1. (a) (i) Cp = 2 + 4 = 6 μF A1
(ii) 1/C
= 1/2 + ¼ C1
Cs = 4/3 =1.33 μF A1
(b) (i) 6.0 V A1
(ii) Q
= CpV C1
= 6 × 6 = 36 μC A1
(c) E
= ½ CsV2 C1
= 24 × 10–6 A1
(d) (i) The capacitors discharge through
the voltmeter. B1
(ii) V
= V0e–t/CR
1/4 =e–t/(6×12) C1
ln 4 = t / 72 C1
t = 72 ln 4 ≈ 100 s A1
[12]
2. (a) Qo = CV = 1.2 × 10–11 × 5.0 × 103; = 6.0 × 10–8;
C (3) 3
(b) (i) RC = 1.2 × 1015 × 1.2 × 10–11 or = 1.44 × 104 (s) (1) 1
(ii) I = V/R = 5000/1.2 × 1015 or = 4.16 × 10–12 (A) (1) 1
(iii) t
= Qo/I; = 6 × 10–8 / 4.16 × 10–12 = 1.44 × 104 (s) 2
(iv) Q
= Qoe–1; Q = 0.37Qo so Q lost = 0.63Qo 2
(c) (i) capacitors in parallel come to same
voltage (1)
so Q stored α C of capacitor (1)
capacitors in ratio 103 so only 10–3 Qo left on football (1) 3
(ii) V
= Q/C = 6.0 × 10–8 /1.2 × 10–8 or 6.0 × 10–11 /1.2 × 10–11 or only 10–3
Q left so 10–3 V left; =
5.0 (V) 2
[14]
3. (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
(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
[11]
4. (a) 29; 34 2
(b) λ = 0.693/T = 0.693/(120 × 3.2 × 107)
= (1.8 × 10–10 s–1)
accept ln 2 1
(c) (i) Q = CV = 1.2 × 10–12 × 90; evidence of calculation (= 1.1 × 10–10 C) 2
(ii) n
= Q/e = 1.1 × 10–10/1.6 × 10–19; = 6.9 × 108 allow sig. fig. variations 2
(iii) A = λN; N = 6.9 × 108/1.8 × 10–10;
= 3.8 × 1018 using 7.0 gives 3.9 3
(iv) 1
y is less than 1% of 120 y so expect to be within 1%/
using e–λt gives exactly 1% fall/
problem of random emission
or other relevant statement 1
[11]
5. (a) (i)
|
capacitor
|
capacitance
/ µF
|
charge
/ µC
|
p.d.
/ V
|
energy
/ µJ
|
|
X
|
5
|
30
|
=
Q/C
=
6 (V) (1)
|
=
½ CV 2(1)
=
½ × 5 × 62
=
90 (1)
|
|
Y
|
25
|
=
CV
=
25 × 6
=150
(µC) (1)
|
=
6 (V) (1)
|
=
450 (1)
|
|
Z
|
10
|
30
+ 150 =
180
(µC) (1)
|
=
Q/C
=
180/10
=
18 (V) (1)
|
=
1620 (1)
|
Each box correctly calculated scores
(1) + (1) for ½ CV2 9
(ii) 1 18 V + 6 V = 24 (V) (1)
2 180 (µC) (1)
3 180 / 24 = 7.5 (1)
4 90 + 450 + 1620 = 2160 (µJ) (1) 4
(b) (i) Kirchhoff’s second law OR conservation of
energy (1) 1
(ii) Kirchhoff’s first law OR conservation of
charge (1) 1
(c) (i) time constant = CR (1)
= 7.5 × 10–6 × 200 000 = 1.5 (s) (1) 2
(ii)
(1)
Q/Qo = e–4 = 0.0183 (1) 2
[19]
6. (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
(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
(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
[11]
7. (i) C = Q/V or gradient of graph / = 24 μC/3V;
= 8.0 (μF) 2
(ii) E = ½ CV2 / = ½ × 8 × 32; = 36 (μJ) ecf
a(i) 2
or ½ QV / = ½ × 24 × 3; = 36 (μJ)
(iii) T
= RC = (0.04); R = 0.04/8.0μ = 5.0 × 103 (Ω) ecf a(i) 2
(iv) idea of exponential/constant ratio in equal
times; which is independent of
initial value/AW or argued mathematically in terms of Q/Qo = e–t/RC
give 1 mark for statement that time depends only on time constant/RC 2
[8]
8. (i) Cp = C + C = 6 μF; 1/Cs = 1/2C + 1/C; = 3/2C giving Cs = 2C/3 = (2 μF) 3
(ii) 2 sets of (3 in series) in parallel/ 3 sets
of (2 in parallel) in series 2
[5]
