1. (a) appropriate
shape; lines perpendicular to and touching plate and sphere; (2)
arrows towards negative sphere (1) 3
arrows towards negative sphere (1) 3
(b) (i) By moments, e.g F cos 20 = W sin 20 / by
triangle of forces /
by resolution of forces / other suitable method; i.e. justification needed (1)
F = 1.0 × 10–5 tan 20; = 1.0 × 10–5 × 0.364; (= 3.64 × 10–6 N) (2)
triangle of forces gives W/F = tan 70, etc (1) 3
by resolution of forces / other suitable method; i.e. justification needed (1)
F = 1.0 × 10–5 tan 20; = 1.0 × 10–5 × 0.364; (= 3.64 × 10–6 N) (2)
triangle of forces gives W/F = tan 70, etc (1) 3
(ii) E
= F/Q; = 3.64 × 10–6 / 1.2 ×
10–9 = 3.0 × 103;N
C–1 / V m–1 3
(c) E
= (1/4πεo)Q/r2; 3.0 × 103 = 9 × 109 × 1.2 × 10–9/r2;
(2)
or use F = (1/4πεo)Q2/r2; r2 = 3.6 × 10–3 giving r = 6 × 10–2 (m) (1) 3
or use F = (1/4πεo)Q2/r2; r2 = 3.6 × 10–3 giving r = 6 × 10–2 (m) (1) 3
(d) field
line sketch minimum of 5 lines symmetrical about line joining
centres with arrows; (1)
Fig 1 sketch matches RHS of Fig 2/plate analogous to mirror/AW
relating to symmetry (1) 2
centres with arrows; (1)
Fig 1 sketch matches RHS of Fig 2/plate analogous to mirror/AW
relating to symmetry (1) 2
[14]
2. (a) (i) Q
= VC; W = ½ VC.V ( = ½ CV2) (2)
(ii) parabolic shape passing through origin (1)
plotted accurately as W = 1.1 V2 (1) 4
plotted accurately as W = 1.1 V2 (1) 4
(b) (i) T = RC; = 6.8 × 103 × 2.2 = 1.5 × 104 s = 4.16 h 2
(ii) ΔW = ½ C(V12 –V22) = 1.1(25 – 16) ; =
9.9 (J) 2
(iii) 4
= 5 exp(–t/1.5 × 104) ; giving t = 1.5 × 104 × ln 1.25 = 3.3 × 103 (s) 2
(iv) P = ΔW/Δt = 9.9/3.3 × 103 = 3.0 mW ecf
b(ii) and (iii) 1
allow P = Vav2 /R = 4.52/6.8 × 103 = 2.98 mW
allow P = Vav2 /R = 4.52/6.8 × 103 = 2.98 mW
[11]
3. nature
and features:
α-particle is 2p + 2n/ mass 4 u (1)
charge of +2e (1)
very short range/heavily ionizing/absorbed by paper (1)
spontaneous; and random nature of radioactive decay (2)
energetically more favourable to eject four particles together than a single
one/other comment about energy minimisation/mainly occurs from higher A
nuclei/AW (1)
small mass decrease/loss provides kinetic energy of α-particle (1)
particle energy of a few MeV; particular decay is monoenergetic (2)
α-particle scattering:
suitable diagram and/or description to illustrate experiment up to 2 marks (2)
most particles have little if any deflection (1)
large deflection of very few shows nucleus is small; and very massive (2)
(Coulomb’s law enables closest approach to) estimate nuclear size
(in case of α-particle back scattering with conservation of energy argument) max 7
Quality of Written Communication 2
α-particle is 2p + 2n/ mass 4 u (1)
charge of +2e (1)
very short range/heavily ionizing/absorbed by paper (1)
spontaneous; and random nature of radioactive decay (2)
energetically more favourable to eject four particles together than a single
one/other comment about energy minimisation/mainly occurs from higher A
nuclei/AW (1)
small mass decrease/loss provides kinetic energy of α-particle (1)
particle energy of a few MeV; particular decay is monoenergetic (2)
α-particle scattering:
suitable diagram and/or description to illustrate experiment up to 2 marks (2)
most particles have little if any deflection (1)
large deflection of very few shows nucleus is small; and very massive (2)
(Coulomb’s law enables closest approach to) estimate nuclear size
(in case of α-particle back scattering with conservation of energy argument) max 7
Quality of Written Communication 2
[9]
4. (a) (i) 5.0 (V) (1) 1
(ii) 10.0 (V) (1) 1
(b) (i) Q = CV;= 1.0 × 10–3 (C) (2) 2
(ii) The total capacitance of each circuit is
the same (namely 100 μF); (1)
because capacitors in series add as reciprocals/ in parallel add/
supply voltage is the same and Q = VC, etc. (1) max 2 marks 2
because capacitors in series add as reciprocals/ in parallel add/
supply voltage is the same and Q = VC, etc. (1) max 2 marks 2
(c) (i) A1 will give the same reading as A2;
because the two ammeters are (1)
connected in series /AW (1) 2
answer only in terms of exponential decrease for a maximum of 1 mark
connected in series /AW (1) 2
answer only in terms of exponential decrease for a maximum of 1 mark
(ii) A4
will show the same reading as A2 at all times; (1)
A3 will show half the reading of A2 initially;
and at all subsequent times (2) 3
A3 will show half the reading of A2 initially;
and at all subsequent times (2) 3
[11]
5. (a) Positive as E-field is downwards/top plate
is positive/like charges repel/AW (1) 1
(b) (i) k.e. = QV; = 300 × 1.6 × 10–19 = (4.8 × 10–17 J) (2) 2
(ii) 1/2mv2 = 4.8 × 10–17; = 0.5 × 2.3 × 10–26 × v2 so v2 = 4.17 × 109;
(giving v = 6.46 × 104 m s–1) (2) 2
(giving v = 6.46 × 104 m s–1) (2) 2
(c) E
= V/d; so d = V/E = 600/4 × 104 = 0.015 m (2) 2
(d) (i) semicircle to right of hole (1) ecf(a);
(a) and d(i) to be consistent 1
(ii) mv2/r; = BQv; (2)
giving r = mv/BQ = 2.3 × 10–26 × 6.5 × 104/(0.17 × 1.6 × 10–19); (1)
r = 55 mm;so distance = 2r = 0.11 m (2) 5
giving r = mv/BQ = 2.3 × 10–26 × 6.5 × 104/(0.17 × 1.6 × 10–19); (1)
r = 55 mm;so distance = 2r = 0.11 m (2) 5
[13]
6. (i) leptons; 1
(ii) neutrino / muon / tau(on); 1
[2]
