pd across the plates.

(Total 1 mark)


The force between two charged particles

A is always attractive

B can be measured in C2 F–1 m–1

C is directly proportional to the distance between them

D is independent of the magnitude of the charges(Total 1 mark)


When a negative ion is projected into an electric field

A the field can change the magnitude of the velocity but not its direction

B the field can change the direction of the velocity but not its magnitude

C the field can change both the magnitude and the direction of the velocity

D the ion will accelerate in the direction of the field

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(Total 1 mark)

Q14.Two fixed parallel metal plates X and Y are at constant potentials of + 100 V and + 70 V respectively. An electron travelling from X to Y experiences a change of potential energy ΔEp.

Which line, A to D, in the table shows correctly the direction of the electrostatic force F on the electron and the value of ΔEp?

Direction of F ΔEp

A towards X + 30 eV

B towards Y – 30 eV

C away from X + 30 eV

D away from Y – 30 eV

(Total 1 mark)

Q15.(a) Figure 1 shows a negative ion which has a charge of –3e and is free to move in a uniform electric field. When the ion is accelerated by the field through a distance of 63 mm parallel to the field lines its kinetic energy increases by 4.0 × 10sup class="xsmall">–16 J.

Figure 1

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(i) State and explain the direction of the electrostatic force on the ion.

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(ii) Calculate the magnitude of the electrostatic force acting on the ion.

magnitude of electrostatic force ........................................ N(2)

(iii) Calculate the electric field strength.

electric field strength .................................. NC–1

(1)

(b) Figure 2 shows a section of a horizontal copper wire carrying a current of 0.38 A.A horizontal uniform magnetic field of flux density B is applied at right angles to the wire in the direction shown in the figure.

Figure 2

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(i) State the direction of the magnetic force that acts on the moving electrons in the wire as a consequence of the current and explain how you arrive at your answer.

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(ii) Copper contains 8.4 × 1028 free electrons per cubic metre. The section of wire in Figure 2 is 95 mm long and its cross-sectional area is 5.1 × 10–6 m2.Show that there are about 4 × 1022 free electrons in this section of wire.

(1)

(iii) With a current of 0.38 A, the average velocity of an electron in the wire is5.5 × 10–6 m s–1 and the average magnetic force on one electron is 1.4 × 10–25 N.Calculate the flux density B of the magnetic field.

flux density ......................................... T(2)

(Total 10 marks)

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Q16.Two horizontal parallel plate conductors are separated by a distance of 5.0 mm in air. The lower plate is earthed and the potential of the upper plate is +50 V.

Which line, A to D, in the table gives correctly the electric field strength, E, and the potential, V, at a point midway between the plates?

electric field strength E / Vm−1 potential V / V

A 1.0 × 104 upwards 25

B 1.0 × 104 downwards 25

C 1.0 × 104 upwards 50

D 1.0 × 104 downwards 50

(Total 1 mark)

Q17.Which path, A to D, shows how an electron moves in the uniform electric field represented in the diagram?

(Total 1 mark)

Q18.The diagram shows a negative ion at a point in an electric field, which is represented by

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the arrowed field lines.

Which one of the following statements correctly describes what happens when the ion is displaced?

When the negative ion is displaced

A to the left the magnitude of the electric force on it decreases.

B to the right its potential energy increases.

C along the line PQ towards Q its potential energy decreases.

D along the line PQ towards P the magnitude of the electric force on it is unchanged.(Total 1 mark)

Q19.The diagram below shows an arrangement to demonstrate sparks passing across an air gap between two parallel metal discs. Sparks occur when the electric field in the gap becomes large enough to equal the breakdown field strength of the air. The discs form a capacitor, which is charged at a constant rate by an electrostatic generator until the potential difference (pd) across the discs is large enough for a spark to pass. Sparks are then produced at regular time intervals whilst the generator is switched on.

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(a) The electrostatic generator charges the discs at a constant rate of 3.2 × 10–8 A on a day when the minimum breakdown field strength of the air is 2.5 × 106 V m–1. The discs have a capacitance of 3.7 × 10–12 F.

(i) The air gap is 12 mm wide. Calculate the minimum pd required across the discs for a spark to occur. Assume that the electric field in the air gap is uniform.

pd ......................................... V(1)

(ii) Calculate the time taken, from when the electrostatic generator is first switched on, for the pd across the discs to reach the value calculated in part (a)(i).

time .......................................... s(2)

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(b) The discs are replaced by ones of larger area placed at the same separation, to give a larger capacitance.

State and explain what effect this increased capacitance will have on:

(i) the time between consecutive discharges,

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(ii) the brightness of each spark.

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(Total 7 marks)

Q20.(a) Describe how a beam of fast moving electrons is produced in the cathode ray tube of an oscilloscope.

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(b) The figure below shows the cathode ray tube of an oscilloscope. The details of how the beam of electrons is produced are not shown.

The electron beam passes between two horizontal metal plates and goes on to strike a fluorescent screen at the end of the tube. The plates are 0.040 m long and are separated by a gap of 0.015 m. A potential difference of 270 V is maintained between the plates.An individual electron takes 1.5 × 10–9 s to pass between the plates.The distance between the right-hand edge of the plates and the fluorescent screen is 0.20 m.

(i) Show that the vertical acceleration of an electron as it passes between the horizontal metal plates is approximately 3.2 × 1015 ms–2.

(3)

(ii) Show that the vertical distance travelled by an electron as it passes between the horizontal metal plates is approximately 3.6 mm.

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(2)

(iii) Show that the vertical component of velocity achieved by an electron in the beam by the time it reaches the end of the plates is approximately 4.7 × 106 m s–1.

(2)

(iv) Calculate the vertical displacement, y, of the electron beam from the centre of the screen. Give your answer in m.

vertical displacement ................................................ m(3)

(Total 13 marks)

Q21.Which one of the following statements about a parallel plate capacitor is incorrect?

A The capacitance of the capacitor is the amount of charge stored by the capacitor when the pd across the plates is 1V.

B A uniform electric field exists between the plates of the capacitor.

C The charge stored on the capacitor is inversely proportional to the pd across the plates.

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D The energy stored when the capacitor is fully charged is proportional to the square of the pd across the plates.

(Total 1 mark)

Q22.Which line, A to D, in the table correctly describes the trajectory of charged particles which enter separately, at right angles, a uniform electric field, and a uniform magnetic field?

uniform electric field uniform magnetic field

A parabolic circular

B circular parabolic

C circular circular

D parabolic parabolic

(Total 1 mark)

Q23. (a) Figure 1 shows an electron at a point in a uniform electric field at an instant when it is stationary.

Figure 1

(i) Draw an arrow on Figure 1 to show the direction of the electrostatic force that acts on the stationary electron.

(1)

(ii) State and explain what, if anything, will happen to the magnitude of the electrostatic force acting on the electron as it starts to move in this field.

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(b) Figure 2a shows a stationary electron in a non-uniform electric field. Figure 2b shows a stationary proton, placed in exactly the same position in the same electric field as the electron in Figure 2a.

Figure 2a Figure 2b

(i) State and explain how the electrostatic force on the proton in Figure 2b compares with that on the electron in Figure 2a.

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(ii) Each of the particles starts to move from the positions shown in Figure 2a and Figure 2b. State and explain how the magnitude of the initial acceleration of the proton compares with that of the electron.

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(iii) Describe and explain what will happen to the acceleration of each of these particles as they continue to move in the electric field.

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(c) The line spectrum of neon gas contains a prominent red line of wavelength 650 nm.

(i) Show that the energy required to excite neon atoms so that they emit light of this wavelength is about 2 eV.

(3)

(ii) An illuminated shop sign includes a neon discharge tube, as shown in Figure 3.A pd of 4500 V is applied across the electrodes, which are 180 mm apart.

Figure 3

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Assuming that the electric field inside the tube is uniform, calculate the minimum distance that a free electron would have to move from rest in order to excite the red spectral line in part (c).

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

(Total 15 marks)

Q24. The distance between two point charges of + 8.0 nC and + 2.0 nC is 60 mm.

At a point between the charges, on the line joining them, the resultant electric field strength is zero. How far is this point from the + 8.0 nC charge?

A 20 mm

B 25 mm

C 40 mm

D 45 mm(Total 1 mark)

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Q25. Which one of the following cannot be used as a unit for electric field strength?

A J m–1 C–1

B J A–1 s–1m–1

C N A–1 s–1

D J C m–1

(Total 1 mark)

Q26. (a) Define the electric potential at a point in an electric field.

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(b) Figure 1 shows part of the region around a small positive charge.

Figure 1

(b) (i) The electric potential at point L due to this charge is + 3.0 V. Calculate the magnitude Q of the charge. Express your answer to an appropriate number of significant figures.

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

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(ii) Show that the electric potential at point N, due to the charge, is +1.0 V.

(1)

(iii) Show that the electric field strength at point M, which is mid-way between L and N, is 2.5 Vm–1.

(1)

(c) R and S are two charged parallel plates, 0.60 m apart, as shown in Figure 2.They are at potentials of + 3.0 V and + 1.0 V respectively.

Figure 2

(i) On Figure 2, sketch the electric field between R and S, showing its direction.(2)

(ii) Point T is mid-way between R and S.Calculate the electric field strength at T.

answer = .......................... Vm–1

(1)

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(iii) Parts (b)(iii) and (c)(ii) both involve the electric field strength at a point mid-way between potentials of + 1.0 V and + 3.0 V. Explain why the magnitudes of these electric field strengths are different.

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(Total 12 marks)

Q27.

An ion carrying a charge of +4.8 × 10–19C travels horizontally at a speed of 8.0 × 105ms–1. It enters a uniform vertical electric field of strength 4200 V m–1, which is directed downwards and acts over a horizontal distance of 0.16m. Which one of the following statements is not correct?

A The ion passes through the field in 2.0 × 10–7s.

B The force on the ion acts vertically downwards at all points in the field.

C The magnitude of the force exerted on the ion by the field is 1.6 × 10–9 N.

D The horizontal component of the velocity of the ion is unaffected by the electric field.(Total 1 mark)

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Q28. (a) The equation F = BQv may be used to calculate magnetic forces.

(i) State the condition under which this equation applies.

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(ii) Identify the physical quantities that are represented by the four symbols in the equation.

F .........................................................................................................

B .........................................................................................................

Q..........................................................................................................

v...........................................................................................................(1)

(b) The figure below shows the path followed by a stream of identical positively charged ions, of the same kinetic energy, as they pass through the region between two charged plates. Initially the ions are travelling horizontally and they are then deflected downwards by the electric field between the plates.

While the electric field is still applied, the path of the ions may be restored to the horizontal, so that they have no overall deflection, by applying a magnetic field over the same region as the electric field. The magnetic field must be of suitable strength and has to be applied in a particular direction.

(i) State the direction in which the magnetic field should be applied.

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(ii) Explain why the ions have no overall deflection when a magnetic field of the required strength has been applied.

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(iii) A stream of ions passes between the plates at a velocity of 1.7 x 105ms–1. The separation d of the plates is 65 mm and the pd across them is 48 V. Calculate the value of B required so that there is no overall deflection of the ions, stating an appropriate unit.

answer = ....................................(4)

(c) Explain what would happen to ions with a velocity higher than 1.7 x 105ms–1 when they pass between the plates at a time when the conditions in part (b)(iii) have been established.

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(Total 11 marks)

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Q29. What is the acceleration of an electron at a point in an electric field where the field strength is 1.5 × 105 V m–1?

A 1.2 × 106 m s–2

B 1.4 × 1013 m s–2

C 2.7 × 1015 m s–2

D 2.6 × 1016 m s–2

(Total 1 mark)

Q30. Figure 1 shows a small polystyrene ball which is suspended between two vertical metal plates, P1 and P2, 80 mm apart, that are initially uncharged. The ball carries a charge of –0.17 μC.

Figure 1

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(a) (i) A pd of 600 V is applied between P1 and P2 when the switch is closed. Calculate the magnitude of the electric field strength between the plates, assuming it is uniform.

answer = ...........................V m–1

(2)

(ii) Show that the magnitude of the electrostatic force that acts on the ball under these conditions is 1.3 mN.

(1)

(b) Because of the electrostatic force acting on it, the ball is displaced from its original position. It comes to rest when the suspended thread makes an angle θ with the vertical, as shown in Figure 2.

Figure 2

(i) On Figure 2, mark and label the forces that act on the ball when in this position.

(2)

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(ii) The mass of the ball is 4.8 × 10–4 kg. By considering the equilibrium of the ball, determine the value of θ.

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

(Total 8 marks)

Q31. A small negatively charged sphere is suspended from a fine glass spring between parallel horizontal metal plates, as shown in the figure below.

(a) Initially the plates are uncharged. When switch S is set to position X, a high voltage dc supply is connected across the plates. This causes the sphere to move vertically upwards so that eventually it comes to rest 18 mm higher than its original position.

(i) State the direction of the electric field between the plates.

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(ii) The spring constant of the glass spring is 0.24 N m–1. Show that the force exerted on the sphere by the electric field is 4.3 × 10–3 N.

(1)

(iii) The pd applied across the plates is 5.0 kV. If the charge on the sphere is–4.1 × 10–8 C, determine the separation of the plates.

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

(b) Switch S is now moved to position Y.

(i) State and explain the effect of this on the electric field between the plates.

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(ii) With reference to the forces acting on the sphere, explain why it starts to move with simple harmonic motion.

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(Total 10 marks)

Q32. Which one of the following statements about a charged particle in an electric field is correct?

A No work is done when a charged particle moves along a field line.

B No force acts on a charged particle when it moves along a field line.

C No work is done when a charged particle moves along a line of constant potential.

D No force acts on a charged particle when it moves along a line of constant potential.(Total 1 mark)

Q33. Two parallel metal plates separated by a distance d have a potential difference V across them.What is the magnitude of the electrostatic force acting on a charge Q placed midway between the plates?

A

B

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C

D (Total 1 mark)

Q34. Which one of the following statements about electric field strength and electric potential is incorrect?

A Electric potential is a scalar quantity.

B Electric field strength is a vector quantity.

C Electric potential is zero whenever the electric field strength is zero.

D The potential gradient is proportional to the electric field strength.(Total 1 mark)

Q35. (a) (i) Define the electric field strength, E, at a point in an electric field.

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(ii) State whether E is a scalar or a vector quantity.

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(b) Point charges of +4.0 nC and –8.0 nC are placed 80 mm apart, as shown in the figure below.

(i) Calculate the magnitude of the force exerted on the +4.0 nC charge by the –8.0 nC charge.

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(ii) Determine the distance from the +4.0 nC charge to the point, along the straight line between the charges, where the electric potential is zero.

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(c) Point P in the figure above is equidistant from the two charges.

(i) Draw two arrows on the figure above at P to represent the directions and relative magnitudes of the components of the electric field at P due to each of the charges.

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(ii) Hence draw an arrow, labelled R, on the figure above at P to represent the direction of the resultant electric field at P.

(3)(Total 10 marks)

Q36. Two parallel metal plates separated by a distance d have a potential difference V across them. What is the magnitude of the electrostatic force acting on a charge Q placed midway between the plates?

A

B

C

D (Total 1 mark)

Q37. (a) An electron travels at a speed of 3.2 × 107 ms–1 in a horizontal path through a vacuum. The electron enters the uniform electric field between two parallel plates, 30 mm long and 15 mm apart, as shown in the figure below. A potential difference of 1400 V is maintained across the plates, with the top plate having positive polarity. Assume that there is no electric field outside the shaded area.

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(i) Show that the electric field strength between the plates is 9.3 × 104 Vm–1.

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(ii) Calculate the time taken by the electron to pass through the electric field.

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(iii) Show that the acceleration of the electron whilst in the field is 1.6 × 1016 m s–2 and state the direction of this acceleration.

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(b) Determine the magnitude and direction of the velocity of the electron at the point where it leaves the field.

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(Total 8 marks)

Q38. An electron travelling at constant speed enters a uniform electric field at right angles to the field. While the electron is in the field it accelerates in a direction which is

A in the same direction as the electric field.

B in the opposite direction to the electric field.

C in the same direction as the motion of the electron.

D in the opposite direction to the motion of the electron.(Total 1 mark)

Q39. The electrical field strength, E, and the electrical potential, V, at the surface of a sphere of radius r carrying a charge Q are given by the equations

A school van de Graaff generator has a dome of radius 100 mm. Charge begins to leak into the air from the dome when the electric field strength at its surface is approximately 3 × 106 V m–1.What, approximately, is the maximum potential to which the dome can be raised without leakage?

A 3 × 104 V

B 3 × 105 V

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C 3 × 106 V

D 3 × 107 V(Total 1 mark)

Q40. Which line, A to D, correctly describes the trajectory of charged particles which enter, at right angles, (a) a uniform electric field, and (b) a uniform magnetic field?

(a) uniform electric field (b) uniform magnetic field

ABCD

circularcircularparabolicparabolic

circularparaboliccircularparabolic

(Total 1 mark)

Q41. (a) Complete the table of quantities related to fields. In the second column, write an SI unit for each quantity. In the third column indicate whether the quantity is a scalar or a vector.

quantity SI unit scalar or vector

gravitational potential

electric field strength

magnetic flux density

(3)

(b) (i) A charged particle is held in equilibrium by the force resulting from a vertical electric field. The mass of the particle is 4.3 × 10–9 kg and it carries a charge of

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magnitude 3.2 × 10–12 C. Calculate the strength of the electric field.

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(ii) If the electric field acts upwards, state the sign of the charge carried by the particle

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(Total 6 marks)

Q42.

Two parallel metal plates of separation a carry equal and opposite charges. Which one of the following graphs, A to D, best represents how the electric field strength E varies with the distance x in the space between the plates?

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(Total 1 mark)

Q43. A small charged sphere of mass 2.1 × 10–4 kg, suspended from a thread of insulating material, was placed between two vertical parallel plates 60 mm apart. When a potential difference of 4200 V was applied to the plates, the sphere moved until the thread made an angle of 6.0° to the vertical, as shown in the diagram below.

(a) Show that the electrostatic force F on the sphere is given by

F = mg tan 6.0°

where m is the mass of the sphere.

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(b) Calculate

(i) the electric field strength between the plates,

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(ii) the charge on the sphere.

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(Total 6 marks)

Q44. Two horizontal parallel plate conductors are separated by a distance of 5.0 mm in air. The lower plate is earthed and the potential of the upper plate is + 50 V.

Which line, A to D, gives correctly the electric field strength, E, and the potential, V, at a point midway between the plates?

electric field strength E/V m–1 potential V/V

A 1 × 104 upwards 25

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B 1 × 104 downwards 25

C 1 × 104 upwards 50

D 1 × 104 downwards 50

(Total 1 mark)

Q45. The mass of the nucleus of an isolated copper atom is 63 u and it carries a charge of +29 e. The diameter of the atom is 2.3 × 10–10 m.

P is a point at the outer edge of the atom.

(a) Calculate

(i) the electric field strength at P due to the nucleus,

.............................................................................................................

.............................................................................................................

.............................................................................................................

(ii) the gravitational potential at P due to the nucleus.

.............................................................................................................

.............................................................................................................

.............................................................................................................(5)

Page 41


(b) Draw an arrow on the above diagram to show the direction of the electric field at the point P.

(1)(Total 6 marks)

Q46.

The diagram shows how the electric potential varies along a line XX’ in an electric field. What will be the electric field strength at a point P on XX’ which is mid−way between R and S?

A 5.0 V m−1

B 10 V m−1

C 20 V m−1

D 30 V m−1

(Total 1 mark)

Q47.If the potential difference between a pair of identical, parallel, conducting plates is known, what is the only additional knowledge required to determine the electric field strength between the plates?

A the permittivity of the medium between the plates

Page 42


B the separation and area of the plates

C the separation and area of the plates and the permittivity of the medium between the plates

D the separation of the plates(Total 1 mark)

Q48.Which one of the following statements about electric field strength and electric potential is incorrect?

A Electric potential is a scalar quantity.

B Electric field strength is a vector quantity.

C Electric potential is zero whenever the electric field strength is zero.

D The potential gradient is proportional to the electric field strength.(Total 1 mark)

Q49.(a) An electron moves parallel to, but in the opposite direction to, a uniform electric field, as shown in Figure 1.

Figure 1

(i) State the direction of the force that acts on the electron due to the electric field.

...............................................................................................................

Page 43


(ii) What is the effect of this force on the motion of the electron?

...............................................................................................................

...............................................................................................................(2)

(b) An electron, which is travelling in a horizontal path at constant speed, enters a uniform vertical electric field as shown in Figure 2.

Figure 2

(i) Sketch on Figure 2 the path followed by the electron.

(ii) Explain the motion of the electron whilst in this field.

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................(3)

(Total 5 marks)

Q50.Both gravitational and electric field strengths can be described by similar equations written in the form

Page 44


(a) Complete the following table by writing down the names of the corresponding quantities, together with their SI units, for the two types of field.

symbol gravitational fieldquantity SI unit

electrical fieldquantity SI unit

a gravitationalfield strength

b m F–1

c

d

(4)

(b) Two isolated charged objects, A and B, are arranged so that the gravitational force between them is equal and opposite to the electric force between them.

(i) The separation of A and B is doubled without changing their charges or masses. State and explain the effect, if any, that this will have on the resultant force between them.

...............................................................................................................

...............................................................................................................

(ii) At the original separation, the mass of A is doubled, whilst the charge on A and the mass of B remain as they were initially. What would have to happen to the charge on B to keep the resultant force zero?

...............................................................................................................

Page 45


...............................................................................................................(3)

(Total 7 marks)

Q51.(a) (i) Define electric field strength, and state whether it is a scalar quantity or a vector quantity.

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

(ii) Complete the diagram below to show the electric field lines in the region around two equal positive point charges. Mark with a letter N the position of any point where the field strength is zero.

(6)

(b) Point charges A, of +2.0 nC, and B, of –3.0 nC, are 200 mm apart in a vacuum, as shown by the figure. The point P is 120 mm from A and 160 mm from B.

Page 46


(i) Calculate the component of the electric field at P in the direction AP.

...............................................................................................................

...............................................................................................................

...............................................................................................................

(ii) Calculate the component of the electric field at P in the direction PB.

...............................................................................................................

...............................................................................................................

...............................................................................................................

(iii) Hence calculate the magnitude and direction of the resultant field at P.

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................(6)

(c) (i) Explain why there is a point X on the line AB in part (b) at which the electric potential is zero.

...............................................................................................................

...............................................................................................................

Page 47


...............................................................................................................

...............................................................................................................

(ii) Calculate the distance of the point X from A.

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................(4)

(Total 16 marks)

Q52.Which one of the following statements about electric potential and electric field strength is correct?

A Electric potential is zero whenever the electric field strength is zero.

B Electric field strength is a scalar quantity.

C Electric potential is a vector quantity.

D Electric potential due to a point charge varies as where r is the distance from the point charge.

(Total 1 mark)

Q53.

Page 48


A positive ion with a mass of 3.4 × 10–26 kg and a charge of 3.2 × 10–19 C is initially at rest at a point P, midway between two parallel conducting plates, A and B which are separated by 40 mm. The ion is accelerated and passes through a hole Q in plate B. It enters a magnetic field of uniform flux density 0.10 T at R. After following a circular path the ion leaves the field at S. Assume that the magnetic field is uniform everywhere within the dotted rectangle and that the space within the solid rectangle is evacuated.

(a) (i) Calculate the electric field strength between the plates AB.

...............................................................................................................

...............................................................................................................

...............................................................................................................

(ii) Calculate the force on the ion due to the electric field.

...............................................................................................................

...............................................................................................................

...............................................................................................................

(iii) Show that the speed of the ion just after it has passed through the hole at Q is 3.1 × 105 m s–1.

...............................................................................................................

...............................................................................................................

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...............................................................................................................

...............................................................................................................(5)

(b) (i) State the direction of the magnetic field.

...............................................................................................................

(ii) Explain why the ion follows a circular path in the magnetic field.

...............................................................................................................

...............................................................................................................

...............................................................................................................

(iii) Show that the radius of the circular path is proportional to the momentum of the ion and calculate the value of the radius.

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................

...............................................................................................................(6)

(c) Explain how, if at all, the trajectory would be different for an ion with a slightly greater mass but carrying the same charge.

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................(2)

(Total 13 marks)

Page 50


Q54. (a) Figure 1 shows the electron gun that accelerates electrons in an electron microscope.

Figure 1

(i) Draw, on Figure 1, electric field lines and lines of equipotential in the region between the anode and cathode. Assume that there are no edge effects and that the holes in the plates do not affect the field. Clearly label your diagram.

(3)

(ii) Calculate the kinetic energy, speed and momentum of an electron as it passes through the hole in the anode.

mass of an electron = 9.1 × 10–31 kgcharge of an electron = –1.6 × 10–19 C

(4)

(b) By calculating the de Broglie wavelength of electrons coming through the anode of this device, state and explain whether or not they will be suitable for the investigation of the crystal structure of a metal.

Page 51


Planck constant = 6.6 × 10–34 J s

........................................................................................................................

........................................................................................................................

........................................................................................................................

........................................................................................................................(4)

(Total 11 marks)

Q55. The Earth’s surface and the base of a charged cloud can be considered to be two plates of a parallel-plate capacitor.

(a) Calculate the capacitance of an Earth-cloud system when the base of the cloud has an area of 1.4 × 106 m2 and is 800 m above the Earth’s surface.

= 8.9 × 10–12 F m–1

for air = 1.0(2)

(b) A potential difference of 3.0 × 106 V across each metre of air will cause the air to break down and allow the cloud to discharge to the Earth.

(i) Show that the average breakdown p.d. for the 800 m layer of air between the Earth and the base of the cloud is about 2.5 × 109 V.

(1)

Page 52


(ii) Calculate the maximum energy that the charged Earth – cloud system can store.

(2)

(iii) Calculate the maximum charge stored by the system before breakdown commences.

(1)

(c) By considering the cloud discharge to be modelled by a resistor connected across a capacitor, calculate the resistance that would allow a cloud to discharge 99% of its charge to Earth in a time of 0.25 s.

(3)

(Total 9 marks)

Q56.(a) The diagram below shows part of a precipitation system used to collect dust particles in a chimney. It consists of two large parallel vertical plates maintained at potentials of +25 kV and –25 kV.

The diagram below also shows the electric field lines between the plates.

Page 53


(i) Add arrows to the diagram to show the direction of the electric field.(1)

(ii) Explain what is meant by an equipotential surface.

...............................................................................................................

...............................................................................................................

...............................................................................................................(1)

(iii) Draw and label on the diagram equipotentials that correspond to potentials of –12.5 kV, 0 V, and +12.5 kV.

(2)

(b) A small dust particle moves vertically up the centre of the chimney, midway between the plates.

Page 54


(i) The charge on the dust particle is +5.5 nC. Show that there is an electrostatic force on the particle of about 0.07 mN.

(2)

(ii) The mass of the dust particle is 1.2 × 10–4 kg and it moves up the centre of the chimney at a constant vertical speed of 0.80 m s–1.

Calculate the minimum length of the plates necessary for this particle to strike one of them. Ignore air resistance.

(4)(Total 10 marks)

Q57. The diagram below shows a diagram of a mass spectrometer.

(a) The magnetic field strength in the velocity selector is 0.14 T and the electric field strength is 20 000 V m–1.

(i) Define the unit for magnetic flux density, the tesla.

...............................................................................................................

...............................................................................................................(2)

(ii) Show that the velocity selected is independent of the charge on an ion.

Page 55


(2)

(iii) Show that the velocity selected is about 140 km s–1.

(1)

(b) A sample of nickel is analysed in the spectrometer. The two most abundant isotopes

of nickel are Ni and Ni. Each ion carries a single charge of +1.6 × 10–19 C.

mass of a proton or neutron = 1.7 × 10–27 kg

The Ni ion strikes the photographic plate 0.28 m from the point P at which the ion beam enters the ion separator.

Calculate:

(i) the magnetic flux density of the field in the ion separator;

(3)

(ii) the separation of the positions where the two isotopes hit the photographic plate.

(2)

(Total 10 marks)

Page 56


Q58.The Earth has an electric charge. The electric field strength outside the Earth varies in the same way as if this charge were concentrated at the centre of the Earth. The axes in the diagram below represent the electric field strength E and the distance from the centre of the Earth r. The electric field strength at A has been plotted.

(a) (i) Determine the electric field strength at B and then complete the graph to show how the electric field strength varies with distance from the centre of the Earth for distances greater than 6400 km.

(3)

(ii) State how you would use the graph to find the electric potential difference between the points A and B.

...............................................................................................................

...............................................................................................................

Page 57


(1)

(b) The permittivity of free space ε0 is 8.9 × 10–12 F m–1 .

(i) Calculate the total charge on the Earth.(2)

(ii) The charge is distributed uniformly over the Earth's surface. Calculate the charge per square metre on the Earth's surface.

(2)(Total 8 marks)

Q59.Which one of the following arrangements of charge will produce zero electric field strength and zero electric potential at the point labelled P?

(Total 1 mark)

Page 58


Q60.In the cathode ray tube illustrated below, electrons are accelerated by a potential difference of 1.8 kV between the cathode (C) and the anode (A).

(a) (i) Calculate the kinetic energy, in J, of the electrons after they have passed the anode.

charge on an electron, e = –1.6 × 10–19 C(2)

(ii) Calculate the velocity of the electrons after they have passed the anode.

Mass of an electron = 9.1 × 10–31 kg(2)

(b) The plates P and Q are 8.0 cm long and are separated by a gap of 4.0 cm.

(i) Define electric field strength.

...............................................................................................................

...............................................................................................................(1)

(ii) Calculate the force acting on an electron when it is between P and Q and state the direction of the force.

Page 59


Direction ..............................................................................................................

(3)

(iii) Calculate the time taken for an electron to pass between the plates.(1)

(iv) Calculate the vertical component of velocity at the time the electron leaves the electric field between P and Q.

(2)

(v) Calculate the additional vertical displacement of the electron between the time it leaves the electric field between P and Q and when it reaches the screen.

(1)(Total 12 marks)

Page 60


M1.C[1]

M2.A[1]

M3.C[1]

M4.A[1]

M5.(a) t = or 4.5 = × 9.81 × t 2 ✓

t = 0.96 s✓2

(b) Field strength = 186000V m–1✓

Acceleration = Eq / m

or 186 000 × 1.2 × 10–6 ✓

0.22 m s–2 ✓3

(c) 0.10(3)m (allow ecf from (i))✓1

Page 61


(d) Force on a particle = mg and

acceleration = F / m so always = g✓

Time to fall (given distance) depends (only) on the distance and acceleration✓

OR:

g = GM / r2 ✓

Time to fall = √2s / g

so no m in equations to determine time to fall✓2

(e) Mass is not constant since particle mass will vary✓

Charge on a particle is not constant✓

Acceleration = Eq / m or (V / d) (q / m) or Vq / dm✓

E or V / d constant but charge and mass are ‘random’ variables so q / m will vary (or unlikely to be the same)✓

4[12]

M6.D[1]

M7.B[1]

M8.D[1]

Page 62


M9.D[1]

M10.A[1]

M11.C[1]

M12.B[1]

M13.C[1]

M14.A[1]

M15.(a) (i) force acts towards left or in opposite direction to field lines ✓ because ion (or electron) has negative charge (∴ experiences force in opposite direction to field) ✓

Mark sequentially. Essential to refer to negative charge (or force on + charge is to right) for 2nd mark.

2

Page 63


(ii) (use of W = F s gives) force F = ✓

= 6.3(5) × 10–15 (N) ✓If mass of ion m is used correctly using algebra with F = ma, allow both marks (since m will cancel). If numerical value for m is used, max 1.

2

(iii) electric field strength = 1.3(2) ✓ 104 (N C-1) ✓

[or (833 V)

= 1.3(2) ✓ 104 (V m-1) ✓ ]

Allow ECF from wrong F value in (ii).

1

(b) (i) (vertically) downwards on diagram ✓ reference to Fleming’s LH rule or equivalent statement ✓

Mark sequentially. 1st point: allow “into the page”.

2

(ii) number of free electrons in wire = A × l × number density = 5.1 × 10–6 × 95 × 10–3 × 8.4 × 1028 = 4.1 (4.07) × 1022 ✓

Provided it is shown correctly to at least 2SF, final answer alone is sufficient for the mark. (Otherwise working is mandatory).

1

(iii) ✓ = 0.16 (0.159) (T) ✓

[or ✓ = 0.16 (0.158) (T) ✓ ]

Page 64


In 2nd method allow ECF from wrong number value in (ii).2

[10]

M16.B[1]

M17.A[1]

M18.D[1]

M19.(a) (i) required pd ( = 2.5 × 10 6 × 12 × 10−3 ) = 3.0(0) × 104 (V) 1

(ii) charge required Q (= CV) = 3.7 × 10−12 × 3.00 × 104

( = 1.11 × 10−7 C)Allow ECF from incorrect V from (a)(i).

2

(b) (i) time increases

(larger C means) more charge required (to reach breakdown pd)Mark sequentially i.e. no explanation mark if effect is wrong.

or or time ∝ capacitance

Page 65

Colonel Frank Seely School2

(ii) spark is brighter (or lasts for a longer time)

more energy (or charge) is stored or current is largerMark sequentially.

or spark has more energy 2

(Total 7 marks)

M20.(a) thermionic emission / by heating

B1

cathode heated / heating done by electric current / overcoming work function

B1Must mention anode for third mark

anode which is positive wrt cathode / accelerated by electric field between anode and cathode

B13

(b) (i) one relevant equation seen: E = V/d / F = Ee / a = F/m

B1Equation should be in symbols

B1Substitution may be done in several stages

3.16 × 1015 (m s−2)

B1Must be more than 2 sf

3

Page 66


(ii) s = (ut)+ at2 or v = u + at and s = vavt OR s = vt used

B1Appropriate symbol equation seen and used for 1st mark

3.56 × 10−3m

B1Expect at least 3 sf but condone 3.6 for candidates who use a = 3.2 × 1015

2

(iii) v = u + at / v = at v2 = u2 + 2as used

B1May also use eV= ½mv2

4.74 × 106 m s−1 to at least 3 sf

B1Allow 4.8 (2 or more sf) – consistent with use of a = 3.2 × 1015

2

(iv) t = 7.5 × 10−9 s seen or used

C1

May use ratios for 1st 2 marks: C1

3.53 × 10−2 (m) A1

3.53 × 10−2 (m) ecf for wrong t

A1

adds 3.56 × 10−3 (m) to their 3.53 × 10−2

B1

clipped with b(i) and b(ii)Allow reasonable rounding

3[13]

M21.C[1]

Page 67


M22.A[1]

M23. (a) (i) horizontal arrow to the left 1

(ii) the electrostatic force is unchanged 2

because electric field strength is constant

(b) (i) forces are equal in magnitude but opposite in direction

(E is the same for both and) Q has same magnitude but opposite sign 2

(ii) acceleration of proton is (much) smaller (than acceleration of electron)

because mass of proton is (much) greater (than mass of electron) 2

(iii) acceleration of proton increases and acceleration of electron decreases

correct reference to changing strength of electric field (for either or both) 2

(c) (i) energy of photon

= 3.06 × 10–19 (J)

Page 68


energy required = = 1.91 (eV) 3

(ii) electric field strength = = 2.50 × 104 (V m–1)

distance =

= 7.64 10–5(m) 3

[15]

M24. C[1]

M25. D[1]

M26. (a) work done [or energy needed] per unit charge[or (change in) electric pe per unit charge]

on [or of] a (small) positive (test) charge

in moving the charge from infinity (to the point)

[not from the point to infinity] 3

Page 69


(b) (i) gives Q (= 4πε0rV) = 4π × 8.85 × 10–12 × 0.30 × 3.0

= 1.0 × 10–10 (C)

to 2 sf only 3

(ii) use of V ∞ gives VM = (= (+) 1.0 V)1

(iii) = (= 2.50 V m–1)1

(c) (i) uniformly spaced vertical parallel lines which start and end on plates

relevant lines with arrow(s) pointing only downwards 2

(ii) = 3.3(3) (V m–1) 1

(iii) part (b) is a radial field whilst part (c) is a uniform field

[or field lines become further apart between L and M but are equally spaced between R and S]

1[12]

M27. C[1]

Page 70


M28. (a) (i) magnetic field (or B) must be at right angles to velocity (or v) 1

(ii) F = (magnetic) force (on a charged particle or ion)

B = flux density (of a magnetic field)

Q = charge (of particle or ion)

v = velocity [or speed] (of particle or ion)all four correct

1

(b) (i) into plane of diagram 1

(ii) magnetic force = electric force [or BQv = EQ]

these forces act in opposite directions [or are balancedor resultant vertical force is zero]

2

(iii) BQv = EQ gives flux density B =

(= 738 V m–1)

= 4.3 × 10–3 T 4

(c) ions would be deflected upwards

magnetic force increases but electrostatic force isunchanged [or magnetic force now exceeds electrostatic force]

2[11]

Page 71


M29. D[1]

M30. (a) (i) (1)

= 7.5 × 103 (V m–1) (1)2

(ii) force F (= EQ) = 7500 × 0.17 × 10–6 (1) (= 1.28 × 10–3 N)1

(b) (i) correct labelled arrows placed on diagram to show the threeforces acting;

• electric force F (or 1.3 mN) horizontally to left (1)

• W (or mg) vertically down and

• tension T upwards along the thread (1)2

(ii) F = T sinθ and mg = T cosθ give F = mg tanθ (1)(or by triangle or parallelogram methods)

tanθ = (1)

gives θ = 15(.2) (°) (1)3

[8]

Page 72


M31. (a) (i) (vertically) downwards [or top to bottom, or down the page] (1)1

(ii) force on sphere F (= kx) = 0.24 × 18 × 10−3 (1) (= 4.32 × 10−3 N)1

(iii) use of F = EQ gives E = (1) (= 1.05 × 105 V m−1)

use of E = gives separation d = (1)

= 4.8 × 10−2 (m) (1) (4.76 × 10−2)3

(b) (i) electric field becomes zero (or ceases to exist) (1)

flow of charge (or electrons) from one plate to the other[or plates discharge] (1)

(until) pd across plates becomes zero [or no pd across plates,or plates at same potential] (1)

max 2

(ii) net downward force on sphere (when E becomes zero)[or gravitational force acts on sphere, or force is weight] (1)

this force extends spring (1)

force (or acceleration) is proportional to (change in)extension of spring (1)acceleration is in opposite direction to displacement(or towards equilibrium) (1)

for shm, acceleration (−) displacement[or for shm, force (−) displacement] (1)

max 3[10]

Page 73


M32. C[1]

M33. B[1]

M34. C[1]

M35. (a) (i) force per unit charge (1)acting on a positive charge (1)

(ii) vector (1)3

(b) (i) (1)

=4.5(0) × 10–5N (1)

Page 74


(ii) (use of gives)

or (1)

x = 26.7mm (1)4

(c) correct directions for E4 and E8 (1)E8 approx twice as long as E4 (1)correct direction of resultant Rshown (1)

3[10]

M36. D[1]

M37. (a) (i) (1)

(ii) s (1)

Page 75


(iii) may = Ee (1)

ay = (1) (= 1.64 × 1016 m s−2)

acceleration is upwards [or towards + plate](1)5

(b) vy (= ayt) = 1.64 × 1016 × 9.38 × 10−10 (1) (= 1.54 × 107 m s−1)

= 3.55 × 107m s–1(1)

at tan−1 above the horizontal (1)3

[8]

M38. B[1]

M39. B[1]

Page 76


M40. C[1]

M41. (a)

quantity SI unit

(gravitational potential) J kg–1 or N m kg –1 scalar

(electric field strength) N C–1 or V m–1 vector

(magnetic flux density T or Wb m–2 or N A–1 m–1 vector

6 entries correct (1) (1) (1)4 or 5 entries correct (1) (1)2 or 3 entries correct (1)

3

(b) (i) mg = EQ (1)

= 1.32 × 104 (V m–1) (1)

(ii) positive (1)3

[6]

M42. A[1]

Page 77


M43. (a) mg = T cos 6 (1)F = T sin 6 (1)hence F = mg tan 6 (1)[or correct use of triangle: (1) for sides correct, (1) for 6°, (1) for tan 6 = F/mg

or FΔx = mg Δh , tan θ = tan 6° = 3

(b) (i) (use of E = gives) E = = 7.0 × 104 V m–1 (1)

(ii) (use of Q = gives) Q =

= 3.1 × 10–9 C

(allow C.E. for value of E from (i))3

[6]

M44. B[1]

M45. (a) (i) E (= ) = (1)

= 3.15 × 1012Vm-1 (or (NC-1) (1)

Page 78


(ii) V(= – ) = (–) (1)

= (–) 6.07 × 10-26 (1) – sign and J kg-1

5

(b) arrow pointing to the right (1)1

[6]

M46.C[1]

M47.D[1]

M48.C[1]

M49.(a) (i) (force) to the right (1)

(ii) electrons accelerate or speed increases (1)2

(b) (i) sketch to show path curving upwards in the field (must not become vertical) (1)

Page 79


(ii) horizontal component of velocity is unchanged (1) vertical or upwards acceleration (or force) (1) parabolic path described (or named) (1)

max 3

The Quality of Written Communication marks are awarded for the quality of answers to this question.

[5]

M50.(a)

________ N kg–1 electric

field strength

N C–1

or V m–1

(1)

gravitational constant

N m2 kg–

2_______

________

_(1)

mass kg charge C (1)

distance (from mass to point) m

distance (from charge to point)

m

(1)

(4)

(b) (i) none (1)

both FE and FG ∝ (hence both reduced to [affected equally] (1)

(ii) charge on B must be doubled (1)(3)

[7]

M51.(a) (i) force per unit positive charge (1)(1)[force on a unit charge (1) only]

Page 80


vector (1)

(ii)

overall correct symmetrical shape (1)outward directions of lines (1)spacing of lines on appropriate diagram (1)neutral point, N, shown midway between charges (1)

6

(b) (i) (1)

= 1250 V m–1 (1)

(ii) EPB = = 1050Vm–1 (1)

(iii)

allow e.c.f. from wrong numbers in (i) and (ii)

E = (1) 1630Vm–1 (1)

Page 81


θ = tan-1 = 50.0° to line PB and in correct direction (1)max 6

(c) (i) potential due to A is positive, potential due to B is negative (1)at X sum of potentials is zero (1)

(ii) = 0 (1)

gives AX (= x) = 0.080m (1) (only from satisfactory use of potentials)4

[16]

M52.D[1]

M53.(a) (i)

(ii) F = Eq = 2.5 × 105 × 3.2 × 10–19 (1) = 8.0 × 10–14 N (1)

(iii)

(= 3.1 × 105 m s–1)(max 5)

(b) (i) field into paper (1)

Page 82


(ii) force ⊥r motion in B-field (1) ∴ directed towards centre of a circular path (1)

(iii)

for B, q constant r ∝ mυ, momentum (1)

r = = 0.33 m (1)(6)

(c) greater mass, so smaller speed at Q (1)

greater momentum justified e.g. E = (E const) (1)

∴ greater radius (1)(max 2)

[13]

M54. (a) (i) Lines of equipotential parallel to the plates

B1

Field lines perpendicular to plates, evenly spaced and with arrows upwards

B1

Lack of clear labelling of at least one of the typesof line loses 1 markEither field shown to be uniform

B13

(ii) KE = 8.8 × 10–17 J

Page 83


B1

Use of ½ mv2

C1

Speed = 1.4 × 107 m s–1 ecf

A1

Momentum =1.27 × 10–23 kg m s–1 ecf

B14

(b) Use of de Broglie wavelength = h/mv

C1

5.2 × 10–11 m ecf

A1

diffraction of electrons necessary

M1

will work because wavelength is of same order as atomicseparation (not just wavelength is too small)/argumentconsistent with their (a) (ii).

A14

[11]

M55. (a) C = A/d

C1

15.6 nF or 16 nF

A12

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(b) (i) 2.4 × 109 (V)

B11

(ii) ½ CV2 (or ½QV if attempt to calculate Q made)

C1

4.3–5.0 × 1010 J

A12

(iii) 36–40 C

B11

(c) recognition that 1% of charge or voltage remains

C1

any appropriate form of decay equation (either exponentialor logarithmic)

C1

3.48 × 106 Ω cao (but do not allow if physics error)

A13

[9]

M56.(a) (i) shows arrows from + to –Bl

(ii) surface of constant potential / no work done in moving chargeon surface OWTTE

Bl

(iii) 3 correct lines between plates, straight, labelled, +12.5 kV on left

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Colonel Frank Seely SchoolBl

outwards curvature at edge of platesBl

(b) (i) F = Vq / d or 50000 × 5.5 × 10–9 / 4Bl

= 0.0690 [mN] [0.0688]Bl

(ii) a = F / m = 0.069 × 10–3 / 0.12 × 10–3

= 0.575 / 0.573 m s–2

Cl

use of appropriate kinematic equationCl

t = √2 × 2 / 0.575 = (2.63) sCl

so length must be 0.8 × 2.63 = 2.11 m [gets mark ecf fromthird mark if number quoted]allow alternative energy approach

Bl[10]

M57. (a) (i) 1 N per A per mor 1 Wb m–2

or quotes: B = F/IL with terms definedor induced EMF = ΔBAN/t with terms definedor a slightly flawed attempt at the definition instatement form

C1

It is the flux density (perpendicular to a wire) thatproduces a force of 1N per m on the wire whenthe current is 1AorB = F/IL and 1 T is flux density when F = 1N; I = 1Aand L = 1 mor induced EMF = ΔBAN /t and 1 T is the flux changewhen emf = 1V for A=1 N =1 and t =1 or similar

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A12

(ii) force on charge due to E field , FE= Eq or Vq/dandforce due to B field, FB = Bqvor Eq=Bqv

B1

= Bqv; cancels q and states explicitly v =

or v =

B12

(iii) v = 20000/0.14 (seen) or 143 × 103 m s–1

B11

(b) (i) Bqv =mv2/r or r = mv/Bq ( allow e instead of q)mass of ion = 1.7 × 10–27 × 58 (may be in equation)or (9.86 × 10–26 kg seen)

C1

orradius = 0.14 m (may be in equation)

C1

Substitutes and arrives at 0.62 to 0.63 T

A13

(ii) Calculates new radius (0.145 m) or diameter (0.288 m)using r m or otherwise allowing ecf

C1

0.010 m (condone 0.01 m) or 0.0096 – 0.0097 m(Allow 0.0079 m or 0.008 m due to use of differentsfs for B and v )

A12

[10]

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M58.(a) (i) E at 2R = 20 to 21 (NC–1) i.e. no upB1

(i.e. have used inverse square law possibly misreading the E axis)

correct curvature with line through given point must not increase near tail (ignore below 6400 km)

B1

no intercept on distance axis and through correctly calculated point

B1

(ii) determine the area under the graphB1

between A and B or between the points ignore any reference to V = Ed)

B1

(b) (i) E = q/4πε0r2 (Q = 84 × 4π 8.9 × 10–12 (6400 000)2

C1

(3.8–3.9) × 105 CA1

(ii) surface area of the Earth = 5.15 × 1014 (m2)C1

or: charge per square metre = total charge/ surface area of Earth (may be seen as a numerical substitution with wrong area)

738 – 760 pC (m–2) ecf for Q from (b)(i)A1

NB

(i) answer is the same when unit is left in km since r2 cancels so condone

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(ii) Use of E = q / 4πε0r followed by area = 4πr gives correct value but no marks

[8]

M59.C[1]

M60.(a) (i) E = eVC1

2.9 (2.88) × 10–16 JA1

(2)

(ii) KE = 0.5 mv2

C1

v = 2.5(2) × 107 m s–1

A1

allow e.c.f. for (i) ie 1.5 × 1015 × √(their (i)(2)

(b) (i) force acting per unit charge or F / q with symbols definedB1

(1)

(ii) F = eE or F = eV / d or E = V / dC1

2.4 × 10–15 NA1

downwards / towards QB1

(3)

(iii) 3.2 (3.17) × 10–9 s

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e.c.f. for their (ii)A1

(1)

(iv) a = F / m or v = Ft / mC1

8.4 (8.36) × 106 m s–1

ecf their (ii) × their (iii)A1

(2)

(v) 4.0 cm (3.98 cm)

do not allow e.c.f.B1

(1)[12]

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E1.This question tested students’ understanding of the effect of the descent towards a planet on the speed and orbit time. 74% of them knew that the speed would increase and the orbit time would decrease, because v ∝ r−1/2 whilst T ∝ r3/2.

E2.This question concerned the strength and direction of an electric field in which a positive ion of given charge-to-mass ratio was held at rest. Application of mg = EQ was the key to finding the field strength. This question was re-used from an earlier year and its facility improved only slightly, from 46% to 53%. The most popular wrong answers were distractor B (wrong direction), which was chosen by 18%, and distractor C (wrong strength), chosen by 19%.

E3.This question proved to be somewhat easier, despite the rather abstract phrasing of the stem. 85% of the students knew that only alternative C gave a consistent expression i.e. g = GM / r2.

E4.This question required students to understand the trajectory of an electron moving in electric and / or magnetic fields. 73% gave a correct response.

E12.This question was on Coulomb’s law. In this question the incorrect statements were readily identifiable by those who had understood the topic. The correct one could then be found by a process of elimination, without necessarily working out that the electrostatic force could be measured in C2 F−1m−1. 80% of the answers were correct; perhaps it is remarkable that only 51% of the responses were correct when this question was pre-tested.

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E13.In this question, 66% of the students knew that an electric field can change both the magnitude and the direction of a moving charge, in the same way that a gravitational field can for a moving projectile. Confusion with the effect of a magnetic field on a moving charge may have caused 20% of the responses being for distractor B (direction can be changed but magnitude is unaffected).

E14.This question, with a facility of 37%, turned out to be the most demanding question in the test. Understanding that plate X was at a higher potential than plate Y should have enabled students to deduce that the electric field direction was from X to Y, and that an electron would experience an electrostatic force directed towards X. Hence an electron travelling from X to Y would be decelerated; its Ek would decrease and its Ep would increase. Therefore ∆Ep would be +30eV. Each of the distractors B and D received about 25% of the responses.

E15.The direction of the force on the negatively charged ion in part (a)(i) was mainly correct. Explanations of the direction of the force were good and marks for this part were high.

At A2 level the students are expected to have retained a comprehensive knowledge of earlier work, which is tested by synoptic components within the questions. Part (a)(ii) was an example of this, because the simplest solution followed directly from “energy gained = work done = force × distance”. This eluded most students, many of whom were completely defeated. Many of them became successful after taking a very roundabout route, involving calculation of the pd from V = W / Q, the field strength from E = V / d and the force from F = EQ. Others produced answers based on the uniform acceleration equations and F = ma. When this was done using algebra the mass m of the ion could be cancelled and the answer was accepted. When a numerical value was chosen for m the mark that could be awarded was limited to 1 out of 2. There were fewer difficulties in part (a)(iii), where E = F / 3e gave the most direct answer but where E = V / d offered an alternative method. Incorrect force values from part (a)(ii) were permitted for full credit in part (a)(iii).

In part (b)(i) the successful application of Fleming’s left hand rule readily showed the majority that the magnetic force on the wire would be downwards (or into the page). Most students gained both marks. It seemed that some students who thought too deeply about the direction of this force eventually got it wrong: their line of thought was that the force due to the current was downwards, but electrons carry a negative charge so the force on them must act upwards. It had not occurred to them that the force on the complete wire

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has to act in the same direction as the force on all of the free electrons within it.

The simple calculation that led to the number of free electrons in the section of wire was almost always worked out correctly in part (b)(ii). This was a “show that” question, so students should have realised that a more precise answer than 4 × 1022 (such as 4.07 × 1022) would be expected before the mark became available. Part (b)(iii) offered two approaches to the value of the flux density B. By considering the average force on a single electron, F = BQv could be used. Alternatively, by considering the force on the section of wire, F = BIl could be used. In the latter method many got into difficulty by forgetting to consider the number of electrons in the wire.

E16.This question was another re-used question, and its facility of 43% this time was a slight improvement over the previous occasion. Candidates’ responses showed that 75% of them recognised that the electric field acts downwards, but those who chose distractor D were evidently under the impression that there is a constant value of potential at all points between two parallel plates. Similar weaknesses in understanding the properties of the field between parallel plates were also evident in Section B of this Unit 4 test.

E17.In this question, once it has been appreciated that this force acts in the opposite direction to the field direction, the choice is narrowed by the elimination of distractors C and D. The correct answer cannot be B, because the force continues to move the electron upwards whilst it remains in the field. 65% of the responses were correct, but 15% chose C and 13% chose B.>

E18.This question was answered correctly by two-thirds of the candidates. No doubt it was misunderstanding of the direction of the force that acts on a negative ion that caused 19% of the candidates to select distractor B.

E19.Very few candidates experienced any difficulty in (a)(i), where the product of field strength

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and plate separation readily led to 30,000 V. In the other parts of Question 2 the principal failing of many of the candidates’ attempts was to resort to time variations that were exponential. Part (a) puts this question clearly in the context of a charging current that is constant, so any references to exponential changes or time constants showed misunderstanding and were irrelevant. Arithmetical slips sometimes caused the loss of marks in part (a)(ii), but Q = CV and t = Q / I were usually applied correctly to arrive at 3.5 s.

In part (b) it was essential for candidates to realise that both the charging current and the breakdown pd remain constant at their original values when the capacitance is changed. The majority of candidates could see in part (b)(i) that the time between discharges would increase. Many also gave an acceptable explanation, either by stating that the charge stored would have to be larger before the breakdown pd was reached, or by reference to t = CV / I, where V and I are both unchanged. A common misconception in part (b)(ii) was to think that the brightness of the spark would be unchanged because the breakdown pd would be the same as it had been originally. It was expected that candidates would know that increased capacitance at the same pd would mean that the energy stored by the capacitor would be greater, so each spark would transfer more energy and would therefore be brighter. Alternatively, explanations in terms of the greater charge stored were also accepted.

E20.Most of the candidates made significant progress in part (a) and there were some very good explanations of thermionic emission. The most common omission was in not describing the role of the electric field between the anode and cathode in accelerating the electrons in the beam.

Many candidates calculated the acceleration correctly and a significant number set out their work clearly and well. Some attempted to use the work done by the electric field between and others tried to use just the equations of motion.

Parts (b) (ii) and (b) (iii) were well done by many candidates.

In part (b) (iv), a significant number of candidates did not know how to proceed but the majority made some progress either by using ratios or equations of motion. For those that made significant headway, the most frequent mistake was to omit to add on the vertical distance that had already been travelled by the time the beam reached the end of the plates.

E21.The Capacitors were the topic tested by this question which needed knowledge of how to apply Q = I t for a constant current, C = Q/V and energy stored = ½ CV2.

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E22.This question had appeared in an examination previously; it tested the fairly familiar knowledge of the trajectory of charged particles in electric and magnetic fields and this time had a facility of 71%.

E23. Parts (a) and (b) of this question, about forces and accelerations of charged particles in electric fields, was more demanding than expected. The direction of the arrow in part (a)(i) was frequently wrong, and was often shown in a vertical direction. A clear force arrow, starting on the electron and directed horizontally to the left, was expected. However, in part (a)(ii) most students recognised that this situation produces a constant force; the explanation that the force is caused by a constant electric field strength was sometimes less clear.

More careful consideration of the wording of the questions, and subsequent review of the answers they had written, would have benefitted many students when answering part (b). For example, it was common for a student’s attempt at part (i) to try to answer the issues raised in part (iii). Students should know that a force has two attributes, magnitude and direction. Both had to be addressed in satisfactory answers to part (i), where both particles are at exactly the same point in an electric field of exactly the same field strength. In part (ii) the relationship between forces and accelerations was usually well understood and both marks were usually awarded. Failure to discuss the accelerations of the proton and electron separately was common in part (iii), leading to many wrong answers. Parabolic and circular motion were often referred to here. Some of the less able students stated that the electron would decelerate, probably because they were unable to distinguish between deceleration and decreasing acceleration.

Students who had revised their AS Physics, and who presented their working fully, were well rewarded in part (c)(i). Part (c)(ii), although it is basically a simple question, proved to be much more challenging. The common acceptable approach was to calculate the strength E of the uniform electric field and then divide 1.91 (or 2) V by E. A slightly longer procedure was divide the energy calculated in part (i) (3.06 × 10-19 J) by the force on the electron – which had to be calculated. However the question could be answered using simple ratios: (2 / 4500) of 180 mm is 0.080 mm.

E24. This question was also on electrostatics but, with a facility of 48%, was much more demanding. At first sight it appears necessary to solve a quadratic equation to answer the question, but this difficulty can be overcome by taking the square root of the expression obtained. Incorrect distractor D was chosen by 35% of the students and consequently the

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discrimination of the question was poor.

E25. This question was the most demanding question on the paper, with only 39% of the students giving the correct answer. In order to identify the correct combinations of units to give V m-1, it was necessary to remember that 1 V = 1 J C-1 and that 1 C = 1 A s. Distractor C was the choice of over a quarter of the students.

E26. The definition of electric potential in part (a) was generally well known. Where students did not score all three marks this was down to oversight; typically either omitting to mention that the charge involved in the definition is positive or that the definition involves the work done per unit charge.

In part (b) (i), most students successfully applied V = Q /4πε0r in order to determine the magnitude of the charge (1.0 × 10–10 C) from the value of V when r = 0.30 m. As the data in the question is given to two significant figures, an answer was expected to two significant figures. Some students need to appreciate that the number of significant figures they should quote in an answer needs to be limited to the least precise data they are working with, not the most (in the Data Sheet (see Reference Material) ε0 is given to three significant figures). At the same time, in these circumstances the answer should never be abbreviated to one significant figure(1 × 10–10 C), as was the case in many answers.

The mark in part (b) (ii) was gained easily, usually by applying V = Q/4πε0r, although more perceptive students saw that V ∝ 1/r could lead to a more concise answer. Part (b) (iii) caused a little more difficulty for some students. Application of E = Q/4πε0r2 with r = 0.60 m was the obvious route. The pitfall for many was that, by first finding V at M (by the same method as before), they then had to apply E = V/d to find the field strength. This last equation specifically applies to a uniform field and it therefore cannot be used here. Surprisingly, there were many students who, having obtained an incorrect charge in part (b) (i) as a result of an arithmetical slip, did not revisit part (i) when they could not show either of the required values in parts (ii) and (iii).

Many good attempts to represent the electric field between two plates were seen in part (c) (i), but careless sketching, such as field lines stopping short of the plates, often meant that it was not possible to award both marks. Because this was the field between two plates at different positive potentials, some students were thrown off course both when sketching the field and when the uniform field strength had to be found in part (c) (ii). In part (c) (iii) the respective radial and uniform fields were usually recognised but a precise statement that identified which was which was required to gain the mark.

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E27. This question tested candidates’ understanding of the mechanics of the motion of a positive ion as it passes through a uniform electric field. Quite a lot of calculation was needed to arrive at the correct response, but 57% were successful. Incorrect answers were evenly distributed amongst the other distractors.

E28. In part (a)(i) many candidates were unaware of the condition under which F = BQv applies, which is given clearly in the specification. A common incorrect answer was to state that the force has to be perpendicular to B, without any reference to v. In part (a)(ii) the main difficulty proved to be the meaning of B; magnetic flux density was correct and the loose ‘magnetic field strength’ was not accepted. Some candidates thought that v represents voltage.

Part (b)(i) was a test of Fleming’ left hand rule when applied to a stream of positive ions. Together with the figure, the first paragraph of part (b) defines ‘downwards’ as the direction towards the lower (negative) plate. The correct answer in (b)(i) is ‘into the plane of the diagram’, not downwards.

In part (b)(ii) candidates were expected to consider the force conditions applying to the undeflected ions. A common misconception was that the magnetic field is equal to the electric field. The main errors in part (b)(iii), where the numerical value obtained was often correct, were the omission of clear working and not knowing that the unit of B is T. Some candidates could only quote F = BQv and were at a loss to make further progress without F = EQ and E = V/d.

Many candidates were totally lost in part (c). Others correctly explained that the ions would now be the magnetic force (which is proportional to v) increases whilst the electrostatic force (which is independent of v) remains constant.

E29. This question, requiring a combination of F = EQ and F = ma, was the most discriminating question in the test; its facility was 67%.

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E30. Calculation of the electric field strength in a uniform field by using E = V/d were known well in part (a) (i), as was finding the force on a charge using F = EQ in part (a) (ii). Most candidates therefore achieved full marks in these parts. Answers to the force diagram in part (b) (i) were much less satisfactory. Examiners were expecting to see three clearly labelled force arrows, starting on the ball, showing the electrostatic force to the left, the weight of the ball downwards and the tension acting upwards along the thread. Careless drawing and inadequate labelling caused marks to be lost in a majority of answers. When labelling the downwards force, ‘weight’, ‘W’ or ‘mg’ were acceptable, whereas ‘gravity’, ‘mass’ or ‘g’ were not. The tension force was often omitted, whilst additional horizontal forces such as ‘centripetal force’ were sometimes shown.

In part (b) (ii), some evidence was expected for the appearance of the equation F = mg tan θ. This could be from a consideration of the resolved components of the forces acting, or from a force diagram showing θ clearly. Many good answers were seen, but a large proportion of the candidates could make little or no progress.

E31. Far fewer correct answers were seen to part (a) (i) than might have been expected. Deducing the correct direction for the electric field involved spotting that the electrostatic force on the sphere acted upwards, and that the sphere carried a negative charge. The vast majority of answers to part (a) (ii) showed that students had not forgotten Hooke’s law from Unit 2 of AS Physics; 0.24 × 0.018 readily gave 4.32 × 10–3 N. Part (a) (iii) was also well answered, either by combining F = EQ and E = V/d before inserting numbers, or by working out E, and then d, separately.

Attempts to answer both sections of part (b) showed that many candidates had little understanding of what would happen when switch S was moved to position Y. The fact that the immediate effect would be to short out the plates, causing them to discharge and therefore reduce the field strength to zero, escaped a very large number of candidates. Common answers to part (b) (i) were that the field was reversed, or that the field became an alternating one. Answers which suggested that an electric force would still be acting received no further credit in part (b) (ii). What was required here was an understanding that, when the field was removed, the sphere would fall under its own weight, extending the spring downwards. The resultant force on the sphere would be proportional to the change in the extension of the spring, producing an acceleration that was proportional to the displacement from equilibrium but acted in the opposite direction to the displacement ie the condition for shm.

E32. This question, about a charged particle moving in an electric field, had a facility of 66% and was a good discriminator. Incorrect responses were almost equally distributed

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between the incorrect distractors.

E33. This question required candidates to apply F = EQ in a charged parallel plate situation. 58% of them appreciated that the separation of the plates was d and that the field strength E would be V/d. However, 29% chose distractor A, for which the field strength must have been interpreted as V ÷ (d/2).

E34. This question on electrostatics, was concerned with field strength and potential. This was the worst discriminator in the test, and only 43% of candidates selected the correct response. A principal reason for as many as 24% of them choosing distractor A (potential is a scalar) must be that they had failed to notice that the question asked for the incorrect statement.

E35. Many candidates appreciated that E is defined as the force acting per unit charge, but very few were able to state that it is the force acting per mat positive charge. Consequently in part (a) (i), it was uncommon for more than one of the two available marks to be awarded. Confusion with the definition of electric potential was evident in many candidates’ responses. In part (a) (ii), fewer candidates than expected knew that E is a vector quantity.

The Coulomb’s law equation was usually correctly recalled at the start of candidates’ answers to part (b) (i), and was often followed by an acceptable value for the force. The principal difficulties here included using the wrong constant of proportionality, failing to square the denominator, and not knowing that nano means 10–9. The correct value of 27 mm in part (b) (ii) was usually given after little or no proper explanation, leading to a loss of one of the two marks. Examiners were expecting that something of the form 4/x = 8/(80 – x) would be given as a necessary step in the working.

It was clear from their attempts to answer part (c) that a large number of candidates could not follow simple instructions. The direction of the arrows was often wrong, whilst many arrows were not drawn at P. The 2:1 length ratio was often correct for the second mark. The third mark was awarded to those candidates who drew an arrow, labelled R, along the correct resultant of two correct component vectors. This final mark was not often awarded.

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E36. This question turned out to be the most demanding question on the paper, with only 48% of the candidates able to find correctly the magnitude of the force on a charge in a uniform electric field. The main difficulty appears to have been that of deciding that the

field strength is rather than because35% of the candidates chose theincorrect distractor A. The discrimination of this question was quite weak.

E37. Application of E = V/d and t = d/v brought two straightforward marks for most candidates in the first two parts of (a). Part (iii) caused greater difficulty, often because F = EQ was not known. One incorrect approach, adopted in several scripts, involved assuming a vertical displacement of 7.5 mm, corresponding to half the vertical separation of the deflecting plates. Using t = 9.4 × 10–10 s from part (ii), a = 2s/t2 was then applied, giving a vertical acceleration of 1.7 × 1016 m s–2. The question required candidates to give the direction of the acceleration as well as its magnitude, but this requirement was often overlooked. A few candidates wrote more than they need have done, and in doing so condemned their own answer; ‘the acceleration is upwards in a parabola’. Confusion between the trajectory and the directions of acceleration and velocity are understandable, but cannot be tolerated in examination answers.

Part (b) was either omitted or answered in a descriptive, non-mathematical way in more than half of the scripts. Those who understood the principles of projectile motion usually had little difficulty in gaining all three marks. A very common mistake however, was attempting to find the new velocity by use of v = u + at with u taken to be 3.2 × 107m s−2. In the work of the more able candidates, whether the direction of the calculated angle (about 26°) was ‘up’ or ‘down’ was often clarified by a diagram.

E38. This question, testing understanding of the direction of the acceleration of a charged particle in an electric field, had a facility of 63% and produced average discrimination. Failure to recognise that an electron has a negative charge and/or that field direction relates to force on a positive charge, must be reasons why distractor A attracted a 20% response.

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E39. This question was also concerned with electric fields, set in the context of a van de Graaff generator. Although the facility of this question moved from 57% in the pre-test to 69% in the examination, the discrimination index deteriorated somewhat to a value of 0.38.

E40. Trajectories of charged particles as they pass through electric and magnetic fields ought to be a fairly simple topic, but the facility of this question improved only slightly, from 55% to 57%, between pre-test and examination. Candidates who did not understand these topics were attracted in almost equal numbers to distractors B and D.

E41. Units of the various physical quantities related to fields and the scalar/vector nature of them, are generally not well known by the candidates. Part (a) showed that the 2004 cohort were no better than their predecessors. Six correct entries in the table were required for three marks, and it was very rare for all three to be awarded. The unit of N m kg–1 was accepted as an alternative to J kg–1 for gravitational potential, but candidates regularly put N kg–1 in the table. The unit of electric field strength was known better, and that of magnetic flux density was usually shown correctly. Candidates often resorted to guesswork when completing the second column of the table. Many did not appreciate that the concept of potential arises from energy considerations and that it is therefore a scalar quantity, whilst the other two quantities are force-related and therefore vectors.

Completely correct answers to part (b) were encountered in many of the scripts. Since the unit of E had already been tested in the table in part (a), no penalty was imposed for wrong or missing units in the answer to part (b)(i). A worrying error, made by a significant minority of the candidates, was to equate the electric force on the particle to its mass, rather than to its weight.

E42. In this question three-fifths of the candidates appreciated that the electric field strength is constant between parallel charged plates. Over 20% of them chose distractor C, where the field strength is shown as decreasing to a minimum value midway between the plates.

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E43. It surprised the examiners that only a minority of candidates gained full marks in part (a). Successful solutions were usually based on the triangle of forces. Only the best candidates resolved the tension into components and equated the components to the weight and the electrostatic force respectively. Many candidates incorrectly resolved the weight into components parallel and perpendicular to the thread.

The majority of candidates obtained the correct value of the electric field strength in part (b) and were able to make good progress in part (ii). Candidates who equated g to 10 N kg–1 or rounded off incorrectly at the end were penalised. Other candidates attempted inappropriate solutions involving Coulomb’s law and did not realise that the force = qE. A small minority of candidates attempted incorrectly to relate the gain of gravitational potential energy to an electrostatic energy formula such as ½QV.

E44. Teachers preparing students for these examinations might like to ponder over why almost as many candidates chose distractor D (implying a constant potential of 50V between the parallel plates) as the correct answer. The facility of this question was only 38% and it was a weak discriminator.

E45. There were many pitfalls en route to successful answers to part (a). Most candidates obtained little reward in this question because they could not steer clear of them. Examiners were pleased that so many of the candidates were not put off by the slightly unfamiliar way in which charge was given in part (a)(i), or by the mass given in u in part (a)(ii). This, at least, showed that some learning is taking place across the topic boundaries within Module 4. The really serious problems arose with arithmetic, units and the need to take care in calculations. Typical errors in part (i) were failing to halve the diameter and forgetting to square the denominator. The unit of electric field strength was known by some, yet hardly any of the candidates could give a correct unit for gravitational potential. Carelessness was apparent in the work of all those who omitted the negative sign from the final value for gravitational potential.

The subject area tested in part (a) remains totally confusing for so many candidates, who

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obviously cannot distinguish between the words gravitational and electric or field and potential. Perhaps they did not read the wording of the question correctly. This may be more excusable than the huge number of wrong answers to the electric field direction in part (c): an arrow pointing inwards at P was common, a tangential arrow at P was fairly frequent, and a vague arrow drawn some distance from P was not exceptional.

E49.Many of the attempts to answer this question showed a complete lack of understanding of forces in electric fields and much confusion between magnetic fields and electric fields. Taken as a whole, the question proved to be an excellent discriminator of the proficient physics student.

Perhaps it is understandable that many candidates wrongly chose ‘to the left’ in part (a)(i) and therefore followed it with ‘deceleration’ in part (ii). It is much more difficult to appreciate why the electron might be considered to follow a curved or circular path in part (a) - as stated by some candidates. The parabolic path in part (b) was often shown curving downwards. Part (b)(ii) required an explanation based on an understanding of projectile motion. Quite a large number of candidates preferred to refer to Fleming’s left hand rule!

E50.Although part (a) was relatively novel, most candidates could handle the comparison of gravitational and electric fields. The gaps in the second line of the table could be filled directly by use of the Data Booklet, but most of the other entries required a little more thought. Derived units were sometimes quoted (but not accepted) for the electric field strength: candidates were expected to know that this is N C–1 or V m–1. In the fourth line, distance (or radius) squared was a surprisingly common wrong answer.

In pan (b)(i) quite a large number of candidates did not state that the resultant force would be unchanged, even though they had correctly considered the separate effects of a 1 / r2 relationship on both the gravitational and electric forces. The most frequent wrong response was that the force (presumably the resultant force) would decrease by a factor of four. In part (b)(ii) many candidates stated that the charge should be increased, without indicating that it should be doubled – this was expected for the mark to be awarded.

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E51.Few of the definitions in part (a) included any reference to a positive charge, and “force on a unit charge” seemed more common than the more correct “force per unit charge”. The only field patterns tested in previous PH03 papers have been magnetic ones, so the need here to draw an electric field caught out the vast majority of candidates. Examiners found that almost all of the drawings more closely resembled the field between two current-carrying wires than that between two point charges.

Attempts to answer part (b) were variable and it was not uncommon for candidates having a good understanding of vector addition to score full marks. Misinterpretation of nanocoulomb was a problem for some candidates. The direction of the resultant field caused some difficulty, principally because candidates did not understand the fact that an electric field is directed inwards towards a negative charge.

Only a minority of able candidates showed a clear understanding of the concept of electric potential in part (c). Most answers tried to present arguments which were concerned with field rather than potential. It needs to be more clearly accepted that potential is a scalar quantity and therefore does not have an associated direction; potential values may add to zero but they can never “balance”. In part (c)(ii) there were many intuitive answers of 80 mm, which quoted 3≤ of 200 mm as the reasoning. Credit was given only for answers which were properly reasoned in terms of potential.

E54. (a) (i) Most of the candidates could draw the field using both lines of equipotential and electric lines. A few omitted to label the lines. A more common mistake was to draft the diagram carelessly so that it was not clear that the field was apparently uniform.

(ii) Weaker candidates got very tangled in this calculation, attempting to use ½ mv2 to calculate kinetic energy rather than using it to calculate speed once they had found the kinetic energy by using the potential difference in the field.

(b) The calculation in this part was done quite well. Few candidates could go on to explain whether or not the de Broglie wavelength made the electrons suitable for the investigation of metallic crystal structures. Some had no idea what the typical values for atomic separations are in metallic crystals. More surprisingly, those who did

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know the separations tended to be unclear about whether the wavelength was too big or too small or broadly applicable.

E55. (a) Although the capacitance equation was used well, a surprisingly large number of candidates rounded the capacitance down from 15.6 nF to 15 nF to incur a mark penalty.

(b) (i) The majority of candidates correctly calculated the average breakdown voltage to be 2.4 × 109 V. A sizeable minority were penalised for simply equating 3.0 × 106 × 800 to 2.5 × 109 without comment.

(ii) Those candidates using the ½CV2 formula usually had no difficulty with this part; use of ½QV meant that the charge needed to be calculated first and presented more of a problem to candidates. Errors arose from wrongly selecting the p.d. to be 3.0 × 106 V.

(iii) Most candidates were able to calculate the maximum charge but a surprising number, having calculated this value in the previous part of the question, were floored by this calculation.

(c) A common mistake made by candidates here was to substitute 99% of the charge (or voltage) value into the decay equation, i.e. to calculate the time for 99% left rather than 1% left. The exponential or logarithmic form of the equation presented a significant problem for the less mathematic candidates and a mark of 1 out of 3 for this part was regularly gained.

E56.(a) (i) The vast majority were able to assign the correct direction to the electric field.

(ii) Descriptions of what is meant by an equipotential surface could have been better. There was lack of clarity in the answers and poorer responses were made in terms of potential difference or field strength.

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(iii) Parallel lines for the equipotentials were common and correct for the first mark. However, only about half the candidates were able to go on to show the correct shape of the equipotential at the edge of the plates.

(b) (i) This was a ‘Show that’ question and candidates must endeavour to show complete solutions if they expect to obtain full credit. Although the vast majority of candidates obtained the correct answer, a significant number simply wrote down numbers that occurred in the question, arriving at a number similar to the suggested answer. Examiners wished to see a clear link between the potential gradient between the plates and ratio of force per unit charge. If this link was missing, candidates obtained little credit.

(ii) This question required a sustained effort by candidates to carry out a number of steps in a calculation to arrive at the correct answer. It was done well by many and showed that the traditional skills of the physicist are still accessible by significant numbers of those taking the examination. A common fault was to try to go down an erroneous route involving a spurious centripetal force acting on the dust particle, presumably the candidates thought that magnetic field theory was the appropriate physics here. Additionally, energy arguments, although appropriate, did not score well as candidates often failed to appreciate that they were dealing with a final and not an average velocity. However, despite the common success with the question, the working was usually shown in a poor and scrambled way. Examiners had to work hard to follow a chain of logic, only rarely expressed in clear steps and seldom presented clearly on the page.

E57. (a) (i) Only a small proportion of the candidates was able to provide a correct definition of the telsa. Most gave another form of the unit Wb m–2 or less commonly N A–1 m–1.

(ii) It would have been useful to see some words to support the algebraic argument but this was very rare. Responses were usually a number of formulae with cancellations (including some irrelevant formulae amongst the relevant ones in the poorest answers). In view of the poor explanations candidates were at least expected to make v the subject of the final formula to gain both marks.

(iii) This was often correct but whether the answers were based on an understanding of the physics or just number crunching was often difficult to follow. Candidates were given the benefit of the doubt.

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(b) (i) Many knew the equation r = mv/Bq or were able to derive it from first principles. A common error was to use the diameter instead of the radius so arriving at half the answer. Some did not calculate the mass of the ion and others used 28 × 1.6 × 10–19 C for the charge on the ion.

(ii) Many were able to gain credit in this part using the error carried forward for an incorrect flux density. However, the majority subtracted the given diameter from their new radius so losing the second mark.

E58.(a) This question was generally well answered.

(i) Most candidates knew the correct curvature but many failed to produce the correct graph because they either misread the scale or applied a 1 / r law instead of a 1 / r2 law.

(ii) This was frequently answered correctly but many simply stated ‘find the area under the graph’ without stating any limits.

(b) (i) The process was generally well known and there were many correct answers although too many significant figures cost a number of candidates a mark here. The most common error was however failure to convert the radius of the Earth to m.

(ii) Even though the formulae are now on the formula sheet a large number used an incorrect formula for the surface area of the Earth which, at this level, is matter of some concern. Many gained one mark for dividing the charge by what they thought was the area. Those who used 6400 km again without conversion were able to get full credit. A significant number of candidates had a unit penalty applied here by giving the answer in C m−1.

E60.(a) (i) This calculation was generally well done.

(ii) This too was well done but there were more computational errors here.

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(b) (i) Many candidates defined field strength correctly but referred to point charge rather than unit charge and others used the expression for potential gradient in a uniform field which was not appropriate.

(ii) Good candidates did this calculation well but a lot of weaker candidates introduced expressions relevant to the movement of charged particles in magnetic fields.

(iii) Many performed this calculation correctly but some used an incorrect value for distance.

(iv) Better candidates had no difficulty with this calculation but some weaker candidates omitted this part and the last part of the question.

(v) There was a tendency to use incorrect data in this calculation. Some of those who managed this part were penalised for giving the answer to only one significant figure.

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Animated Science – A Science & Food Blog · Web view2021/03/03 · A positive ion with a mass of 3.4 × 10–26 kg and a charge of 3.2 × 10–19 C is initially at rest at a point