Exam Style Answers 23 Asal Physics CB
March 11, 2023 | Author: Anonymous | Category: N/A
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CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK
Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded may be different.
Coursebook answers Chapter 23 Exam-style questions 1
2
A (B: incorrect calculation of charge (use of V 2), C & D: incorrect calculation of capacitance (use of V not not V 2))
3
[1]
Three in series, 33 µF
[1]
Two in parallel, the third in series with the pair 67 µF [1]
[1]
B (Both 1 and 4 increase the time constant (= CR CR); ); large time constant leads to long decay time) [1] 6
11 Three in parallel, 300 μF
Two in series, the third in parallel over the pair, 150 µF 12 a
1 1 1 1 = 10 + 200 + 600 = 100 C ttotal 600 otal
[1] [1]
3
Q = CV = 470 × 10 × 9.0 = 4.2 × 10 C
[1]
4
.03 033 3 × 10 6 = 15 V V = Q = 02200 C
so, combined capacitance = 600 = 60 µF [1] 10
[1]
charge stored = 1.5 × 60 × 10 6 = 90 µC [1]
5 6
2.0 = 4.0 × 10 4 = 400 µF C = V = 5000 [1] 1 2 6 2 W = 2 CV = 0.5 × 470 × 10 × 12 = 0.034 J [1]
Q1 = +90 µC, Q2 = −90 µC, Q3 = +90 µC, Q4 = −90 µC, Q5 = +90 µC, Q6 = −90 µC
7
W = 12 QV = 0.5 × 1.5 × 10 3 × 50 = 0.0375 J [1]
8
a
−
−
−
Q
−
−
−
b
½CV W = ½ CV 2 = 0.5 × 5000 × 10 6 × 242 = 1.44 J [1]
[1]
energy dissipated in lamp = 1.44 J − 0.36 J = 1.08 J
[1]
[1]
p.d. = 0.90 V across the 100 µF capacitor, capacitor, 0.45 V across the 200 µF capacitor [1]
0.15 V across the 600 µF capacitor
[1]
13 a
Time delay, antisu antisurge, rge, antispa antispark, rk, etc.
[1]
i R = V = 9.0/15 × 10−3 I
[1]
b
R = 600 Ω
[1]
b
Q = CV = 4700 × 10 6 × 12 = 0.056 C
[1]
c
average current = Q/t Q/t = 0.056/2.5 = 0.023 A
[1]
drive current through the resistor iii Evidence of using the area iii
average p.d. = 6.0 V
[1]
R = V/I = 6.0/0.023 = 260 Ω
[1]
iivv C = Q/V = 45 × 10 3/9.0
e
−
−
Current is dependent on p.d., which decreases at a non-uniform rate
ii p.d. decreases across capacitor ii
14 a
[1]
as charge flows off, so less p.d. to
45 ± 5 mC
=
[1] [1] [1]
−
[1]
[2]
When charge is halved, p.d. is halved [1] so energy stored = 0.5 × 5000 × 10 6 × 122 = 0.36 J
Q
Using V = C
W = ½ ½CV CV 2 = 0.5 × 4700 × 10 6 × 122 = 0.34 J [1]
a
d
10
b
−
−
9
−
5.0 × 10 3 µF
[1] [1]
−
i Q = V × 4πε0 r = 5.4 × 103 × 4π × 8.85 12 × 10 × 0.20 −
=
1.2 × 10 7 C −
ii C = Q/V = 2.4 × 10 8 / 5.4 × 103 = 2.2 × 10 11 F = 2.2 pF −
−
1
Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside © Cambridge University Press 2020
CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK
b
or 12 CV 2 = 12 × 1.2 × 10 7 × E = 12 QV or 5.4 × 103 (or equivalent)
[1]
Q × frequency = 804 × 10 6 × 50 iiii I = Qt = Q [1]
initial energy = 3.24 × 10 4 J
[1]
p.d.. after discharge = 15 kV p.d
[1]
where V is is average p.d. iiii ii P = VI where
−
−
−
ttherefore, herefore, energy remaining = ½ ½CV CV 2 11 = ½ × 2.2 × 10 × 15 0002 = 0.0025 J [1]
=
eenergy nergy released 0.0324 0.0025 ≈ 0.03 J c
15 a
−
P = 0.24 W
b
−
=
4500 V
½QV ii E = ½ ii QV = ½ × 5 × 10
8
−
×
4500
Charge stored is halved.
[1]
Current is halved but (av (average) erage) V is is unchanged.
[1]
Power is halved.
[1]
(Maximum [2] if qualitativ qualitative, e, i.e., capacitance reduced, etc.)
17 a
The time constant is the time taken for the charge on a capacitor to fall to 1/e 1/e of its initial value [1]
=
[1] [1]
[1]
[1]
−
[1]
[1]
[1] [1]
12 × 12 × 0.04
=
[1]
Capacitance is halved.
[1]
Not sufficient p.d. between sphere and plate [1] to continue ionising the air Q V = so Q = 4 4π πε0 × V × r 4πε 0 r Q C = = 4πε 0 r V 5 ×10 8 Q V = = i 4 πε 0 r 4π × 8.85 ×10 10 12 × 0.1
c
[1]
−
=
0.040 A
[1]
RC
b
[1]
i
1 1 + 1 = 5 = 500 C ttotal 2000 2000 otal
so, combined capacitance = 2000 = 400 µF 5
[1]
−
c
= 1.1 × 10 4 J [1] i Q is shared, therefore, each sphere has a charge of 2.5 × 10 8 C [1] −
and the p.d p.d.. is halved = 2250 V
thus, ½QV = ½ × 2.5 × 10 8 × 2250 = 2.8 × 10 5 J [1]
2
Q = 20 000 µC = 0.020 C
[1]
iii Q Q0 exp iii =
Capacitance is the charge stored
per unit potential difference across the capacitor plates. [1] i Q = CV = 67 × 10 6 × 12
[1]
Q = 804 ≈ 800 µC
[1]
(CRt ) which leads to −
5
=
[1]
−
− t = exp and ln 0.05 ( = − 3.0 ) 100 CR
ii Energy is lost to heating the spheres, ii spheres, as charge moves from one to the other. [1]
[1]
−
−
b
ii Q = CV = 400 × 10 6 × 50 ii −
[1]
16 a
[1]
t
( 400 ×10 × 250 × 10 )
t = 3.0 s
−6
3
=
t 1.0 0
[1] [1]
Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside © Cambridge University Press 2020
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