PRE-U STPM PHYSICS TERM 2

CONTENTS






Chapter Chapter
12 ELECTROSTATICS 1 16 MAGNETIC FIELDS 150
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
12.1 Coulomb’s Law 2 16.1 Concept of a Magnetic Field 151
12.2 Electric Field 6 16.2 Force on a Moving Charge 152
12.3 Gauss’s Law 9 16.3 Force on a Current-Carrying
12.4 Electric Potential 15 Conductor 156
STPM PRACTICE 12 32 16.4 Magnetic Fields due to Currents 159
QQ1 38 16.5 Force between Two Current-
Carrying Conductors 167
Chapter e
13 CAPACITORS 43 16.6 Determination of the Ratio m 172
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • STPM PRACTICE 16 185
13.1 Capacitance 44 QQ5 191
13.2 Parallel – Plate Capacitors 45
13.3 Dielectrics 47 Chapter
ELECTROMAGNETIC INDUCTION 195
13.4 Capacitors in Series and in Parallel 49 17 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
13.5 Energy Stored in a Charged 17.1 Magnetic Flux 196
Capacitor 56 17.2 Faraday’s Law and Lenz’s Law 197
13.6 Charging and Discharging of a 17.3 Self-Induction 212
Capacitor 63
STPM PRACTICE 13 71 17.4 Energy Stored in an Inductor 215
QQ2 77 17.5 Mutual Induction 216
STPM PRACTICE 17 221
Chapter QQ6 228
14 ELECTRIC CURRENT 81
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
Chapter
14.1 Conduction of Electricity 82 18 ALTERNATING CURRENT CIRCUITS 232

14.2 Dri Velocity 84 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •

14.3 Current Density 85 18.1 Alternating Current rough a
14.4 Electric Conductivity and Resistivity 87 Resistor 233

STPM PRACTICE 14 96 18.2 Alternating Current rough an 239

Inductor
QQ3 99
18.3 Alternating Current rough

Chapter a Capacitor 243
15 DIRECT CURRENT CIRCUITS 101
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 18.4 R-C and R-L Circuits in Series 246
15.1 Internal Resistance 102 STPM PRACTICE 18 253
15.2 Kirchho ’s Laws 108 QQ7 258

15.3 Potential Divider 119
15.4 Potentiometer and Wheatstone STPM Model Paper 262
Bridge 127
STPM PRACTICE 15 140 Summary of Key Quantities and Units 273
QQ4 147

vi







00 PRELIMS PHYSICS T2.indd 6 10/18/18 3:09 PM

CHAPTER
13 CAPACITORS







13 13
13


Concept Map




Capacitors
Dielectric

Parallel-Plate Capacitor d
Capacitance
ε ε ε A
0 0
C =
ε
ε
Q C C = +Q –Q εε A
C
C
C = — d C = r 0 0
= —
= —
= —
V V d
V
Energy Stored in Charging a Capacitor
Capacitor t t t
– —–
– —–
– —–
– —–
– —–
– —–
=
Q
V
V
I
Q
E
(1 – e
E
I
=
=
CR
CR
CR
CR
CR
CR
1 (a) V = E(1 – e – —– ) (b) Q = Q (1 – e – —– ) (c) I = I e – —–
0
0
=
CV

U = CV 2 2
U U
CV
2 V Q I
1
= QV E Q 0 I 0
2
1 Q Q Q 2 2
=
2 V
0 t 0 t 0 t

Discharging a Capacitor
– —–
– —–
– —–
– —–
– —–
– —–
V
=
(c)
=
(a) V = V e – —– t (b) Q = Q e – —– t (c) I = I e – —– t
I
V
I
Q
=
(b)
Q
CR
CR
CR
CR
CR
CR
CR
CR
CR
CR
0 0 0
V Q I
V 0 Q 0 I 0
0 t 0 t 0 t
Bilingual Keywords
Capacitance: Kapasitans Discharging: Menyahcas
Capacitor: Kapasitor Energy stored: Tenaga disimpan
Charging: Mengecas In parallel: Selari
Concentric sphere: Sfera-sfera sepusat In series: Sesiri
Dielectrics: Dielektrik Parallel-plate: Plat selari
43
43
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Physics Term 2 STPM Chapter 13 Capacitors

INTRODUCTION

1. Capacitors are widely used in electronic circuits. Capacitors store electric charge.
2. Electrons flow in and out but not through a capacitor. A capacitor blocks direct current but

alternating current is able to pass through it.

13 3. Basically, a capacitor consists of two parallel metal plates with an insulator known as dielectric 13
in between.
4. Example of dielectrics are air, paper, wax and mica.







Figure 13.1


Figure 13.2

5. Figure 13.1 shows the circuit symbol of a capacitor.
6. Figure 13.2 shows capacitors of various sizes used in electrical and electronic circuits.




13.1 CapacitanceCapacitance
13.1
Learning Outcome 2016/P2/Q3, 2018/P2/Q4

Students should be able to:
• defi ne capacitance

1. Figure 13.3 shows what happens when a capacitor is
connected to a battery. Electrons from the battery charges
the plate X of the capacitor with a charge of –Q and a
charge of +Q is induced on the opposite plate Y. Hence, _ X
the charge on both plates of the capacitor is the same Battery _ + _ _ _ Capacitor
+ +
+
magnitude but opposite in sign. + Y
2. As the charges on the plates of the capacitor increase, the Figure 13.3
potential difference across the capacitor increases until it
is equal to the e.m.f. of the battery.
3. The charges remain in the capacitor even after the battery is disconnected.

4. The quantity of charge a capacitor is able to store depends on its capacitance.
5. The capacitance C of a capacitor is the ratio of the charge on a plate of the capacitor to
the potential difference between the plates.
Charge on either plate of capacitor –Q +Q
Capacitance, C = —————————————————–
Potential difference between the plates
V
Q
C = — Figure 13.4
V
44






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Physics Term 2 STPM Chapter 13 Capacitors

6. Since a conductor is able to store charges, the concept of capacitance can also be applied to a
conductor. The capacitance of a conductor is the ratio of the charge Q on the conductor to the
electric potential V of the conductor.
Charge on conductor
Capacitance of conductor = ————————————––—
Electric potential of conductor
Q
C = —
13 V 13

7. The unit of capacitance is farad (F).
The farad (1 F) is the capacitance of a capacitor that has a charge of one coulomb (1 C)
on each plate when the potential difference between the plates is one volt (1 V).
Capacitors with capacitance of a few µF are used in simple radio receiver circuits and large capacitors
capacitance of a few MF are used in electrical appliances such as washing machines.


Quick Check 1


1. A capacitor stores 24 mC of charge when Sketch a graph to show how the charge Q
the potential difference between the plates is in the capacitor varies with the potential
12 V. What is the capacitance of the capacitor? difference V across the capacitor.

2. A 50 µF capacitor is charged by connecting it 3. The capacitor of variable capacitance is
to a 6.0 V battery. connected to a 12 V supply. Its capacitance
What is the charge in the capacitor when the is 50 µF. After the capacitor is charged,
potential difference across the capacitor is the voltage supply is disconnected and the
(a) 3.0 V? capacitance is changed to 100 µF. What is the
(b) 6.0 V? new potential difference across the capacitor?



13.2 Parallel-Plate Capacitorsapacitors
C
Parallel-Plate
13.2
2016/P2/Q5
Learning Outcomes
Students should be able to:
• describe the mechanism of charging a parallel-plate capacitor
Q ε A
• use the formula C = to derive C = 0 for the capacitance of a parallel-plate capacitor
V d

1. Figure 13.5 shows a parallel-plate capacitor which consists of two
parallel metal plates each of area A and separated by a distance d d
in free space or vacuum. The capacitor is charged to a potential
difference V.
Q Q
E
2. The charge on each plate is Q. Using Gauss’s law,
ε Φ = ΣQ
0
ε (EA) = ΣQ V
0
Q
E = ………
ε A Figure 13.5
0




45






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Physics Term 2 STPM Chapter 13 Capacitors

3. The electric field E between the plates is also given by

V
E = ………
d
Q V
Equating and , =
ε A d
0
–12
–1
Capacitance, C = Q = ε A where ε = 8.85 × 10 F m is the permittivity of free space between
0
0
13 the plates V d 13
Example 1

A capacitor is formed using two parallel metal plates, each measuring 15 cm × 20 cm separated by
a distance of 0.50 cm.
(a) Find the capacitance of the capacitor.
(b) If the plate separation is increased to 1.0 cm, what is the new capacitance?
(c) The new capacitor is connected to a battery of 12 V. What is the charge in the capacitor?

Solution:

ε A
0
(a) Capacitance, C = —–
d
(8.85 × 10 )(0.15 × 0.20)
–12
= F
(0.50 × 10 )
–2
= 5.31 × 10 F
–11
(b) Capacitance, C ∝ 1 , when d’ = 2d,
d
1
C’ = C
2
= 2.66 × 10 F
–11
(c) Charge, Q’ = C’V
= (2.66 × 10 )(12)
–11
= 3.19 × 10 C
–10

Quick Check 2



1. A parallel-plate capacitor has a capacitance of (i) What is the maximum potential
6.0 nF. The plates are separated by a distance difference that can be applied across
of 2.0 mm. What is the area of each plate? the capacitor?
(ii) What is the maximum charge that
2. The plates of a parallel-plate capacitor are can be stored in the capacitor?
separated by a distance of 5.00 mm. The area

2
of each plate is 0.040 m . 3. The capacitor used in the flash-light system
(a) Find the capacitance of the capacitor. of a camera has a capacitance of 24 mF. If the
(b) Insulation of the air between the capacitor is charged by a constant current of
capacitor breaks down if the electric 1.5 mA from a 6.0 V battery, what is the time
field between the plate exceeds taken to charge the capacitor to a potential
–1
6
3.0 × 10 V m . difference of 6.0 V?
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Physics Term 2 STPM Chapter 13 Capacitors

13.3 Dielectrics
13.3 Dielectrics

Learning Outcomes 2012/P1/Q25, 2018/P2/Q3
Students should be able to:
• defi ne relative permittivity ε (dielectric constant)
r
• describe the effect of a dielectric in a parallel-plate capacitor
13 • use the formula C = ε ε A 13
r 0
d

1. The parallel-plate capacitor discussed in section 13.2 has free space or vacuum in between the plates.

2. The capacitance of a capacitor is greatly increased by having an insulator, known as dielectric, in
between the plates.
3. The dielectric constant or relative permittivity ε of an insulator is defi ned as
r
capacitance of a parallel-plate capacitor with the insulator in between the plates
ε =
r capacitance of the parallel-plate capacitor with free space in between the plates
C
=
C 0
Hence, capacitance of a parallel-plate capacitor with an insulator of dielectric constant ε in between
r
the plates is
ε ε A
C = ε C = r 0
r 0 d
4. Typical values of dielectric constant ε are
r

Air: 1.0006, Paper: 3, Mica: 7, Paraffin wax: 2.5
5. Since the dielectric constant for air ε = 1.0006 = 1.00 (to 3 signifi cant fi gures), parallel-plate capacitors
r
with air between the plates can be taken as a good approximation to capacitors with free space or
vacuum between the plates.
6. The action of an insulator is illustrated as in + – Dielectric
Figure 13.6. + – + – + – + –
(a) The molecules of the insulator are polarised + + – + – + – + – – Original field

by the electric field in between the plates. + + – + – + – + – – Reverse field
(b) This results in the surface of the insulator + + – + – + – + – –
facing the positive plate being negatively + + – + – + – + – –
charged, and the other side being positively + + – + – + – + – –
charged. + –
(c) A reverse electric field is set up. The + + – + – + – + – – Polarized
resultant electric field between the plates molecule

is reduced.
V
(d) Since E = —, when E decreases, the potential
d
difference V between the plates decreases. Battery
Figure 13.6
(e) If the capacitor is still connected to the
battery, charges continue to flow into the
capacitor until the potential difference across the
capacitor equals the e.m.f. of the battery.
(f) Since the capacitor is able to store more charge for the same potential difference, its capacitance
increases.





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Physics Term 2 STPM Chapter 13 Capacitors

7. On the other hand, if an air capacitor after being charged is disconnected from the battery, the
insertion of an insulator between the plates
1
(a) reduces the electric field to — its initial value,

ε
r 1
(b) reduces the potential difference to — its initial value,
ε
r
Q
(c) Since C = — , when V decreases for the same charge Q on either plates, the capacitance C
V
13 increases by a factor ε . 13
r
Example 2
(a) A parallel-plate capacitor consists of two metal plates each of area 2.0 m separated by a distance
2
4
of 5.0 mm in air. A potential difference of 1.0 × 10 V is applied across the capacitor. Calculate
(i) the capacitance,
(ii) the charge on each plate,
(iii) the electric field strength between the plates.

(ε for air = 1.00)
r
(b) The charged capacitor is then disconnected from the charging voltage and insulated so that
the charge in the capacitor is constant. A piece of dielectric of thickness 5.0 mm and dielectric
constant 5.0 is inserted in between the plates. Calculate

(i) the electric field strength between the plates,
(ii) the potential difference across the capacitor,
(iii) the capacitance.
Solution:

ε A
0
(a) (i) Capacitance, C = —– Exam Tips
d
–12
8.85 × 10 × 2 1. When a charged capacitor is disconnected from the
= ———————
5 × 10 –3 charging voltage, the charge Q remains constant.
= 3.54 × 10 F 2. If the capacitor remained connected to the
–9
charging voltage, the potential difference V remains
(ii) Charge on each plate unchanged.
Q = CV
= (3.54 × 10 ) × (1 × 10 )
–9
4
= 3.54 × 10 C
–5

(iii) Electric field strength
V
E = —
d
1 × 10 4
= —–——
5 × 10 –3
= 2.0 × 10 V m –1
6
(b) (i) Electric field strength

E
E’ = —
ε
r
2.0 × 10 6
= ————–
5
= 4.0 × 10 V m –1
5
48






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Physics Term 2 STPM Chapter 13 Capacitors

(ii) Potential difference
V
V’ = —
ε
r
1 × 10 4
= —–——
5
= 2.0 × 10 V
3
13 13

(iii) Capacitance C’ = ε C
r
–9
= 5 × 3.54 × 10
–8
= 1.77 × 10 F
Quick Check 3


1. A parallel-plate capacitor is charged in air. It 2. The electric field between the plates of an

is then electrically isolated and lowered into isolated air-spaced parallel-plate capacitor
a liquid dielectric. Which of the following sets is E. Without disconnecting the charging
of changes is correct? voltage, an insulator of relative permittivity
A Both the capacitance and the charges on 2 is inserted in between the plates. What is

the plates increase. the final electric field between the plates?

1
B Both the capacitance and the charges on A ––E C 2E
the plates decrease. 2
C The capacitance increases and the B E D 2E
potential difference across the plates 3. A rolled paper capacitor is made from strips
decreases. of metal foil of dimensions 2 cm × 40 cm
D The capacitance decreases and the separated by paper of relative permittivity
potential difference across the plates 2 and thickness 0.002 cm. Estimate its
increases. capacitance.




Capacitors in Series and in Parallel
13.4
13.4 Capacitors in Series and in Parallel
Learning Outcome 2012/P1/Q24, 2013/P2/Q3, 2014/P2/Q3, 2017/P2/Q4

Students should be able to:
• derive and use the formulae for effective capacitance of capacitors in series and in parallel

1. Figure 13.7 shows three capacitors of capacitance C , C and C 3

1
2
connected in parallel to a battery. Q 1 C 1
2. Charges flow into the capacitors until the potential difference

Q 2 C 2
across each capacitor is equal to the charging voltage V. Hence
the potential differences across the capacitors are the same, V.
Q 3 C 3
3. The charge in each capacitor is given by
Q = C V, Q = C V, Q = C V

1 1 2 2 3 3 V
The total charge, Q = Q + Q + Q
1 2 3
= C V + C V + C V
1 2 3 Battery
= (C + C + C )V
1 2 3
Figure 13.7
49




13 Physic T2.indd 49 10/18/18 3:18 PM

Physics Term 2 STPM Chapter 13 Capacitors

The three capacitors can be replaced by a single capacitor of capacitance C, if the charge in the
equivalent capacitor is
Q = CV
Hence CV = (C + C + C )V
1 2 3 Exam Tips
Equivalent capacitance, C = C + C + C 3
1
2
For capacitors in parallel
13 This formula for the equivalent capacitance can be • same potential difference across all capacitors 13
extended for any number of capacitors connected • charge in each capacitor ∝ capacitance.
in parallel. • equivalent capacitance, C = sum of capacitance

Capacitors in Series

1. Figure 13.8 shows the three capacitors of capacitance C , +Q C 1 –Q +Q C 2 –Q +Q C 3 –Q
1
C , and C connected in series to a charging voltage V. A B M N X Y
2 3
2. Electrons flow from the negative terminal of the charging
voltage and charge plate Y of the capacitor C negative. V 1 V 2 V 3
3
3. The charge –Q on plate Y induced a charge +Q on the V
plate X which in turn causes the plate N to be charged
–Q and so on.
Hence the charges on all the capacitors are the same,
i.e. Q. Figure 13.8
Q
Q
Q
4. The potential difference across each capacitor is given by V = — , V = — , V = —
1 C 2 C 3 C
1 2 3
The total potential difference, V = V + V + V
1 2 3
Q Q Q
= — + — + —
C C C
1 2 3
C )
1 1 1
= Q — + — + —
( C C
1 2 3
5. Since only a charge of Q is transferred from the charging voltage to the capacitors, the equivalent
capacitor would have a charge of Q when connected to the same charging voltage V, and
Q
V = —
C
C = equivalent capacitance
C )
Q 1 1 1
Hence, — = Q — + — + —
C ( C C
1 2 3
1 1 1 1
— = — + — + — Exam Tips
C C C C
1 2 3
When two capacitors C and C are connected in series,
1
2
• the charges in both capacitors, and in the equivalent capacitor
are equal.
1 ( C + C )

• potential difference across C : V = C 2 V
1
1 2
2 ( C + C )
potential difference across C : V = C 1 V
2
1 2
50





13 Physic T2.indd 50 10/18/18 3:18 PM

Physics Term 2 STPM Chapter 13 Capacitors

Example 3


A capacitor made from two thin, flat metal sheets
separated by an insulating material has a capacitance C.
Each metal sheet is then cut into four identical sheets,
which are used to make the capacitor as shown. The
13 separation between the interleaved sheets is filled with 13

the same insulating material.
Find in terms of C, the capacitance of the reconstructed
capacitor.
Solution:
Exam Tips
If A = area of each large metal sheet
d = separation between sheets To determine the number of capacitors in the
question: Count the number of spaces between
ε ε A two plates
r 0
then C = ——–
d
The reconstructed capacitor consists of 7 capacitors connected in parallel. Four plates are to one
terminal, and another four to the other terminal.
A C
,
Each capacitor is of area — hence of capacitance = —
4 4
C
Hence capacitance of 7 capacitors each of capacitance — in parallel
4
C ’ = 7 × ( —) 7
C
= —C
4 4
Example 4

A parallel-plate capacitor consisting of two metal plates
separated by a distance x has capacitance C. Each metal plate x x x
is cut into three identical small plates and connected to form Metal
the system of capacitors shown in the diagram. The separation plate
between plates remains as x.
Deduce in terms of C, the effective capacitance of the arrangement
shown in the diagram.

Solution: +Q + – –Q
+ –
+ –
– + –
–Q –Q –Q
–Q – + + – – + + – + – + – +4Q –4Q
1 2 3 4 + 5 – +Q + – –Q + –
– + + – – + + – + + – – + – + –
+ – + – + –
– + + – – + + – + + – – + + – –
+ – + + – –
– + + – – + + – + – + – + –
+4Q –4Q +Q + – –Q + –
+Q +Q +Q +Q + + + – –
Figure (a)
When the arrangement is connected to a voltage supply, the +Q + + – – –Q
charges on the plates are as shown in Figure (a). + –
Since the number of spaces between plates is 5, there are 5 + –
capacitors. Figure (b)


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Physics Term 2 STPM Chapter 13 Capacitors

Four of the capacitors have one plate each connected to one common terminal, and the other plate
to another common terminal. Hence these four capacitors are in parallel. The charge on the fifth

capacitor 4Q = sum of charges on the four capacitors.

Hence the fifth capacitor is in series with the four capacitors.
A
C
.
The area of each small plate = — Hence capacitance of each small capacitor = — .
3 3
( 3)
13 The equivalent capacitance of 4 such capacitors in parallel = 4 × — = —C 13
4
C
The effective capacitance of the arrangement C’ is given by 3
1 1 1
— = —— + ——
C’
4
1
—C —C
3 3
3 + 12
= ———
4C
4
C’ = –—C
15
Example 5
Two capacitors of capacitance 2 µF and 4 µF, initially uncharged are connected in series with a
12 V battery. Calculate
(a) the equivalent capacitance,
(b) the charge on each capacitor,
(c) the potential difference across each capacitor.
Solution:
(a) The equivalent capacitance C is given by 12 V
1 1 1
— = — + —
C 2 4 2 µF 4 µF
3
= —
4 V 1 V 2
4
C = — µF
3
(b) Charges on both capacitors are equal
= charge on equivalent capacitor
( 3 )
4
= — µF × 12 V
= 16 µC
(c) Potential difference across 2 µF capacitor
Q
V = —
1 C
1
16 µC
= ———
2 µF
= 8 V
Potential difference across 4 µF capacitor
V = V – V
2 1
= (12 – 8) V
= 4 V


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Physics Term 2 STPM Chapter 13 Capacitors

ANSWERS




1 3. D 4. A 5. C 6. D
C
7. A: Q = ( )V and Q = (2C)V
Q 24 mC 1 2 2
1. C = = = 2 mF
13 V 12 V 8. (a) (i) 0.5 µF (ii) 6 µC 13
2. (a) Q = CV = (50 µF)(3.0 V) = 150 µC (b) (i) 2 µF (ii) 12 µC
(b) Q = (50 µF)(6.0 V) = 300 µC (c) (i) 8 µF (ii) 24 µC
(d) (i) 0.8 µF (ii) 9.6 µC, 4.8 µC
Q
9. (a) Two 2 µF capacitors in series with 2 × 10 V.
(b)


0
V
20 V
3. Q = (100)V = (50)(12) 20
1 10. (a) — µC
V = 6 V 3
1
(b) 44.4 V 2Q = (C + C ) V
2 1 2
11. (b) (C + C )V = C V + C V
1 2 1 1 1 2 2
ε A (C V + C V )
1. C = 0 V = ——–———
1
1
2
2
d 1 (C + C )
–3
Cd (6.0n) (2.0 × 10 ) 1 2
A = = = 1.36 m 2
ε 8.85 × 10 –12
0 5
–12
ε A (8.85 × 10 )(0.04)
2. (a) C = 0 = –3 = 7.08 × 10 F 1. C: Same charge Q on all capacitors.
–11
d 5 × 10
Q
V Potential difference across capacitor C , V = —–

(b) (i) E = , 1 1 C
d 1 1 1
2. A: U = —CV , U’ = —(ε C)V = ε U
2
2
V = E d 2 2 r r
max max
–3
6
d = (3.0 × 10 )(5 × 10 ) 3. D 4. C 5. A 6. B 7. B 8. C
4
= 1.50 × 10 V
9. (a) 5 µC Q = CV
–6
(ii) Q = CV = 1.06 × 10 C
max max 1
–4
(b) 1.25 × 10 J U = –– CV 2
3. Final charge, Q = CV = (24 m)(6.0 V) 2
–4
= 144 mC (c) 2.50 × 10 J W = QV
Q (d) 1.25 × 10 J (iii) – (ii)
–4
Time taken, t = = 96 s
I 1
10. (a) 1.25 J U = –– CV 2
2
3 1 Q 2
(b) 0.25 J U = –– ——–—
2 (C + C )
1 2
1. C 2. B Energy dissipated as heat = 1.00 J
3. 1.42 × 10 F C = 2 × —–––)
(
ε ε A
r 0
–8
2
1
1
1
2
d 11. (a) (C + C ) V = C V + C V 2
(C V + C V )
1
1
2
2
4 V = ——–——–
(C + C )
1 2
–2
(b) 2.0 × 10 J
1. A
–7
2. D: Capacitor X: Q = CV 12. (a) 1.00 ×10 C Q = CV
Q
′
Capacitors X and Y in parallel. (b) (i) 1 000 V V = —
C
Same V, same Q. 1 1
–5
2
(ii) 4.0 × 10 J W = –– C V – –– CV 2
2 1 1 2
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Physics Term 2 STPM Chapter 13 Capacitors

ε ε A –t t
r 0
RC
13. (a) C = —––– V = V e , ln V = ln V – –––
d 0 0 RC
1
0.8
(b) C increases Gradient = – ––– = – ––– = – 0.02
RC 40
–8
14. (a) (i) 1.20 × 10 C Q = CV 1
1 C = –––––––––––––– F = 25.0 mF
3
(ii) 6.00 × 10 J U = ––QV (2.0 × 10 )(0.02)
–7
2
1
1
(b) (i) 70 pF C = 10 pF (air) 4. C: U = ––CV , U = ––(4.0C)V = 4U 0
2
2
0
1
1
13 C = 60 pF (dielectric) 2 –t 2 t 13
2
RC
C = C + C 5. D: V = V e , ln V = ln V – –––
0
0
1 2 RC
1 0.4
1 Q 2 Gradient = – ––– = ––––, time constant = RC = 50 s
(ii) 1.03 × 10 J U = –– — RC 20
–6
2 C 1
6. A: Effective capacitance: C = ––C, C = 0.4 C,
Work done on system against electrostatic A 4 B
forces. C = C, C = C
C
D
Q
15. (a) (i) (8 + 6) V = (8)(6.0) + (6)(12.0) 7. C: V= V + V = IR + ––
C
C

R
V = 8.6 V ε ε A
r 0
8. D: C = —–––
(ii) Energy loss d
9. B: Charge is conserved.
1 2 1 2 1 2
= [––C V + ––C V ] – ––(C + C )V (4.0 × 12.0 + 6.0 × 6.0) = (4.0 + 6.0)V
2 1 1 2 2 2 2 1 2

= 0.058 J V = 8.4 V
10. C: With dielectric, capacitance is increased.
(b) (i) (8 + 6)V = (6)(12.0) – (8)(6.0)
Q 2
V = 1.7 V 11. B: C = ε C , U = –––
1 r 0 2C
Q

(ii) Energy loss = 0.56 J 12. B: Charge is conserved, new V = –––—— , some
C + C
1 2
6 energy dissipated as heat.
13. A 14. A 15. B
1. D 2. B 3. C 4. C 5. D 16. C: Induced charges on opposite faces of dielectric
6. B 7. C 8. C 9. D 10. A sets up a reverse field that lowers the electric field
11. (a) Before ball strikes the bat: V constant = 6.0 V in the dielectric.
When ball strikes the bat: V decreases 17. D 18. A 19. C 20. B 21. C
exponentially. 22. C
1
1
1
When ball leaves the bat: V constant. 23. (a) ––– = ––– + ––– , C= 1.71 µF
(b) (i) 0.292 V = 6.0 V, V = 1.75 V C 3.0 4.0
0 e
– —– t (b) Same charge on both capacitors,
–3
(ii) 2.96 × 10 s, V = V e CR
0
12. (a) 6.0 V Q = CV = (1.71 µF)(12.0 V)

(b) V decreases exponentially: Capacitor = 20.5 µC

discharges through R. (c) With dielectric between the plates,
(c) C = 1 µF, R = 5 kΩ capacitance increases.
–3
τ = CR = 5.0 × 10 s is in the same order as time Equivalent capacitance increases.
taken for bullet to travel 10 cm. Charge stored in both capacitors increases.
( —— = 2.5 × 10 s ) 24. (a) (i) 50 µF and 100 µF capacitors are in series.
0.10
–3


Charge of 50 µF

40
(d) V = voltmeter reading after Y breaks = charge on 100 µF capacitor
-t = 700 µC
= (6.0)e . Calculate t.
CR
(ii) Equivalent capacitance of 200 µF capacitor
(e) Discharge of capacitor through voltmeter
and 150 µF capacitor (in parallel) is 350 µF
negligible.
Charge on 350 µF capacitor = (350 µF)V
STPM PRACTICE 13 = 700 µC
Hence V = 2.0 V
1. A: Charge is conserved,
Charge on 200 µF capacitor
Q = constant
= (200 µF)(2.0 V) = 400 µC
2. D: As V across the capacitor increases, V, across R 1 Q 1 1 1
decreases to zero. (b) — = V = — = (700 µC)(—– + —–– + —––)
C XY C 50 350 100
1
1
1
3. C: –– = ––– + –––, R = 2.0 kΩ = 23.0 V
R 3.0 6.0
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