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A-Level Physics · guided topic map

Electric fields and potential for Cambridge International AS & A Level Physics

Electric fields and potential for A-Level Physics, organized into 1 syllabus topic and 7 mapped concept guides.

Syllabus topics
1
Mapped concept guides
7
Educational level
Cambridge International AS & A Level

Choose the exact concept

Work in order or jump to the concept named in your specification, course outline, or assignment.

18

Electric fields

A Level extension

7 guides
  1. 01Electric field strength and forceMapped lesson
  2. 02Electric field linesMapped lesson
  3. 03Uniform fields and parallel platesMapped lesson
  4. 04Coulomb's lawMapped lesson
  5. 05Point-charge fieldsMapped lesson
  6. 06Point-charge potentialMapped lesson
  7. 07Potential gradient and equipotentialsMapped lesson

Diagrams

Electric fields and potential as A-Level Physics draws it

The figures from the A-Level Physics practice papers that sit on these syllabus points — the apparatus, circuits and graphs an exam question actually puts in front of you.

01Fig. 10.1Electric fields · CapacitanceA Level
Two parallel metal plates in a vacuum, connected to a 2.0 kV supply2.0 kV+vacuum5.0 mm

Figure comment

Fig. 10.1Two long horizontal metal plates are drawn one directly above the other and connected by wires at their left-hand ends to a battery labelled 2.0 kV; the upper plate carries a plus sign and the lower plate a minus sign. The space between the plates is empty and labelled vacuum, and a dimension at the right-hand end gives the separation of the plates as 5.0 mm. No field lines are drawn between the plates.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

Guided questions 5 parts

Reading cue. 5.0 mm is the gap, not a plate length, and no field lines are drawn — the plus and minus signs on the plates are the only thing in the figure that fixes the field direction.

  1. aState State the direction of the electric field in the gap in Fig. 10.1, and the direction of the force it exerts on an electron placed there.

    recall2 marks

    Check answer 2 marks
    1. the field points vertically downwards, from the positive upper plate to the negative lower plate
    2. the force on an electron is vertically upwards, towards the positive plate
  2. bCalculate Calculate the work done on an electron that moves from the lower plate to the upper plate, and the speed with which it arrives.

    routine3 marks

    Check answer 3 marks
    1. W = eV = 1.60 × 10⁻¹⁹ × 2.0 × 10³ = 3.2 × 10⁻¹⁶ J
    2. ½mv² = 3.2 × 10⁻¹⁶ J
    3. v = 2.7 × 10⁷ m s⁻¹
  3. cDetermine A charged dust particle of weight 1.28 × 10⁻¹³ N is held at rest midway between the plates. Determine the magnitude and sign of its charge, and the number of excess electrons it carries.

    demanding4 marks

    Check answer 4 marks
    1. E = V/d = 2.0 × 10³ / 5.0 × 10⁻³ = 4.0 × 10⁵ V m⁻¹
    2. for equilibrium qE = weight, so q = 1.28 × 10⁻¹³ / 4.0 × 10⁵ = 3.2 × 10⁻¹⁹ C
    3. the electric force must act upwards while the field points downwards, so the charge is negative
    4. 3.2 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 2 excess electrons
  4. dDeduce The battery p.d. is suddenly reduced to 1.0 kV while the particle is still midway between the plates. Deduce the acceleration of the particle, and calculate the time it takes to reach a plate.

    top of the paper4 marks

    Check answer 4 marks
    1. new field = 2.0 × 10⁵ V m⁻¹, so the electric force = 3.2 × 10⁻¹⁹ × 2.0 × 10⁵ = 6.4 × 10⁻¹⁴ N upwards
    2. resultant = 1.28 × 10⁻¹³ − 6.4 × 10⁻¹⁴ = 6.4 × 10⁻¹⁴ N downwards
    3. mass = 1.28 × 10⁻¹³ / 9.81 = 1.30 × 10⁻¹⁴ kg, so a = 4.9 m s⁻² downwards, that is g/2
    4. falling the 2.5 mm to the lower plate: t = √(2 × 2.5 × 10⁻³ / 4.9) = 3.2 × 10⁻² s

Transfer challenge

An electron travelling at 2.0 × 10⁷ m s⁻¹ enters midway between two parallel plates 5.0 cm long, moving parallel to them, in a uniform field of strength 1.2 × 10⁴ V m⁻¹. Calculate the sideways deflection of the electron as it leaves the plates.

Check answer 3 marks
  1. a = eE/m = (1.60 × 10⁻¹⁹ × 1.2 × 10⁴) / 9.11 × 10⁻³¹ = 2.1 × 10¹⁵ m s⁻²
  2. time between the plates t = 0.050 / 2.0 × 10⁷ = 2.5 × 10⁻⁹ s
  3. deflection = ½at² = 6.6 × 10⁻³ m