1.	(a)	Direction of field lines
	Downwards (1)	1
(b)	(i)	Calculation of force
	Use of V/d i.e. 250 V/0.05 m [if 5 used mark still awarded] (1)
Use of  e [Mark is for correct use of 1.6 × 10–19 C] (1)
= 8.0 × 10–16 N (1)	3
(ii)	Direction and explanation
	(Vertically) upwards / towards AB (1)
	No (component of ) force in the horizontal direction OR because (1)	2
(the force) does no work in the horizontal direction
(c)	Calculation of p.d.
Use of ∆EK = ½ mυ2 / ½ 9.11 × 10–31 (kg) × (1.3 × 107)2 (1)
Use of Ve / V × 1.6 × 10–19 (C) (1)
= 480 V (1)	3
(d)	Beam of electrons
	Diagram showing:
	Spreading out from one point (1)
fastest electrons labelled (1)	
	2
[11]

2.	Diagram
	Electric pattern:
	Straight, parallel, reasonably perpendicular to plates and equispaced
[Minimum 3 lines] (1)
Correct direction labelled on one line [Downwards arrow]  (1)	2
Equipotential lines:
Any two correct equipotentials with any labelling to identify
potentials (rather than field lines) (1)	1
[Arrows on electric field lines – none on equipotential being
sufficient labelling]

	Force
	E =  [Correct substitution] (1)
Use of F = Ee even if value of “e” is incorrect (1)
F = 120 ×(103) V m–1 × 1.6 × 10–19 C
= 1.9 (2) × 10–14 (N) (1)	3
	Graph
	Straight horizontal line [Even if extending beyond 25 mm] (1)
Value of F marked [e.c.f. their value] provided graph begins on
force axis and is marked at this point (1)	2
	Speed
Use (1)
eV = ½ mv2
v2 = 2 eV/m	v2 = 2s	Fd = ½ mv2
v2 = 2Fd/m	
	Substitution (1)
V2 = 
= 2  × 25 10–3 m
= 
Answer: V = 3.2 × 107 ms–1 (1)	3
	[If F= 2 × 10–14 N, then V= 3.3 × 107 ms–1]
[11]

3.	Charge distribution on spheres marked on diagram
Left hand sphere marked positive	1
Right hand sphere marked negative	1
Arrow on field line to indicate direction
Direction on one field line correctly drawn
Arrow left to right [Any contradictory arrows lose the mark]	1
Work done
(i)	0	1
(ii)	–50
	One mark for 50	1
One mark for negative sign	1
[6]