y13 assessment mark scheme

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) (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]



3 (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 (µC) (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

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4. (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

(c) E = V/d; so d = V/E = 600/4 × 104 = 0.015 m (2) 2