PRA U STPM PHYSICS PENGGAL 2

CONTENTS

Chapter Chapter

12 • ELECTROSTATICS • • • • • • • • • • • • • • • • • • • • • •1• • 16 • MAGNETIC FIELDS • • • • • • • • • • • • • • • • • • •1• 5• •0•

•••••••••••••••••••• ••••••••••••• ••••••••

12.1 Coulomb’s Law 2 16.1 Concept of a Magnetic Field 151
12.2 Electric Field 6 152
12.3 Gauss’s Law 9 16.2 Force on a Moving Charge
12.4 Electric Potential 15 156
16.3 Force on a Current-Carrying 159
Conductor
167
STPM PRACTICE 12 32 16.4 Magnetic Fields due to Currents 172
185
QQ1 38 16.5 Force between Two Current-
Carrying Conductors
Chapter
16.6 Determination of the Ratio e
13 • C• •A• •P•A• •C•I•T•O• • R• •S• • • • • • • • • • • • • • • • • • • • • • • • • • •4• 3• • m
STPM PRACTICE 16
13.1 Capacitance 44
13.2 Parallel – Plate Capacitors 45 QQ5 191
13.3 Dielectrics 47
13.4 Capacitors in Series and in Parallel 49 Chapter
13.5 Energy Stored in a Charged
56 17 • ELECTROMAGNETIC INDUCTION • • •1• 9• •5•
Capacitor
13.6 Charging and Discharging of a 63 ••••••••••••••••••••••• ••••••••••••••

Capacitor 17.1 Magnetic Flux 196
17.2 Faraday’s Law and Lenz’s Law 197
STPM PRACTICE 13 71 17.3 Self-Induction 212
17.4 Energy Stored in an Inductor 215
QQ2 77 17.5 Mutual Induction 216

Chapter STPM PRACTICE 17 221

14 • •E•L• E• •C•T• •R•I•C• •C• U• •R• •R•E• N• •T• • • • • • • • • • • • • • • • • • • •8• 1• • QQ6 228

14.1 Conduction of Electricity Chapter

82 18 • ALTERNATING CURRENT CIRCUITS •2• 3• •2•

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

14.3 Current Density 85 18.1 Alternating Current rough a 233
Resistor
14.4 Electric Conductivity and Resistivity 87 239
18.2 Alternating Current rough an
STPM PRACTICE 14 96 Inductor 243
246
QQ3 99 18.3 Alternating Current rough
a Capacitor
Chapter
18.4 R-C and R-L Circuits in Series
15 • DIRECT CURRENT CIRCUITS • • • • • • • • •1•0• 1• •

••••••••• ••••••••••• •••••••••••

15.1 Internal Resistance 102 STPM PRACTICE 18 253
15.2 Kirchho ’s Laws 108
15.3 Potential Divider 119 QQ7 258
15.4 Potentiometer and Wheatstone
127 STPM Model Paper 262
Bridge Summary of Key Quantities and Units 273

STPM PRACTICE 15 140

QQ4 147

vi

CHAPTER CAPACITORS

13 13

Concept Map

Capacitors Parallel-Plate Capacitor Dielectric
C = ε0A
Capacitance d d
C = Q—V
+Q –Q C = εrε0A
d

V

Energy Stored in Charging a Capacitor
Capacitor
—t– —t– —t–
1 (a) V = E(1 – e– CR ) (b) Q = Q0(1 – e– CR ) (c) I = I0e– CR
2
U = CV 2 V Q I
E
= 1 QV Q0 I0
2

= 1 Q2
2 V
0 t0 t0 t

(a) V = V0e–—Ct–R Discharging a Capacitor

V (b) Q = Q0e–—Ct–R (c) I = I0e–—Ct–R
V0
Q I
Q0 I0

0 t0 t0 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

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
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 Capacitance

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

Students should be able to:
• define capacitance

1. Figure 13.3 shows what happens when a capacitor is _ X_ _ _ _ Capacitor
connected to a battery. Electrons from the battery charges Battery ++++
the plate X of the capacitor with a charge of –Q and a Y
charge of +Q is induced on the opposite plate Y. Hence, +
the charge on both plates of the capacitor is the same
magnitude but opposite in sign.

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.

Capacitance, C = —Po—Cteh—natr—igael—doi—nffee—riet—hnec—rep—blea—ttwe—eoef—nca—tpha—ecpi—tloa—rtes– –Q +Q
C = —QV
V
Figure 13.4

44

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.

Capacitance of conductor = —El—ect—Crihc—apr—ogte—eon—ntia—cloo—nfd—cuoc—ntod—ruc–t–o—r

C = —QV 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 Sketch a graph to show how the charge Q
in the capacitor varies with the potential
1. A capacitor stores 24 mC of charge when difference V across the capacitor.
the potential difference between the plates is
12 V. What is the capacitance of the capacitor? 3. The capacitor of variable capacitance is
connected to a 12 V supply. Its capacitance
2. A 50 µF capacitor is charged by connecting it is 50 µF. After the capacitor is charged,
to a 6.0 V battery. the voltage supply is disconnected and the
What is the charge in the capacitor when the capacitance is changed to 100 µF. What is the
potential difference across the capacitor is new potential difference across the capacitor?
(a) 3.0 V?
(b) 6.0 V?

13.2 Parallel-Plate Capacitors 2016/P2/Q5

Learning Outcomes

Students should be able to:
• describe the mechanism of charging a parallel-plate capacitor
Q ε0A
• use the formula C = V to derive C = d for the capacitance of a parallel-plate capacitor

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

ε0Φ = ΣQ

ε0(EA) = ΣQ
Q
E = ε0A ………‫ܨ‬

45

Physics Term 2 STPM Chapter 13 Capacitors

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

E= V ………‫ܩ‬
d

Equating ‫ ܨ‬and ‫ܩ‬, Q = V
ε0A d

Capacitance, C= Q = ε0A where ε0 = 8.85 × 10–12 F m–1 is the permittivity of free space between
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:

(a) Capacitance, C = —εd0A–

= (8.85 × 10–12)(0.15 × 0.20) F
(0.50 × 10–2)

= 5.31 × 10–11 F

(b) Capacitance, C ∝ 1 , when d’ = 2d,
d
1
C’ = 2 C

= 2.66 × 10–11 F

(c) Charge, Q’ = C’V
= (2.66 × 10–11)(12)
= 3.19 × 10–10 C

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?

2. The plates of a parallel-plate capacitor are (ii) What is the maximum charge that
separated by a distance of 5.00 mm. The area can be stored in the capacitor?
of each plate is 0.040 m2.
(a) Find the capacitance of the capacitor. 3. The capacitor used in the flash-light system
(b) Insulation of the air between the of a camera has a capacitance of 24 mF. If the
capacitor breaks down if the electric capacitor is charged by a constant current of
field between the plate exceeds 1.5 mA from a 6.0 V battery, what is the time
3.0 × 106 V m–1. taken to charge the capacitor to a potential
difference of 6.0 V?

46

Physics Term 2 STPM Chapter 13 Capacitors

13.3 Dielectrics

Learning Outcomes 2012/P1/Q25, 2018/P2/Q3

Students should be able to: 13
• define relative permittivity εr (dielectric constant)
• describe the effect of a dielectric in a parallel-plate capacitor
εrε0A
• use the formula C = 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 εr of an insulator is defined as

εr = capacitance of a parallel-plate capacitor with the insulator in between the plates
capacitance of the parallel-plate capacitor with free space in between the plates

He=ncCeC0, capacitance of a parallel-plate capacitor with an insulator of dielectric constant εr in between
the plates is
εrε0A
C = εrC0 = d

4. Typical values of dielectric constant εr are
Air: 1.0006, Paper: 3, Mica: 7, Paraffin wax: 2.5

5. Since the dielectric constant for air εr= 1.0006 = 1.00 (to 3 significant figures), parallel-plate capacitors
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. + – Original field
(a) The molecules of the insulator are polarised + –+ –+ –+ – Reverse field
by the electric field in between the plates.
(b) This results in the surface of the insulator + –+ –+ –+ – Polarized
facing the positive plate being negatively +– molecule
charged, and the other side being positively + –+ –+ –+ –
charged. +–
(c) A reverse electric field is set up. The + –+ –+ –+ –
resultant electric field between the plates +–
+ –+ –+ –+ –
+ –
+ –+ –+ –+ –
+ –
+ –+ –+ –+ –

+–

is reduced.
(d) Since E = —Vd , when E decreases, the potential
difference V between the plates decreases.
(e) If the capacitor is still connected to the Battery
Figure 13.6

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.

47

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

(a) reduces tt=hhee—QVpe,loewtcethnrietcinafliVedliddffeteocrre—eε1narcsieetsstoifno—iε1rtriatihtlsevianslaiutmiea,el value, Q on either plates, the capacitance C
(b) reduces charge
(c) Since C

13 increases by a factor εr.

Example 2

(a) A parallel-plate capacitor consists of two metal plates each of area 2.0 m2 separated by a distance
of 5.0 mm in air. A potential difference of 1.0 × 104 V is applied across the capacitor. Calculate
(i) the capacitance,
(ii) the charge on each plate,
(iii) the electric field strength between the plates.
(εr for air = 1.00)

(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:

(a) (i) Capacitance, C = —εd0A– Exam Tips
= —8.8—55—××—110—0–1–32—×—2
= 3.54 × 10–9 F 1. When a charged capacitor is disconnected from the
charging voltage, the charge Q remains constant.
(ii) Charge on each plate
Q = CV 2. If the capacitor remained connected to the
= (3.54 × 10–9) × (1 × 104) charging voltage, the potential difference V remains
= 3.54 × 10–5 C unchanged.

(iii) Electric field strength
E = —Vd
= —51 ×–×—1100—–43
= 2.0 × 106 V m–1

(b) (i) Electric field strength

E’ = —Eεr

= —2.—0 ×5—1—06–
= 4.0 × 105 V m–1

48

(ii) Potential difference Physics Term 2 STPM Chapter 13 Capacitors
V’ = —Vεr
= —1 –×5—10—4 13
= 2.0 × 103 V

(iii) Capacitance C’ = εrC
= 5 × 3.54 × 10–9
= 1.77 × 10–8 F

Quick Check 3 2. The electric field between the plates of an

1. A parallel-plate capacitor is charged in air. It isolated air-spaced parallel-plate capacitor
is then electrically isolated and lowered into
a liquid dielectric. Which of the following sets is E. Without disconnecting the charging
of changes is correct?
A Both the capacitance and the charges on voltage, an insulator of relative permittivity
the plates increase.
B Both the capacitance and the charges on 2 is inserted in between the plates. What is
the plates decrease.
C The capacitance increases and the the final electric field between the plates?
potential difference across the plates A –21–E
decreases. C ͱ⒓2E
D The capacitance decreases and the
potential difference across the plates BE D 2E
increases.
3. A rolled paper capacitor is made from strips
of metal foil of dimensions 2 cm × 40 cm
separated by paper of relative permittivity
2 and thickness 0.002 cm. Estimate its
capacitance.

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 C1, C2 and C3 Q 1 C1
connected in parallel to a battery.
Q2 C2
2. Charges flow into the capacitors until the potential difference
across each capacitor is equal to the charging voltage V. Hence Q3 C3
the potential differences across the capacitors are the same, V. V

3. The charge in each capacitor is given by Battery
Figure 13.7
Q1 = C1V, Q2 = C2V, Q3 = C3V
The total charge, Q = Q1 + Q2 + Q3

= C1 V + C2 V + C3V
= (C1 + C2 + C3)V

49

  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 = (C1 + C2 + C3)V Exam Tips
Equivalent capacitance, C = C1 + C2 + C3
For capacitors in parallel
13 This formula for the equivalent capacitance can be • same potential difference across all capacitors
extended for any number of capacitors connected • charge in each capacitor ∝ capacitance.
• equivalent capacitance, C = sum of capacitance
in parallel.

Capacitors in Series

1. Figure 13.8 shows the three capacitors of capacitance C1, C1 C2 C3
C2, and C3 connected in series to a charging voltage V. +Q –Q +Q –Q +Q –Q

2. Electrons flow from the negative terminal of the charging AB MN XY
voltage and charge plate Y of the capacitor C3 negative.
V1 V2 V3
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. Figure 13.8

Hence the charges on all the capacitors are the same,
i.e. Q.

4. The potential difference across each capacitor is given by V1 = —CQ1 , V2 = —CQ2 , V3 = —CQ3
The total potential difference, V = V1 + V2 + V3

= —CQ1 + —CQ2 + —CQ3

( )= Q —C11 + —C12 + —C13

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
V = —QC


C = equivalent capacitance

( ) Hence, —QC = Q —C11 + —C12 + —C13

—C1 = —C11 + —C12 + —C13 Exam Tips

When two capacitors C 1 and C 2 are connected in series,

• the charges in both capacitors, and in the equivalent capacitor

are equal.

( )• potential difference across C 1: V1 =
C2 V
C1 + C2

( )
potential difference across C 2: V 2 = C1 V
C1 + C2

50

Physics Term 2 STPM Chapter 13 Capacitors

Example 3

A capacitor made from two thin, flat metal sheets 13
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
separation between the interleaved sheets is filled with
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 To determine the number of capacitors in the
d = separation between sheets question: Count the number of spaces between
two plates
then C = —εrεd—0A–
The reconstructed capacitor consists of 7 capacitors connected in parallel. Four plates are to one

terminal, and another four to the other terminal.

Each capacitor is of area —A4 , hence of capacitance = —C4

Hence capacitance of 7 capacitors each of capacitance —C4 in parallel

( )C ’ = 7 × —4C = —47 C

Example 4

A parallel-plate capacitor consisting of two metal plates

separated by a distance x has capacitance C. Each metal plate x xx

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

–+ +– –+ +– +++++++++ 5 ––––––––– +Q + – –Q +4Q –4Q
–1+ +2– –3+ +4– + – +–
+– +–
–+ +– –+ +– +– +–
+–
–+ +– –+ +– + +– –
+–
+ 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)

51

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.
The area of each small plate = —A3 . Hence capacitance of each small capacitor = —3C .

( )13 —C3 = —43 C
The equivalent capacitance of 4 such capacitors in parallel = 4 ×
The effective capacitance of the arrangement C’ is given by

—C1’ = ——341—C + ——311—C

= —3 +4—C1—2
C’ = –1—45 C

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
—C1 = —21 + —14
= —43 2 µF 4 µF
C = —34 µF V1 V2

(b) Charges on both capacitors are equal
= charge on equivalent capacitor

( )= —34 µF × 12 V

= 16 µC

(c) Potential difference across 2 µF capacitor
V 1 = —CQ1
= —126—µµF—C
= 8V

Potential difference across 4 µF capacitor
V 2 = V – V1

= (12 – 8) V
= 4V

52

Physics Term 2 STPM Chapter 13 Capacitors

ANSWERS

1 3. D 4. A 5. C 6. D

1. C = Q = 24 mC = 2 mF 7. A: Q1 = ( C )V and Q2 = (2C)V
V 12 V 2
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

Q (d) (i) 0.8 µF (ii) 9.6 µC, 4.8 µC

9. (a) Two 2 µF capacitors in series with 2 × 10 V.
(b)

0 V 20 V

3. Q = (100)V1 = (50)(12) 10. (a) —20 µC
V1 = 6 V 3
2 (b) 44.4 V 2Q = (C1 + C2) V
11. (b)
(C1 + C2)V1 = C1V1 + C2V2

1. C = ε0A V1 = —(C—1V1–+—C—2V—2)
d (C1 + C2)
Cd (6.0n) (2.0 × 10–3)
A = ε0 = 8.85 × 10–12 = 1.36 m2 5

2. (a) C = ε0A = (8.85 × 10–12)(0.04) = 7.08 × 10–11 F 1. C: Same charge Q on all capacitors.
(b) d 5 × 10–3
—CQ1–
(i) E = Vd , Potential difference across capacitor C1, V1 =
A: U = —21 CV2, U’ = —21 (εrC)V2 = εrU
Vmax = Emax d 2.

d = (3.0 × 106)(5 × 10–3) 3. D 4. C 5. A 6. B 7. B 8. C

= 1.50 × 104 V 9. (a) 5 µC Q = CV
(ii) Qmax = CVmax = 1.06 × 10–6 C
(b) 1.25 × 10–4 J U = –21– CV2
3. Final charge, Q = CV = (24 m)(6.0 V) (c) 2.50 × 10–4 J W = QV
= 144 mC

Time taken, t = Q = 96 s (d) 1.25 × 10–4 J (iii) – (ii)
I
10. (a) 1.25 J U = –12– CV2
(b) 0.25 J U = –12– ——Q2–—
3 (C1 + C2)

1. C 2. B Energy dissipated as heat = 1.00 J

( )3. 1.42 × 10–8 F C = 2 × —εrεd–0–A– 11. (a) (C1 + C2) V = C1V1 + C2V2
(b)
4 V = (—C1—V1–+—C—2V–2)
(C1 + C2)

1. A 2.0 × 10–2 J
2. D: Capacitor X: Q = CV
12. (a) 1.00 ×10–7 C Q = CV
Capacitors X and Y in parallel. (b) (i) 1 000 V V′ = —QC
Same V, same Q.
(ii) 4.0 × 10–5 J W = –12– C 1V 2 – –21– CV2
1

78

Physics Term 2  STPM  Chapter 13 Capacitors 

13. (a) C = —εr εd–0–A– –t = ln V0 – R––tC–
(b) C increases –04–.08– = – 0.02
V = V0 eRC, ln V
14. (a) (i) 1.20 × 10–8 C Q = CV Gradient = – R––1C– = –
C = –(–2–.0––×–1–10–3–)–(0–.–0–2–) F = 25.0 mF
(ii) 6.00 × 10–7 J U = –12–QV
UVEGf0 =fr=eaVcd–t210–iieveCnRe–VCttc ,2=a,lpUn–a1R–Vc–1=iC–t=a–12=–nl(nc–4e–0.V2:0–.0C04C––A,)Vt=iR–m2–tC––=41e–Cc4oU,nC0sBta=n0t =.4RCC,
(b) (i) 70 pF C1 = 10 pF (air) 4. C: 13
C2 = 60 pF (dielectric)
C = C1 + C2 5. D:

(ii) 1.03 × 10–6 J U = –21–  Q—C2 = 50 s
Work done on system against electrostatic 6. A:

forces. CC = C, CD = C –QC–
7. C: V = VR + VC = IR
15. (a) (i) (8 + 6) V = (8)(6.0) + (6)(12.0) C = —εr ε–d0–A– +
8. D:
V = 8.6 V

(ii) Energy loss 9. B: Charge is conserved.

= [–12–C1V 12 + –12–C2V 22] – –12–(C1 + C2)V 2 (4.0 × 12.0 + 6.0 × 6.0) = (4.0 + 6.0)V
= 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)
B: C1 = εrC0, U = –2Q–C–2
V = 1.7 V 11. B: Charge is conserved, new –C––1 Q—+ C—2
12. energy dissipated as heat.
(ii) Energy loss = 0.56 J V = , some

6

1. D 2. B 3. C 4. C 5. D 13. A 14. A 15. B
16. C: Induced charges on opposite faces of dielectric
sets up a reverse field that lowers the electric field
6. B 7. C 8. C 9. D 10. A in the dielectric.
11. (a) Before ball strikes the bat: V constant = 6.0 V
When ball strikes the bat: V decreases 17. D 18. A 19. C 20. B 21. C

exponentially. 22. C
When ball leaves the bat: V constant. 23. (a) –C–1–e = –3–1.0– + –4–1.0– , C = 1.71 µF
(b) (i) 0.292 V0 = 6.0 V, V = 1.75 V (b) Same charge on both capacitors,

(ii) 2.96 × 10–3 s, V = V0 – —t–

12. (a) 6.0 V e CR Q = CV = (1.71 µF)(12.0 V)

(b) V decreases exponentially: Capacitor = 20.5 µC
(c) With dielectric between the plates,
discharges through R. capacitance increases.
(c) C = 1 µF, R = 5 kΩ
τ = CR = 5.0 × 10–3 s is in the same order as time Equivalent capacitance increases.
Charge stored in both capacitors increases.
taken for bullet to travel 10 cm.

( ) —0.1—0 24. (a) (i) 50 µF and 100 µF capacitors are in series.
40 = 2.5 × 10–3 s Charge of 50 µF

(d) V = voltmeter reading after Y breaks = charge on 100 µF capacitor
-t = 700 µC
(ii) Equivalent capacitance of 200 µF capacitor
= (6.0)eCR. Calculate t.

(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 (b) —C1 = VXY = —QC = (700 µC)(—510– + —351–0– + —110–0–)
decreases to zero. = 23.0 V

3. C: –R1– = –3–1.0– + –6–1.0– , R = 2.0 kΩ

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