SP025 Module Mutual - Perak Matriculation College

1

TABLE OF CONTENTS 1
15
1.0 ELECTROSTATICS 31
2.0 CAPACITOR AND DIELECTRICS 44
3.0 ELECTRIC CURRENT AND DIRECT CURRENT CIRCUITS 57
5.0 ELECTROMAGNETIC INDUCTION 74
6.0 ALTERNATING CURRENT 83
7.0 GEOMETRICAL OPTICS 93
8.0 PHYSICAL OPTICS 95
9.0 QUANTIZATION OF LIGHT 96
10.0 WAVE PROPERTIES OF PARTICLE
11.0 NUCLEAR AND PARTICLE PHYSICS

Siti Farahiyah & Nur ‘Ainiza
Electrostatic

CHAPTER 1.0 ELECTROSTATICS

1.1 Coulomb’s law
L/O 1.1 (c) :Apply Coulomb’s law for a system of point charge

= 1 2 where,
2 F = magnitude of electrostatic force

Q1Q2 = magnitude of charges
k = electrostatic (Coulomb) constant = 9.0x109 N m2 C-2
r = distance two point charges

Example 1:
Two point charges, Q1= 85 C and Q2= −50 C are separated by a distance of 3.5 cm as shown in Figure
below. Calculate the magnitude and direction of the electric force on Q1 due to Q2

Solution:

( )( )( )F = kQ1Q2 = 9.0 109 85 10−6 50 10−6
( )r 2 3.510−2 2

= 3.12 104 N

Direction : to the right (towards Q2)

Exercise 1-1: Exercise 1-2:
When the point charge Q =+8.4 µC and test charge
Two point charges, Q1= -65 C and Q2= −40 C are qo=+ 5.6 µC are brought near each other, each
experiences a repulsive force of magnitude 0.66 N.
separated by a distance of 2.0 cm. Calculate the Determine the distance between the charges. [0.8 m]

magnitude and direction of the electric force on Q1 due

to Q2 [5.85 x 104 N]

1

Siti Farahiyah & Nur ‘Ainiza
Electrostatic

L/O 1.1 (c) :Apply Coulomb’s law for a system of point charge

= 1 2 where,
2 F = magnitude of electrostatic force

Q1Q2 = magnitude of charges
k = electrostatic (Coulomb) constant = 9.0x109 N m2 C-2
r = distance two point charges

Example 2:
Two equal positive point charges q1 = q2 = 2.0 C are located at x = 0, y = 0.3 m and x = 0, y = –0.30 m
respectively. What is the magnitude and direction of the resultant electric force that these charges exert on a

third point charge Q = 4.0 C at x = 0.40 m, y = 0?

Solution:
Step 1: Draw the direction of forces

q1= 2.0 C F2

0.3 Q = 4.0 C

Ө = tan-1(0.3/0.4) = 36.87o

0.4

q2 = 2.0 C F
1

Step2: Calculate magnitude of forces
( )( )( )F1
= F2 = kQq1 = 9 109 2 10 −6 2 10 −6 = 0.288 N
r2
(0.5)2

Step 3: Resolve into x-component and y-component

Force x-component y-component

F1 0.288 cos 36.87 -0.288 sin 36.87

F2 0.288 cos 36.87 0.288 sin 36.87

0.46 N 0

Step 4: Calculate the resultant

( )FR = (Fx )2 + Fy 2 = 0.46 N to the right

Exercise 2-1:

Calculate the resultant force acting on q1 from Example 2. [0.36 N, 49.9o]
Step 1:

Step 2:

Step 3:

Step 4:

2

Siti Farahiyah & Nur ‘Ainiza
Electrostatic

1.2 Electric field
L/O 1.2a): Use electric field strength

= where:
E= Electric field strength/intensity
F= electrostatic force
qo= test charge

Example 3:

When a test charge q = 2 nC is placed at the origin, it experiences a force of 8.0 x 10 -4 N. Calculate the

magnitude of electric field at the origin.

Solution: = = 8×10−4 = 4 × 105 −1
2×10−9

Exercise 3-1: Exercise 3-2:

A test charge qo= 10 µC is placed at the origin, it A test charge qo= 15 µC is placed at the origin, it
experiences a force of 12.0 x 10 -4 N. Calculate the experiences an electric field strength of 12.0 x 10 4
magnitude of electric field at the origin. [120 NC-1]
NC-1. Calculate the magnitude of electrostatic force at

origin. [1.8 N]

3

Siti Farahiyah & Nur ‘Ainiza
Electrostatic

L/O1.2b) : Use

for a point charge where:
2 E= Electric field strength/intensity
= Q= point charge
r = distance two point charges
k = electrostatic (Coulomb) constant= 9.0x109 N m2 C-2

*The direction of the electric field strength, E depends on the sign of the point charge only.

L/O1.2b) : Use E = kQ for a point charge Where:
r2 E= Electric field strength/intensity

qo= test charge

Q= point charge

*The direction of the electric field strength, E depends on the sign of the point charge only.

Example 4:
Determine the magnitude and the direction of electric field at point B.

8 cm

Q1 = −30 n C B

Solution:
( ( )( ) )E
= kQ = 9 109 30 10−9
r2 8 10−2 2

= 4.22 104 N C−1(direction to the left)

Exercise 4-1: Exercise 4-2:
A test charge is placed near to the point charge Q. If
Determine the magnitude and the direction of electric the test charge experience the electric field strength
3.5 x 105 N C-1. Determine the distance between Q and
field at point R. [1.8 x 105 N C-1] qo [0.481 m]

5 cm

R Q2 =+5×10-8 C qo = -2x10-12 C Q =+9 µC

4

Siti Farahiyah & Nur ‘Ainiza
Electrostatic

L/O 1.2d) Determine electric field strength E for a system of charges
*Simple configuration of charges with a maximum of four charges in 2D

Example 5:
Two point charges, Q1= -3.0 µC and Q2= -5.0 µC, are placed 12 cm and 30 cm from the point P
respectively as shown in Figure below

Calculate the magnitude and direction of the electric field intensity at P,

Solution:
( )( )E
kQ E1P = 9.00 109 3.0  10 −6
r2 12 10−2
= = (towards Q1) 2 E1P = 1.88 106 N C −1
( )Direction
: to the left
( ( )( ) )E2P =
9.00 109 5.0 10−6 E2P = 5.00 105 N C −1
30 10−2 2
Direction : to the right (towards Q )
EP = −1.38 106 N C −1
EP = − E1P + E2P EP 2 − 1 .8 8  1 0 6 + 5.00 105

=

Direction : to the left (towards Q1)

Exercise 5-1: Exercise 5-2:

Determine the magnitude and the direction of electric The figure shows two situations in which charges are

field at point P. [9105 NC-1] placed on the x and y axes. They are all located at the

5 cm 5 cm same distance from the origin O. For the situations in
the drawing, the magnitude of electric field is

Q1 = −20×10-8 C P Q2 =+5×10-8 C 1.77107 N C-1 at the origin. Determine the distance
from the origin.

[ ]

Q1 =

Q4 = Q2 =

Q3 =

5

Siti Farahiyah & Nur ‘Ainiza
Electrostatic

1.3 Electric Potential where:
L/O 1.3 c) Use V= Electric potential
Q= point charge
r = distance two point charges
= k = electrostatic (Coulomb) constant= 9.0x109 N m2 C-2

for a point charge and system of charges. Maximum four charges in 2D.

Example 6:

A point charges, Q1= −40 C is placed distance of 15 cm from point A. Calculate the electric potential at

point A.

Q1= −40 C A

10 cm

( ( )( - ) )VASolution:

= kQ1 = 9 109 − 40 10−6
r1A 10 10−2

= −3.6 106 V

Exercise 6-1: Exercise 6-2:
The electric potential at point A is – 10.0 x106 V.
A point charges, Q2= 50 C is placed distance of 13

cm from point B . Calculate the electric potential at Calculate the distance between Q1 and point A.

point B. [3.46 x106 V] State the sign of charge Q1 [6.75x10-2m]

B Q1 A

B13 cm -75 µC

-Q2= 50 C

6

Siti Farahiyah & Nur ‘Ainiza
Electrostatic

L/O 1.3 c) Use where:
V= Electric potential
Q= point charge
= r = distance two point charges
k = electrostatic (Coulomb) constant= 9.0x109 N m2 C-2

for a point charge and system of charges. Maximum four charges in 2D.

Example 7:

Two point charges, Q1= −40 C and Q2= −30 C are separated by a distance of 15 cm as shown in Figure

Calculate the electric potential at point A B

B

13 cm

Q1= −40 C Q2= −30 C
A

5 cm 10 cm
- (A ) -VA
= VA1 + VA2 =  kQ1  +  kQ2  = 9 109  − 40 10−6 + − 30 10−6 
r1A r2A 5 10−2 10 10−2

= −9.89 106 V

Exercise 7-1:
Calculate the electric potential at point B from Example 7.

[ = − . × V]

7

Siti Farahiyah & Nur ‘Ainiza
Electrostatic

L/O 1.3(d) Calculate potential difference between two points: where:
V= potential difference
= final − initial Vfinal = final electric potential
Vinitial= intial electric potential

Example 8:

Two point charges, Q1= −40 C and Q2= −30 C are separated by a distance of 15 cm as shown in Figure

Calculate the potential difference between point A and B B

B13 cm

Q1= −40 C Q2= −30 C
A

5 cm 10 cm
- A( ) -VASolution:

= VA1 + VA2 =  kQ1  +  kQ2  = 9 10 −9  − 40 10 −6 + − 30 10−6 
r1A r2A 5 10 −2 10 10−2

= −9.89 106 V

( )VB  kQ1   kQ2  =  − 40 10−6 − 30 10−6 
= VB1 + VB2 = r1B + r2B 9 10 −9 19.85 10−2 + 10 10 −2

= −3.89 106 V

( )V = VB − VA = −9.89 106 − − 3.89 106 = −6 106 V

Exercise 8-1: Exercise 8-2:
Figure shows two point charges q1 and q2 placed 15 Given the potential difference between point A and
cm apart. If the charges q1 and q2 are +10 nC and point B is 1.09  107 V. The distances from point A
−10 nC respectively. Calculate the potential and point B are 20 cm and 15 cm to the point charge,
difference between point A and B. Q respectively. Calculate the value of charge Q.

[1350 V or -1350 V] [7.27  10-4 C]

A

15 cm 15 cm

+ 15 cm -
B 5 cm q1 q2

8

Siti Farahiyah & Nur ‘Ainiza
Electrostatic

L/O 1.3(d) Calculate potential difference between two points: where:
V= potential difference
W = work done
= 0 qo= test charge

Example 9:
A test charge q0=+1.2 C is at point A from a point charge Q. If a work done of −20 J is required in bringing
the test charge q0 to point B from the charge Q, calculate the potential difference between A and B.

Solution:


∆ =

− 20
∆ = 1.2 × 10−6

∆ = −1.67 × 107 V

Exercise 9-1: Exercise 9-2:

A test charge q0=+2.1 C is at point A from a point Calculate the work required to transfer a charge of
charge Q. If a work done of −30 J is required in
bringing the test charge q0 to point B from the charge 5µC against a potential difference of 120 V?
Q, calculate the potential difference between A and [6.0× − ]
B. [− . × ]

9

Siti Farahiyah & Nur ‘Ainiza
Electrostatic

L/O 1.3 f): Calculate potential energy of a system of point charges:

= ( 1 1 2 2 + 1 3 + 2 3 ) where:
13 23 V= Electric potential
q1, q2, q3, q4 = point charge
up to maximum three charges. r12, r13, r23 = distance two point charges
k = electrostatic (Coulomb) constant

= 9.0x109 N m2 C-2

Example 10:

Given three point charges, Q1= +2.0 C , Q2= −6.0 C and Q3= +3.0 C . Calculate the electric

potential energy for the system of charges.
Q2 -

-Q1 + 5.0 m

4.0 m + Q3

PThe total electric potential energy for the system of three charges is given by
+Solution:
U = k  Q1Q2 + Q1Q3 + Q2Q3 
r12 r13 r23

( )( ) ( )( ) ( )( )U = 9.00 109
2.0 10−6 − 6.0 10−6 + 2.0 10−6 + 3.0 10−6 + − 6.0 10−6 + 3.0 10−6
3.0 4.0 5.0

U = −5.49 10−2 J

Exercise 10-1: Exercise 10-2:

Given three point charges, Q1= -3.0 nC , Q2= +8.0 Three point charges of q1 = +2µC, q2 = -5µC and q3
nC and Q3= +2.0 nC . Calculate the electric
potential energy for the system of charges =-3µC are arranged as shown in figure below. They

[5.67x10-8J] are all located at the same distance of 2 m from the

Q2 + origin O. Calculate electric potential energy of this

system. [ . × − J]

-Q1 - 5.0 m + Q3 q2=
4.0 m q1= q3=
P
+

10

Siti Farahiyah & Nur ‘Ainiza
Electrostatic

1.4 Charge in a uniform electric field where:
L/O 1.4 b) Use E = electric potential
V= potential difference
d = distance between parallel plates
=

for uniform E in:
i. stationary charge

Example 11:
If the potential difference from point A to B is 60 V, calculate the electric field intensity of the uniform
electric field?

Solution:

E = V = 60 = 240 V m−1
d 0.25

Exercise 11-1: Exercise 11-2:
If the potential difference from point A to B is 50 V The electric field intensity of uniform electric field is
and the distance between two points is 20 cm, 150 V m-1 . Determine the distance between two
calculate the electric field intensity of the uniform points if the potential difference is 35 V.
electric field?
[0.233 m]
[250 V m-1]

11

Siti Farahiyah & Nur ‘Ainiza
Electrostatic

L/O 1.4 b) Use where:
E = electric potential
V= potential difference
= d = distance between parallel plates

for uniform E in:
ii. charge moving perpendicularly to the field

Example 12:

An electron enters a region of uniform electric field E as in figure below. If the acceleration of the electron is
3.51 × 1013 m s-2, calculate the magnitude of electric field.

Solution:
= = ⇒ =

=

=

= = 9.11 × 10−31(3.51 × 1013) = 200 N C-1
1.6 × 10−19

Exercise 12-1: Exercise 12-2:
A proton enters a region of uniform electric field E
An electron enters a region of uniform electric field between two parallel plates. If the acceleration of
proton and distance between plates are 1.05 ×
E as in figure below. If the acceleration of the 1010 m s-2 and 10 mm respectively, calculate the
electron is 3.86 × 1013 m s-2, calculate the potential difference of the electric field.

magnitude of electric field. [ -1] [ . ]

12

Siti Farahiyah & Nur ‘Ainiza
Electrostatic

L/O 1.4 b) Use where:
E = electric potential
V= potential difference
= d = distance between parallel plates

for uniform E in:
iii. charge moving parallel to the field

Example 13:
An electron moving parallel to a uniform electric field increases its speed from 2.0 × 107 m s-1 to 4.0 × 107 m
s-1 over a distance of 2.0 cm. Calculate the electric field strength.

Solution:
2 = 2 + 2
(4 × 107)2 = (2 × 107)2 + 2 (0.02)
= 3 × 1016 m s-2

= = 9.11 × 10−31(3 × 1016) = 171 kN C-1
1.6 × 10−19

Exercise 13-1: Exercise 13-2:

An electron moving parallel to a uniform electric An electron enters a uniform electric field between

field increases its speed from 2.2 × 107 m s-1 to 4.4 × two parallel plates with acceleration 2.5 × 1016 m s-2.

107 m s-1 over a distance of 2.3 cm. Calculate the If the potential difference is 200 V, calculate the

electric field strength. [180 kN C-1] distance between the two plates. [1.41 mm]

13

Siti Farahiyah & Nur ‘Ainiza
Electrostatic

L/O 1.4 b) Use where:
E = electric potential
V= potential difference
= d = distance between parallel plates

for uniform E in:
iv. charge in dynamic equilibrium

Example 14:
A 4 µC particle of mass 1.22 g is travelling horizontally at constant velocity 5 ms-1. It then enters an upward

uniform electric field. If the distance between plate is 2 cm, calculate the electric potential between two

plate.

Solution:

=

=

= = (1.22 × 10−3)(9.81) = 2.992 × 103 V
4 × 10−6

Exercise 14-1: Exercise 14-2:

A 5.4 µC particle of mass 2.11 g is travelling If the distance between plates in Exercise 14-1 is 2

horizontally at constant velocity 6 ms-1. Calculate the cm, calculate the electric potential between two

electric field strength when the particle then enters an plates. [76.6 V]

upward uniform electric field. [3.83103 V m-1]

14

Siti Farahiyah & Wan Idayu Farhana
Capacitor and Dielectrics

CHAPTER 2.0 : CAPACITOR AND DIELECTRICS
2.1 Capacitance and Capacitors in Series and Parallel

L/O 2.1a): Use capacitance, Where,
C= capacitance
C=Q Q= magnitude of charge on each plate
V V= potential difference across 2 plate

Example 15:
When the potential difference between the plates of a capacitor is increased by 3.25 V, the magnitude of the
charge on each plate increases by 13.5 C. Calculate the capacitance of this capacitor

Solution:

C = Q = 13.510−6 = 4.15 µF
V 3.25

Exercise 15-1 : Exercise 15–2 :
A capacitor stores 100 pC of charge when it is The capacitance of this capacitor 1000 F. The
connected across a potential difference of 20 V. magnitude of the charge on each plate increases by 80
Calculate the capacitance of the capacitor. C. Calculate the potential difference between the
plates of a capacitor.
[5 pF]
[0.08 V]

15

Siti Farahiyah & Wan Idayu Farhana
Capacitor and Dielectrics

L/O 2.1 b): Determine the effective capacitance of capacitor in series.

1 = 1 + 1 + 1 + .... + 1
Ceff C1 C2 C3 Cn

Example 16: Calculate the effective capacitance between A and B

Solution:

C1=15 F C2=3 F

A B 1 = 1 + 1 = 1 +1= 4
Ceff C1 C2 15 3 15

C eff = 15 μF
4

Exercise 16-1: = 2.5 µF
Calculate the effective capacitance between A and B Exercise 16-2:
The effective capacitance between A and B is 1.2 µF
C1=12 F C2=4 F Calculate the value of C1

AB

[3 µF] C1 C2=4 F
C3=2 F
A

B
[12 µF]

16

Siti Farahiyah & Wan Idayu Farhana
Capacitor and Dielectrics

L/O 2.1 b): Determine the effective capacitance of capacitor in series.

1 = 1 + 1 + 1 + .... + 1 Q = Q1 = Q2 = Q3 V = V1 +V2 +V3
Ceff C1 C2 C3 Cn

Example 17 : Calculate the charge on each capacitor if the circuit is connected to 24 V.

C1=15 F C2=3 F
A
Solution: B

= 15

4 15
4
= = (24) = 60

1 = 2 = = 60

Exercise 17-1: Exercise 17-2:
Calculate the charge on each capacitor if the circuit Calculate the charge on each capacitor if the circuit
is connected to 12 V. is connected to 6 V.

[36 µC] [72 µC]

C1=12 F C2=4 F C1 C2=4 F
A A
B
C3=2 F

B

17

Siti Farahiyah & Wan Idayu Farhana
Capacitor and Dielectrics

2.1 b): Determine the effective capacitance of capacitor in parallel

Ce ff = C1 + C2 + C3 + ...Cn

Example 18: Calculate the effective capacitance between A and B.

C1= 15 F

A B

Solution: C2=6 F

Ce ff = C1 + C2
= 15 + 6
= 21μ F

Exercise 18-1: Exercise 18-2:
Calculate the effective capacitance between A and B FIGURE shows an arrangement of capacitors. If the
equivalent capacitance is 1 F, calculate the value of
C1= 5 F C.

C

AB

C2= 12 F

[17 µF] 0.6 F

[0.4 µF]

18

Siti Farahiyah & Wan Idayu Farhana
Capacitor and Dielectrics

2.1 b): Determine the effective capacitance of capacitor in parallel

V = V1 = V2 = V3 Ce ff = C1 + C2 + C3 + ...Cn Q = Q1 + Q2 + Q3

Example 19: Calculate the charge on each capacitor if the circuit is connected to 24 V.

C1= 15 F

A B

Solution: C2=6 F

V = V1 = V2 = 24V

Q1 = C1V = (15)(24) = 3.6 10−4 C = 360 μC

Q1 = C2V = (6)(24) = 144 μC

Exercise 19-1: Exercise 19-2:
Calculate the charge on each capacitor if the circuit Calculate the charge on each capacitor if the circuit
is connected to 15 V between A and B is connected to 9 V between A and B

C1= 5 F C1=0.4 µF

AB C2=0.6 F
C2= 12 F [ 3.6 µC, 5.4 µC]

[ 75 µC, 180 µC]

19

Siti Farahiyah & Wan Idayu Farhana
Capacitor and Dielectrics

2.1 b): Use energy stored in a capacitor. Where,
C= capacitance
U = 1 CV 2 U = energy stored
2 V= potential difference across 2 plate

Example 20:
Calculate the energy stored in 2000 mF capacitor charged to a potential difference of 10 V.

Solution:

U = 1C V2
2

( )= 1 2000 10−3 (10)2
2
= 100 J

Exercise 20-1: Exercise 20- 2:

A battery of emf 9 V is connected to two capacitors If the energy stored 6 mF capacitor is 6 J, calculate
as in figure below. Calculate the total energy stored the voltage across its plates.
in the capacitors.
[ 44.72 J ]
C1= 5 F

AB
C2= 10 F
[ 6.0810−4 J ]

20

Siti Farahiyah & Wan Idayu Farhana
Capacitor and Dielectrics

2.1 b): Use energy stored in a capacitor. Where,
U = energy stored
U = 1 QV Q= magnitude of charge on each plate
2 V= potential difference across 2 plate

Example 21:

Calculate the energy stored if the charge in the capacitor is 4 C and connected to 9 V battery.

Solution:

U = 1 QV
2

= 1 (4)(9)

2
= 18 J

Exercise 21-1: Exercise 21-2:

Calculate the energy stored if the charge in the If the charge in a capacitor is 4 C and the energy

capacitor is 1.2 C and connected to 12 V battery. stored in it is 4 J, calculate the voltage across its

plates.

[ 7.210−6 J ] [2 V]

21

Siti Farahiyah & Wan Idayu Farhana
Capacitor and Dielectrics

2.1 b): Use energy stored in a capacitor.

U == 1 Q2 Where,
2C C= capacitance
Q= magnitude of charge on each plate
U = energy stored

Example 22:
If the charge stored in a capacitor is 4 C and the value of capacitance is 2 F, calculate the energy stored in it.

Solution:

U = 1 Q2
2C

= 1 (4)2
2 (2)

=4 J

Exercise 22-1: Exercise 22- 2:

Calculate the energy stored if the charge in the If the charge in a capacitor is 4 C and the energy
capacitor is 1.2 µC and the capacitance in the stored in it is 4 J, find the value of capacitance.
capacitor is 19.5 pF.
[2 F]
[ 3.7 10−2 J ]

22

Siti Farahiyah & Wan Idayu Farhana
Capacitor and Dielectrics

2.2 Charging and Discharging of Capacitors Where,
L/O 2.2a) Use τ = time constant

τ = RC R = resistance

C = Capacitance

Example 23:
A 40 µF capacitor has a charge of 250 µC is connect to a 30 kΩ resistor. Calculate the time constant.

Solution:

τ = RC

( )( )= 30 103 40 10−6

= 1.2 s

Exercise 23-1: Exercise 23-2 :
A 20 µF capacitor has a charge of 240 µC is connect The time constant of RC circuit is 100 s. The
to a 50 kΩ resistor. Calculate the time constant. capacitor has a charge is connecting to a 50 kΩ
resistor. Calculate the capacitance.
[1 s]
[2 mF]

23

Siti Farahiyah & Wan Idayu Farhana
Capacitor and Dielectrics

L/O 2.2 c) Use Where:
Charge on discharging capacitor
Q0 : maximum charge
−t R : resistance of the resistor

Q = Q0e RC C : capacitance of the capacitor

Example 24: The 1000 F capacitor has a maximum charge 6.0 mC charge on each plate. The capacitor is
then

discharged through a 100 k resistor. Calculate the charge stored in capacitor at t = 2 s

Solution:

−t

Q = Q0e RC

( )= −2
6  10 −3
e ( )( )100103 100010−6

= 5.88 10 −3 C

= 5.58 mC

Exercise 24-1: Exercise 24-2:
The 100 pF capacitor has a maximum charge 4.0 mC A fully charged 12.0 F capacitor has 6.0 mC charge
charge on each plate. The capacitor is then discharged on each plate. The capacitor is then discharged
through a 100  resistor. Calculate the charge stored through a 100  resistor. Calculate the time taken if
in capacitor at t = 0 s just 20% of charge is left in the capacitor.

[4 mC] [1.93 ms]

24

Siti Farahiyah & Wan Idayu Farhana
Capacitor and Dielectrics

L/O 2.2 c) Use where

Charge on charging capacitor where

Q = Q0 1 − − t  Q0 : maximum charge
RC R : resistance of the resistor
e
C : capacitance of the capacitor

Example 25 : A 20 µF capacitor is connected to a 50 kΩ resistor. If the maximum charge id 240 µC, calculate
the charge induced in capacitor at time, t = 0.1 s

Solution:

Q = Q0 1 − − t 
RC
e

−6 1 − (50103 0.1 ) 

= 240  10 − e )(2010−6

= 2.28 10 −5 C

Exercise 25-1: Exercise 25-2:
50 µF are connected in series with a 15 kΩ resistor. If

the maximum charge is 360 µC, calculate the charge

induced in capacitor at time, t = 0.5 s

[1.7510−4 C ]

Figure shows a circuit which is used to charge a
capacitor of 50 µF which is initially without charge.
Calculate the charge stored in capacitor at t = 1 s.

[1.4810−5 C ]

25

Siti Farahiyah & Wan Idayu Farhana
Capacitor and Dielectrics

2.3 Capacitors With Dielectrics
L/O 2.3 c): Calculate capacitance on air filled parallel plate capacitor,

Co = oA Where,
d ε0 : permittivi ty of free space
(0 = 8.85 10−12 C2 N−1 m−2 )
d :distance between the two plates

A : area of each plate

Example 26:
A parallel-plate capacitor has plates of area 280 cm2 are separated by a distance 0.550 mm. The plates are in
vacuum. If a potential difference of 20.1 V is supplied to the capacitor, calculate the capacitance of the
capacitor.

Solution:

( )( )C = ε0 A =
d
8.85 10−12 280 10−4 = 4.5110−10 F
0.550 10−3

Exercise 26-1: Exercise 26-2:
A circular parallel-plate capacitor with radius of 1.2
A parallel-plate capacitor has plates of area 350 mm2 cm is connected to a 6.0 V battery. If the separation
between plates is 2.5 mm and the medium between
are separated by a distance 0.2 cm. The plates are in plates is air. Calculate the capacitance of the capacitor.

vacuum. If a potential difference of 12 V is supplied [1.6 pF]

to the capacitor, calculate the capacitance of the

capacitor. [1.55 pF]

26

Siti Farahiyah & Wan Idayu Farhana
Capacitor and Dielectrics

L/o 2.3 d) Use dielectric constant (relative permittivity) of the material.

r =  Where:
o ε0 : permittivi ty of free space
= 8.85 x 10−12 F −1
ε : permittivi ty of dielectric material

(the ability of a substance to hold an electrical charge)

Example 27:

The capacitance of an empty capacitor is increased when the space between its plates is completely filled by a
PVC with dielectric constant of 3. Calculate the permittivity of dielectric PVC.

Solution:

=

3 =
8.85 10−12

= 2.66 10−11 −1

Exercise 27-1: Exercise 27- 2:
A capacitor with air between its plates is charged to A parallel plate capacitor filled with benzene has a
150 V and then disconnected from the battery. If a capacitance of 500 pF. If the permittivity of benzene
piece of paper with dielectric constant of 2.0 is placed is 2.02 10−11 −1, calculate the dielectric
between the plates, calculate the permittivity of constant.
dielectric of paper
[2.28]
[ . − − ]

27

Siti Farahiyah & Wan Idayu Farhana
Capacitor and Dielectrics

L/o 2.3 e) Use capacitance with dielectric,

C =  rCo Where:

r = Dielectric constant

C0.= capacitance of the same capacitor with plates in a vacuum,

Example 28:

Paper has a dielectric constant of 3.70. If capacitor originally has a capacitance of 0.04 F. Calculate the
capacitance after paper, a dielectric material, is inserted to completely fill the space between the plates of
capacitor?

Solution:

C =  rCo = 3.70(0.04) = 0.148 F

Exercise 28-1: Exercise-28 2:
A capacitance of capacitor is initially 12 µF. The
Teflon with dielectric constant 2.1 is completely fills A parallel plate capacitor filled with benzene ( r
the volume between the plates. Calculate the
capacitance of capacitor after inserted the dielectric. =2.28) has a capacitance of 500 pF. Calculate the
initial capacitance of capacitor before inserting the
[ 25.2 μF ] dielectric.

[ 219 pF ]

28

Siti Farahiyah & Wan Idayu Farhana
Capacitor and Dielectrics

Example 29:
The capacitance of an empty capacitor is 1.2 F. The capacitor is connected to a 12 V battery and
charged up. With the capacitor still connected to the battery, a slab of dielectric material is inserted between
the plates. As a result, the charges flow from one plate through the battery and on to the
other plate is increase to 40.4 µC What is the dielectric constant of the material?

Solution:

Initial charge, Qo = CoV = (1.2  10-6)(12) = 1.44  10-5 C

r = C  Q  =Q = 4.04 10 −5 = 2.8
Co = V  Qo 1.44 10 −5

 Qo 
V 

Exercise 29-1: Exercise 29-2:
The capacitance of an empty capacitor is 2 F. The The capacitance of an empty capacitor is 28 pF. It is
capacitor is connected to a 15 V battery and charged connected to a 12 V battery. While the battery remains
up. With the capacitor still connected to the battery, a connected, a sheet paper is inserted and completely
slab of dielectric material is inserted between the fills the space between the plates. If the dielectric
plates. As a result, the charges flow from one plate constant is 3.7, calculate the charge storage in the
through the battery and on to the other plate is capacitor because of the dielectric insertion.
increase to 50 µC What is the dielectric constant of
the material? [1.24 nC]

[1.7]

29

Siti Farahiyah & Wan Idayu Farhana
Capacitor and Dielectrics

Example 30:
A capacitor with air between its plates is charged to 100 V and then disconnected from the battery. When a
piece of glass is placed between the plates, the voltage across the capacitor drops to 25 V. What is the dielectric
constant of this glass?

Solution:

C  Q  = Vo = 100
Co V  V 25
r = = = 4.0
 Q 
Vo

Exercise 30-1: Exercise 30-2:
A capacitor with air between its plates is charged to
A capacitor with air between its plates is charged to 150 V and then disconnected from the battery. When
a piece of paper with dielectric constant of 2.0 is
50 V and then disconnected from the battery. When a placed between the plates, the voltage across the
capacitor drops. Calculate the voltage drops.
piece of paper is placed between the plates, the voltage
[75 V]
across the capacitor drops to 20 V. What is the

dielectric constant of this glass?

[2.5]

30

Mohd Khairul Azmi & Noraida
Electric Current and Direct-Current Circuits

CHAPTER 3: ELECTRIC CURRENT AND DIRECT-CURRENT CIRCUITS
3.1 Electrical Conduction
L/O 3.1 (c) : Use electric current

I = dQ , Q = ne where,
dt I = magnitude of current

dQ = net charge that moving passes
through an area
dt = time interval
n = number of electron
e = magnitude of moving charges
(electron)

Example 31:
In an electrical circuit, 5.0 x 1014 electrons moving through a cross section of a wire in 0.04s. Calculate the

magnitude of the current.

Solution:

Q = ne I = dQ
dt
( )( )= 5 x1014 1.6 x10 −19
= 8 x 10 −5
= 8 x10 −5 C 0.04

= 2 mA

Exercise 31-1: Exercise 31-2:
In an electrical circuit, 1 x 1015 electrons moving A current of 5 A flows in the circuit for 3 seconds.
Calculate the total charges flowing during this time.
through a cross section of a wire in 0.032s. Calculate
[15 C]
the magnitude of the current.
[5 mA]

31

Mohd Khairul Azmi & Noraida
Electric Current and Direct-Current Circuits

3.2 Ohm’s law and Resistivity where,
L/O 3.2 (a) : Use Ohm’s Law I = magnitude of current

V = IR V = potential difference or voltage drop
R = electrical resistance

Example 32: A light bulb uses a single 3 V battery to provide a current of 0.4 A in the filament. Calculate the
resistance of the glowing filament.

Solution:

R = V = 3 = 7.5
I 0.4

Exercise 32-1: Exercise 32-2:
A potential difference of 15 V is applied to a 12-Ω
A light bulb uses two 1.5 V batteries to provide a
resistor. Calculate the current flowing through the
current of 0.3 A in the filament. Calculate the
resistor.
resistance of the glowing filament.
[1.25 A]
[10 Ω]

32

L/O 3.2 (b) : Use resistivity Mohd Khairul Azmi & Noraida
Electric Current and Direct-Current Circuits
 = RA
where,
l ρ = resistivity
R = electrical resistance
A = cross sectional area of conductor
l = length of conductor

Example 33:

A copper extension cord has diameter of 0.912 mm and length of 30 m. Calculate the resistance of the copper

extension at 20 ºC.
(ρcopper = 1.67 x 10-8 Ω m)

Solution:

( )A = d 2 =  0.912 x10−3 2 = 6.53 x10−7 m2
44
( )→
 = RA R= l = 1.67 x10−8 (30) = 0.77 
l A 6.53 x10
−7

Exercise 33-1: Exercise 33-2:
Determine the resistivity of a metal wire if the
A tungsten light bulb filament extension cord has resistance is 6 Ω, uniform cross-sectional area of the

diameter of 0.02 mm and length of 4 cm. Calculate wire is 8 mm2and length is 0.5 m.
[9.60 x 10-5 Ω m]
the resistance of the tungsten light bulb filament

extension cord at 20 ºC.
(ρtungsten = 5.4 x 10-8 Ω m)

[6.88 Ω]

33

Mohd Khairul Azmi & Noraida
Electric Current and Direct-Current Circuits

3.3 Variation of resistance with temperature where,

L/O 3.3 (b) : Use R = R 1+  (T − T ) R = electrical resistance at temperature T
(higher than To)
R = R 1+  (T − T ) Ro = electrical resistance at temperature
To (20 ºC)
α = temperature coefficient of resistivity
A = cross sectional area of conductor
l = length of conductor

Example 34:
A copper extension cord has resistance of 10 Ω at 20 ºC. Calculate the resistance at 120 ºC.
(αcopper = 4.05 x 10-3 ºC-1)

Solution:

R = R 1+ (T − T )

 ( )=10 1+ 4.05 x10-3 (120 − 20)

= 14.05 

Exercise 34-1: Exercise 34-2:

A nichrome heating element of a toaster has a At temperature of 20 ºC, a platinum resistance
resistance of 12 Ω at 20 ºC. Calculate the resistance thermometer has a resistance of 225 Ω. When the

of the element at 1200 ºC. thermometer is placed in a furnace, its resistance
(αnichrome = 0.4 x 10-3 ºC-1) rises to 448 Ω. Calculate the temperature of the

[17.66 Ω] furnace.
(αplatinum = 3.64 x 10-3 ºC-1)

[292.28 ºC]

34

Mohd Khairul Azmi & Noraida
Electric Current and Direct-Current Circuits

L/O 3.4 (d) : Use terminal voltage where,

V =  − Ir V = terminal voltage
ε = temperature coefficient of resistivity
I = magnitude of current
r = internal resistance

Example 35:
Calculate the terminal voltage when the 10 A current is drawn from a 12 V battery with 0.010 Ω internal
resistance.

Solution:

V =  − Ir
=12 − (10)(0.01)
=11.9 V

Exercise 35-1: Exercise 35-2:

The current supplied by an alkaline D-cell (1.5 V A resistor is connected in series with a dry cell of
emf, 0.100 Ω internal resistance) in a clock is 50.0
emf 1.5 V. When a current of 3.0 A flows from the
mA. Calculate the terminal voltage of the battery
cell, the voltage across the cell is 0.42 V. Calculate
[1.50 V]
the internal resistance of the dry cell.

[0.36 Ω]

35

Mohd Khairul Azmi & Noraida
Electric Current and Direct-Current Circuits

L/O 3.5 (a) : Determine effective resistance of resistors in series and parallel R1
R2
Resistors in Series: Resistors in Parallel: R3
R1
R2 R3 R3 = 1 Ω
B
Reff = R1 + R2 + R3 1 =1+1+1
Reff R1 R2 R3 R3 = 1 Ω

Example 36: R1 = 2 Ω
Calculate the equivalent resistance between point A and B (R123).
R2 = 2 Ω
Solution: A R12 = 1 Ω

1 = 1 + 1 = 1 + 1 =1 
R12 R1 R2 2 2
R12 = 1 
R123 = R12 + R3 = 1 + 1 = 2 

Exercise 36-1: Exercise 36-2:

Calculate the equivalent resistance between point A Calculate the equivalent resistance between point A

and B (R123). and B.
[4 Ω]
[7.5 Ω]

R1 = 5 Ω R3 = 5 Ω R1 = 6 Ω R3 = 6 Ω
A
AB B
R2 = 6 Ω
R2 = 5 Ω

36

Mohd Khairul Azmi & Noraida
Electric Current and Direct-Current Circuits

L/O 3.6 (b) : Use Kirchhoff’s Rules Kirchhoff’s Voltage Rule
Kirchhoff’s Current Rule At any loop,
At any junction,
 =  IR
 Iin = I out

Voltage arrow:

ε

Voltage arrow and loop, same direction → ‘+’

Example 37:
Generate three equations of unknown currents using Kirchhoff’s Rules.

Solution: I1 A I3

I1 + I3 = I2 (KCLat Junction A) 4Ω I2 2 3Ω
1.5 = 4I1 + 6I2 (KVL Loop1) 1 6Ω
3 = − 6I2 − 3I3 (KVL Loop2)

1.5 V 3V

Exercise 37-1: Exercise 37-2:
Calculate the current in each branch of the circuit.
Generate three equations of unknown currents using
[I1 = - 0.20 A, I2 = 0.24 A, I3 = -0.05 A]
Kirchhoff’s Rules. (KCLat Junction A)]
(KVL Loop1)
[ I3 = I1 + I2 25 V 122 Ω

5 = 22I1 − 56I2

−1 =56I2 − 75I3 (KVL Loop2)

I1 A 2 I3 I1 1
22 Ω I2 75 Ω I2 5.6 Ω
1 56 Ω
A 2
75 Ω
I3

5V 1V 5V

37

Mohd Khairul Azmi & Noraida
Electric Current and Direct-Current Circuits

L/O 3.7 (a) : Use power where,

P = IV P = power (known as power loss)
I = current
V = potential difference or emf

Example 38:
Calculate the power dissipated by a small lamp if the emf is 2.00 V and the current flowing in the lamp is 2A.

Solution:

P = IV

= (2)(2)

=4W

Exercise 38-1: Exercise 38-2:
Calculate the power of a chandelier if the emf is 120 Calculate the current in a 60 W kettle when
V and the current flowing in the chandelier is 5 A.
connected to a 120 V emf.
[600 W]
[0.5 A]

38

Mohd Khairul Azmi & Noraida
Electric Current and Direct-Current Circuits

L/O 3.7 (a) : Use power where,

P = I2R P = power (known as power loss)
I = current
R = resistance

Example 39:
Calculate the power dissipated by the resistor in a small radio if the resistor is 6 Ω and the current flowing is

1A.

Solution:

P= I2R

= (1)2 (6)

=6W

Exercise 39-1: Exercise 39-2:

Calculate the power dissipated by the resistor in a Calculate the current flowing in an electric heater with
printer if the resistor is 0.15 Ω and the current power 1.8 kW and resistance 8 Ω.

flowing is 10A. [15 A]

[15 W]

39

Mohd Khairul Azmi & Noraida
Electric Current and Direct-Current Circuits

L/O 3.7 (a) : Use power

P = V2 where,
R
P = power (known as power loss)
V = potential difference or emf
R = resistance

Example 40:
Calculate the power dissipated by a 5 Ω resistor in a circuit if the emf is 2.00 V.

Solution:

P= V2
R

= 22
5

= 0.8 W

Exercise 40-1: Exercise 40-2:
A bulb is stamped 40 W/120 V. Calculate the
The heating element in an iron has a resistance of 24
Ω. The iron is plugged into a 120-V outlet. Calculate resistance when lighted by a 120-V source.
[360 Ω]
the power delivered to the iron.

[600 W]

40

L/O 3.7 (b) : Use electrical energy Mohd Khairul Azmi & Noraida
Electric Current and Direct-Current Circuits
W = IVt
where,
W = electrical energy
I = current
V = potential difference or emf
t = time taken

Example 41:
An electric motor takes 5 A from a 120 V line. Calculate the electrical energy applied to the motor in 10 s.

Solution:

W = IVt

= (5)(120)(10)

= 6 kJ

Exercise 41-1: Exercise 41-2:
An electric motor takes 0.4 A from a 6 V emf The electrical energy applied to a motor in 30 s is
sources. Calculate the electrical energy applied to the 60 J. The electric motor is connected to a 9 V emf
motor in 20 s. sources. Calculate the current flowing in the electric
motor.
[48 J]
[0.22A]

41

Mohd Khairul Azmi & Noraida
Electric Current and Direct-Current Circuits

L/O 3.8 (b) : Use equation of potential divider ε

V1 = ( R1 + R1 ...Rn )V
R2 +

where, R1 R2 Rn
V1 [12 V]
V1 = potential difference across resistor 1
9V
R1 = resistance of resistor 1
R1
R2 = resistance of resistor 2 1Ω 8Ω
Rn = resistance of resistor n V1
ε = emf

Example 42:
Calculate the potential difference across resistor 1.

Solution:

V1 =  R1 R1 
+ R2

=  1 1 8 9
 + 

=1V

Exercise 42-1: Exercise 42-2:

Calculate the potential difference across resistor 2. Calculate the unknown emf.

[5 V]

6V ε

R2 R1 R2
1Ω 5Ω 2Ω
1Ω
V2 V1
4V

42

Mohd Khairul Azmi & Noraida
Electric Current and Direct-Current Circuits

L/O 3.9 (b) : Use related equations for potentiometer ε1

1 = l1 l2 l1
 2 l2 ε2 G

where,
ε1 = emf of cell 1
ε2 = emf of cell 2
l1 = length of a uniform wire
l2 = balanced length on the wire

Example 43:
A potentiometer consists of a uniform wire of length 100 cm and resistance 5 Ω is connected to a cell which

has an emf, ε1 = 6 V and negligible internal resistance. When a cell of unknown emf, ε2 is connected to the
potentiometer, the balanced length is 40 cm. Calculate emf of the cell, ε2.

Solution: ε1

1 = l1 l1
 2 l2 l2
ε2 G
2 =  l2 1
l1

=  40 6
100 

= 2.4 V

Exercise 43-1: Exercise 43-2:

A potentiometer consists of a uniform wire of length A potentiometer consists of a uniform wire of length
100 cm and resistance 5 Ω is connected to a cell 100 cm and resistance 5 Ω is connected to a cell of
which has an emf, ε1 = 4 V and negligible internal
resistance. When a cell of unknown emf, ε2 is unknown emf and negligible internal resistance.
When a cell of emf, ε2 = 2.4 V is connected to the
connected to the potentiometer, the balanced length
is 75 cm. Calculate emf of the cell, ε2. potentiometer, the balanced length is 20 cm. [12 V]
Calculate emf of the driver cell, ε1.
[3 V]
ε1
ε1

l1 l1
l2 l2

ε2 G ε2 G

43

Ummi Atiah & Siti Farahiyah
Electromagnetic Induction

CHAPTER 5.0 ELECTROMAGNETIC INDUCTION
5.1 Magnetic flux
L/O 5.1 (a) Use Magnetic flux,

 = B.Acos where,

B = magnitude field
A = surface area
θ = angle between normal A and B

Example 44:
A magnetic field of strength 0.3 T is directed perpendicular to a plane circular loop of area 0.196 cm2.
Calculate the magnetic flux through the area enclosed by this loop.

Solution:

 = 0
 = BAcos

= (0.3)(0.196 10−4 )(cos0)
= 5.88 10−6 Wb

Exercise 44-1
A magnetic field of strength 0.5 T is directed perpendicular to a plane circular loop of area 2 cm2. Calculate
the magnetic flux through the area enclosed by this loop.

[110−4 Wb ]

Exercise 44-2 Exercise 44-3
A rectangular loop of area 0.126 m2 is placed in a A magnetic field is directed 30o to a plane circular
loop. The magnetic flux produced in the circular loop
uniform magnetic field of magnitude 0.85 T. If the of radius 25 cm is 5.910−2 Tm2 . Calculate the
strength of the magnetic field.
magnetic flux through the loop is 9.310−2 Tm2 ,
[0.60T]
calculate the angle between the normal to the loop and

the magnetic field. [29.7⁰]

44

Ummi Atiah & Siti Farahiyah
Electromagnetic Induction

L/O 5.1 (b): Use Magnetic flux linkage, where,
N = number of turns
 = N = NBAcos
 = magnetic flux

Example 45:
Calculate the flux linkage in a coil of 15 turns and area 25 cm2 in a field of strength 5 T perpendicular to the

plane.

Solution:
Φ = NBA

= (15) (5) (25 x 10-4 )

= 0.1875 Wb

Exercise 45-1: Exercise 45-2:
Calculate the flux linkage in a coil of 20 turns and area A circular loop is placed in a uniform magnetic field
4 cm2 in a field of strength 8 T perpendicular to the perpendicularly. If the flux linkage through the loop is
plane.
9.3 10−2 T m2 and the flux for each loop is
[0.64 Wb] 4.65 10−4 Wb . Calculate number of turns of the

loop.
[200]

45

Ummi Atiah & Siti Farahiyah
Electromagnetic Induction

5.2 Induced emf where,
L/O 5.2 (b) Use Faraday’s Law N = number of turns

ε = −N d d

dt = rate of change of magnetic flux

dt

Example 46:
A 50-turn coil has a rate of change of magnetic flux of 0.5 Wb s-1 through a uniform magnetic field. Calculate

the emf induced in the coil?

Solution:

ε = −N d
dt

= −(50)(0.5)

= −25 V

Exercise 46-1: Exercise 46-2:
A 60-turn coil has a rate of change of magnetic flux An 150V emf is produced by a coil of 100 turns that
of 0.4 Wb s-1 through a uniform magnetic field. passes through a uniform electric field. Calculate the
Calculate the emf induced in the coil. rate of change of magnetic flux of the coil.

[-24 V] [-1.5 Wb s-1]

Exercise 46-3:
The magnetic flux through one turn of a 60-turn coil of wire is reduced from 35 Wb to 5 Wb in 0.8 s, which
results in an induced current of 360 A. Calculate the total resistance of the wire in the coil.

[6.25  ]

46

Ummi Atiah & Siti Farahiyah
Electromagnetic Induction

L/O 5.2 (d)(i) Use induced emf in a straight conductor, where,

 = Blvsin B = magnetic field strength

l = length of conductor

v = velocity of conductor
θ = angle between B and v

Example 47:

The rod shown above moves to the right on essentially zero-resistance rails at a speed of 3 ms-1. If the
magnetic 0.75 T everywhere in the region, Calculate the induced emf in a straight conductor?

Solution:

 = Blvsin 
= (0.75)(4 10−2 )(3) sin 90o
= 0.09 V

Exercise 47-1:
The rod of 5 cm moves to the right on essentially zero-resistance rails at a speed of 5 ms-1. If the magnetic
0.9 T everywhere in the region, calculate the induced emf in a straight conductor?

[0.225 V]

Exercise 47-2: Exercise 47-3:
A conducting rod of 10 cm is sliding on a metal rail.
The apparatus is in a uniform magnetic field of A 4 cm rod is in a uniform magnetic field which is
strength 0.25 T, which is directly into the page. The
conducting rod is moving to the right by a force. If the directly into the page as shown below. The rod is
induced emf produced is 0.125 V, calculate the pulled to the right at a constant speed of 5 m s-1 by a
velocity of the moving rod.
force . The only significant resistance in the circuit
[5 m s-1] comes from the 2.0 Ω resistor with the 0.025 A

current flowing. Calculate the magnetic field strength.

[0.25 T]

47

Ummi Atiah & Siti Farahiyah
Electromagnetic Induction

L/O 5.2 (d)(ii) Use induced emf in a coil,

Changing Magnetic Field, where
dB : change in magnetic field strength
 = −NA dB A : surface Area
dt N : number of turns
dt :change in time

Example 48:
A coil of 1000 turns encloses an area of 25cm2. The coil is placed in an external magnetic to a magnetic field
of 6 ×10-5T. The coil is then pulled out of the field in 0.3 s, calculate the average emf induced during this

interval?

Solution:

 = −NA dB
dt

= −(1000)(25 10−4 ) (0 − 6 10−5 )
0.3

= 5 10−4 V

Exercise 48-1: Exercise 48-2:
A coil of 900 turns encloses an area of 40cm2. The A 5 ×10-4 V emf is induced in a 2000 turns coil of an
coil is placed in an external magnetic to a magnetic area of 25cm2. This coil is initially placed in a region
field of 9 ×10-5T. The coil is then pulled out of the
field in 0.5s, calculate the average emf induced of magnetic field and later it is moved out of the field
during this interval.
in 0.3 s. Calculate the strength of the magnetic field.
[ 6.4810−4 V ]
[ 310−5 T ]

48