Active Physics Full Solutions to Textbook Exercises

Chapter 20

Electrostatics

3. No
Checkpoint
Metal conducts electricity because it contains free
Checkpoint 1 (p.7) electrons. But the no. of electrons and no. of protons
in a metal could be the same, so a metal could be
1. (a) T Electrically neutral means carrying zero net
neutral.
charge. 4. As metal is a conductor, the excess charges on the
(b) T If the positive charge outnumbers the metal rod would spread to all parts.
negative charge, the object carries a positive
5. No
charge. In the opposite, the object carries a
negative charge. The interaction between charges will not be altered
by merely changing their name.
(c) T The charge carried by a proton and that
−9
carried by an electron are equal in amount but −1.6 × 10
6. (a) Difference = −1.6 × 10−19
= 1010
opposite in sign.
(b) Electrons
(d) F A neutral object does contain positively
charged particles and negatively charged
Checkpoint 3 (p.14)
particles. But the amounts of the opposite
charges are equal.
1. D The balloon could be neutral or positively

2. B The positive and negative signs were chosen charged.

arbitrarily by Benjamin Franklin. Positive charged
2. C and D attract each other as a charged object
particles repel each other too.
attracts a neutral object.
3. Because Tom’s hair is also oppositely charged. The As A attracts B but repels D , we know that B and D
amount of charge on the balloon and on Tom’s hair carry charges of opposite signs. So B and D attract
is the same, but are of opposite signs. So the net each other.
charge remains zero.
3. (a) F When a charged conductor is brought

close, the positive and negative charges on the

Checkpoint 2 (p.10)
neutral object separates, but the total amount
of charge remains unchanged.
1. (a) B As metal is a conductor, the excess charge
(b) F The two electric forces are action–reaction
can move freely on the sphere. So they would
pair.
spread over the surface of the sphere due to
mutual repulsion. (c) T As the charged conductor is brought close,
positive and negative charges are induced on
(b) A As the sphere is made of insulating
the neutral object. These induced charges then
material, the excess charge is not free to move
affect the charge distribution on the charged
around. So they would stay at the original
conductor.
position.
4. No
2. (a) F Some conductors conduct electricity by
movable ions. As a neutron has negligible size, there would not be
any charge separation due to electrostatic induction,
(b) T Insulators are materials that do not have
and hence there is no electric force between a
movable charged particles.
proton and a neutron.
2| Chapter 20 Electrostatics Checkpoint Active Physics Full Solutions to Textbook Exercises

Checkpoint 4 (p.23) 4. (a) The force increases to 4 times of its original

value.
1. (a) X is positive while Y is neutral. (b) The force remains the same.
(b) X is positive, while Y and Z are neutral. (c) The force decreases to 2/9 of its original value.
( ) ( )
2. (a) T The total charge of the two objects is 1 q2 q2 q2
5. (a) F = 4πε 0 r2
+ (2r )2 = 45 4πε
1
0
· r 2
conserved.
(b) F = 0 (due to symmetry)
(b) F The amount of charge shared by the balls ( ) p ( )
1 q2 1 q2
depend on the ratio of their size. (c) F = 2 × 4πε0 · r2
× cos 45° = 2 4πε 0
· r 2

∴ (c) > (a) > (b)

(c) F Electrons will be drawn up from the
ground. Protons are immovable.
Checkpoint 6 (p.34)
3. At the beginning, electrons and protons are evenly
distributed on the surfaces of the spheres.
1. (a) T An electric ield is a region in which a static
When the charged rod approaches A , some charge experiences an electric force.
electrons in A is repelled to B .
(b) T The electric force acting on the test charge
So, when separated, A in short of electrons is is directly proportional to the electric ield (i.e.
positively charged. B having excess electrons is F = qE ∝ E ).
negatively charged.
(c) F The electric ield set up by itself cannot
We cannot remove the charged rod before separating the exert an electric force on it, but the ields set
spheres; otherwise, electrons would evenly distribute themselves up by the other two charges can.
on the two spheres again.
2. (a) T As E = F /q , the electric force acting on a
4. Because some electrons low from the object to the
+1 C charge is the strength of an electric ield.
Earth (or the opposite) during earthing.
(b) F Electric ield strength at a point is indepen-
5. As the hair touches the dome, it is charged by the dent of the test charge.
dome. The like charges on the hair repel each other,
(c) F Electric ield strength points along the
so the hair stand and spread.
electric force acting on positive test charge only.

Checkpoint 5 (p.29) 3. E = Fq = 240.015

× 10−6
= 625 N C−1 (upwards)

1 Qq 1 (1)(1)
1. (a) F = 4πε 0 r2
= 4πε0 12 = 9 × 109 N
Checkpoint 7 (p.42)
(b) Q 2 = 4πε0 r 2 F = 4πε0 (12 )(1)
p
∴ Q = 4πε0 ≈ 10−5 C 1. (a) T E - ield lines are drawn according to the
vectors of the E - ield.
2. (a) F Coulomb’s law is valid for point charges of
any sign. (b) F Electric ield lines do not cross as E - ield
cannot point in two directions at a point.
(b) F Coulomb’s law is valid for spheres and point
charges only. (c) T Electric ield lines point from positive
charge to negative charge.
3. See the diagram below.
2. (a) Point A
(b) See the diagram below.
Active Physics Full Solutions to Textbook Exercises Chapter 20 Electrostatics Exercise |3

(c) See the diagram below. Symbols:

E : E - ield strength at a point
Q : charge carried by the source charge
r : distance between the source charge and that
point
Application: The equation can be applied for a
point source only.

4. (a) Corner C
3. (a) F Electric ield line does not represents the (b) The electric ields at A due to the two charges
path of a charge. cancel out.
(b) F The ield line density at X is lower, so the 1 Q+ 1 Q−
=
electric ield at X is weaker. 2
4πε0 r + 4πε0 r −2
4 × 10−6 Q−
p = p
Checkpoint 8 (p.60) (0.5 × 2 × 5 )
2 2 ( 2 × 52 )2
∴ Q− = 16 μC

1. (a) T The electric ields at different points that

are at the same distance from a point source The sign of the charge is negative.
are the same, and they all point in the radial
5. (a) KE = work done by electric ield
direction.
= F × d = qE × d = eV
(b) F It is valid for point charges only.
(b) KE = work done by electric ield
(c) T The electric ield around a point charge is = F × (d /2) = eV /2
inversely proportional to r 2 .
(c) KE = work done by electric ield
(d) F Electric ield is a vector quantity, so its = F × (d /2) = eV /2
direction has to be considered when inding
the resultant ield.
1 Q
p p Exercise
2. (a) E 0 = 4πε 0 r2
and r a = r 0 12 + 12 = 2r 0
0

1 Q E0
⇒ Ea = 4πε0 r a2 = Exercise 20.1 (p.14)
2
p p E0
1 Q
(b) r b = r 0 12 + 22 = 5r 0 ⇒ E b = 4πε = 1. (a) See the diagram below.
0 r2 5
b

p p 1 Q E0
(c) r c = r 0 12 + 32 = 10r 0 ⇒ E c = 4πε 0 r2
=
c 10
p p 1 Q E0
(d) r d = r 0 12 + 42 = 17r 0 ⇒ E d = 4πε 0 r2
=
d 17
(b) See the diagram below.
3. (a) Description: E - ield strength is the electric force
per unit charge.
Symbols:
E : E - ield strength at a point
F : electric force on a charged particle at that
(c) See the diagram below.
point
q : charge carried by the charged particle
Application: The equation is generally applica-
ble. It de ines the E - ield strength at a point.
(b) Description: E - ield strength around a point
source is directly proportional to Q and is
inversly proportional to r 2 .
4| Chapter 20 Electrostatics Exercise Active Physics Full Solutions to Textbook Exercises

2. (a) (i) See the diagram below. (b) The copper rod is attracted by the glass rod as
the end near the glass rod (the left end) lacks
electrons and becomes positively charged.
Therefore the copper rod rotates anticlockwise
as viewed from the top.
(ii) See the diagram below.
If you consider the far end (the right end) of the copper
rod, you will get the same answer. The far end of the
copper rod is repelled by the glass rod because it gains
electrons.

(b) (i) See the diagram below. 9. (a) See the diagram below.

(b) When the ruler approaches the tissue paper,

the molecules inside the tissue paper are
(ii) See the diagram below.
polarized. As the polarized molecules line up,
net positive charge is induced on the left due
to the attraction of the negative charge on the
ruler, while net negative charge is induced on
the right due to repulsion. The charges in the
Steps: First, determine the sign of charge. Second, middle cancel out each other.
determine the interactions.
10. (a) The electrostatic cloth induces opposite
3. B Unlike charges attract. charges on the end of the dust near the cloth.
No electric force between a charge and a neutral particle. These opposite charges produce an attractive
electric force between the cloth and the dust.
4. (a) No
The like charges induced at the far end would
(b) Yes result in a repulsive force. As these like charges
(c) Yes are further away, the repulsive force is smaller
than the attractive force. So a net attractive
5. C Q repels S , which means that S also carries
force is resulted.
negative charges. As R attracts both positively and
negatively charged spheres, it must be neutral. (b) When the attractive electric force acting on
the dust is larger than its weights, it would be
6. (a) Yes, charge is conserved. stuck to the cloth. Because paper scraps are
(b) The two ions carry opposite charges, so they much heavier than the dust, it would not be
experience attractive electric force and join stuck to the cloth.
together.

7. (a) The pencil contains exactly the same amount of Exercise 20.2 (p.23)
protons (positively charged) which attract the
electrons and hold them together. 1. (a) Electrons move from the sphere to the rod.
−9 Charge distribution:
(b) No. of extra electrons = 1.6010× 10−19 = 6.25 × 109
9
Ratio = 6.2510×2410 = 6.25 × 10−15
So, this amount is very small compared with
the number of existing electrons.
(b) Electrons do not move.
8. (a) The free electrons in the copper rod will move
Charge distribution: (unchanged)
from the end near the glass rod to the end
furthest away.
It is because they are repelled by the negative
charges on the glass rod.
Active Physics Full Solutions to Textbook Exercises Chapter 20 Electrostatics Exercise |5

(c) Electrons move from the ground to the sphere. As the negative charges on B are attracted by the
Charge distribution: positive charges on C , B won’t be discharged and it is still
negatively charged.
(b) Yes, charge is conserved.
When Roy touches B , electrons low from the
ground to A . So no charge is created and thus
(d) Electrons move from the sphere to the ground. charge is conserved.
Charge distribution: (unchanged)
7. The ball is attracted by the dome due to electrostatic
induction.
It touches the dome and becomes positive.
Then, it is repelled by the dome.
Initially there are induced negative charges (electrons It touches the metal plate and discharges.
accumulated) on the sphere. As the rod leaves, the After that, it is attracted by the dome again.
electrons on the sphere move to the ground continuously
It swings back and forth several times and stops
and so the net negative charges on the sphere decreases.
inally due to discharging.
Finally the sphere becomes neutral.
8. First, rub the rod with the cloth to charge it.
2. (a) No; yes
Then let the two spheres touch each other, and put
(b) Yes; no
the charged rod near one of the spheres without
(c) Yes; no touching it.
(d) Yes; no Finally separate the spheres while the rod is still
close. Do not touch the spheres directly when you
3. (a) No
separate them.
(b) No
9. (a) No, it could not. Since the charges on the
(c) Yes
insulator cannot move freely, they cannot be
(d) Yes
neutralized by earthing.
Note that the steel ruler is earthed. Therefore, the ruler can be (b) Yes, it could. The mobile ions in tap water
charged in (c) and (d), but cannot in (a) and (b). If the ruler was would neutralized the charges on the insulator.
not earthed, this time (b) and (d) work, but (a) and (c) fail.
(c) Yes, it could. The ionized gas in the lame
−9
4. (a) −3 × 10 C would neutralized the charges.
(b) Yes, charge is conserved.
Exercise 20.3 (p.30)
5. (a) Each ball carries −0.5 × 10−9 C of charge.
(b) Yes, charge is conserved. 1. By Coulomb’s law,
Just before the balls collide, net charge of the
−7 2
balls = 1 × 10−9 + (−2 × 10−9 ) = −1 × 10−9 C. 1 Q 1Q 2 9 (3 × 10 )
F= = (9 × 10 ) = 9 × 10−3 N
4πε0 r 2 0.32
Just after the separation, net charge of the balls
= −0.5 × 10−9 + (−0.5 × 10−9 ) = −1 × 10−9 C,
2. By Coulomb’s law,
which is the same as before.
1 Q 1Q 2
6. (a) A : Neutral F=
4πε0 r 2
B : Negatively charged (5 × 10−13 )2
9 × 10−9 = (9 × 109 )
Initially positive charges are induced on A while d2
negative charges are induced on B . ∴d = 5 × 10−4 m

When Roy touches B , free electrons from the ground

flow through him and then B to neutralize the charges on 3. (a) Zero. The four forces cancel each other due to
A . So A becomes neutral. symmetry.
6| Chapter 20 Electrostatics Exercise Active Physics Full Solutions to Textbook Exercises

(b) By symmetry, the top-right and bottom-left Dividing (2) by (1), we have
charges cancel the effect of the each other.
1 Q2 1
Since the top-left charge is positive, the tan θ =
4πε0 r 2 mg
resultant force on the central +q points bottom
Q2 1
right. Its magnitude is tan 14.48° = (9 × 109 ) 2 −5
0.1 (10 )(9.81)
1 q2
F = 2 × 4πε (p )2 = 3.6×1012 q 2
0
0.12 +0.12 ×0.5 ∴Q ≈ 5.31 × 10−9 C

(c) All the charges in the upper half are +ve while
all those in the lower half are −ve. Obviously 7. (a) Between the two charges, the electric forces
the resultant force points downwards, by acting on the third charge are in opposite
symmetry. direction, so they would cancel out each other
at a point.
Since all four forces make 45◦ with the vertical
line, the magnitude of the resultant force is Let the third charge carries charge Q and is at
1 q 2 a distance d away from charge +q . As the two
F = 4× 4πε (p )2 ×cos 45° = 5.09×1012 q 2
0
0.12 +0.12 ×0.5 electric forces cancel out each other,
(d) By symmetry, the top-right and bottom-
1  
qQ 1 2
qQ
left charges cancel the effect of each other. =
4πε0 d 2 4πε0 (1 − d )2
This part is effectively the same as (b). So,
(1 − d )2 = 2d 2
the resultant force points bottom right. Its
0 = d 2 + 2d − 1
magnitude is 3.6×1012 q 2 .
∴ d ≈ 0.414 m or −2.41 (rejected)
4. (a) T Because it lifts up with a smaller angle.
(b) F They may carry equal amount of charges. So, it is 0.414 m on the right of charge +q .

(c) F They experience the same magnitude of Note that the equilibrium position is independent of
electric force due to Newton’s third law. the third charge Q .
(b) No, it would not.
5. D The downward force experienced by the sphere
The permittivity of different medium are different from
on the balance equals the sum of the repulsive force
each other. The electric force between each pair will be
experienced and its weight. So the reading is given
1 Q
2 different.
by mg + 4πε 0 d2
.
8. (a) See the diagram below.
6. The free-body diagram of the left ball:

Consider the electric force of X from the charge on the

2
1 q
right of X , we have F = 4πε 0 2
d
.
Consider the electric force of X from the charge below X ,
q2 1 q
2
we have F ′ = 4πε1
0 (2d )2
= 14 × 4πε 0 d2
= F4 .
′
Therefore the resultant of F and F has the direction
shown in the figure.
(b) The magnitude of the force decreases, while
By geometry, sin θ = 0.05
0.2 , so θ = 14.48°.
the direction remains unchanged.
Since the balls are at rest, we have As the permittivity is larger in polythene than
 in air, the magnitudes of the electric forces

 T cos θ = mg (1)
decrease, so as the net force.
 1 Q2
 T sin θ = (2) Since the ratio of the two electric forces does
4πε0 r 2
not depend on medium, the direction of the net
force remains unchanged.
Active Physics Full Solutions to Textbook Exercises Chapter 20 Electrostatics Exercise |7

9. (a) By Coulomb’s law, the electric force on the 11. When A is at the bottom, the electric force on B
−8 μC charge by one +4 μC charge is F = balances the weight of B . By Coulomb’s law,
−6 −6
1 Q1Q2
4πε0 r 2 = (9 × 109 ) (4 × 10 0.25
)(8 × 10
2
)
= 4.608 N.
1 Q AQB
= mB g (1)
4πε0 r 2

Similarly, when B is at the bottom, the electric force

on A balances the weight of A . So,

1 Q AQB
= mAg (2)
4πε0 d 2

Dividing (1) by (2), we get

r 2 mA (3)(0.5)2
= ⇒ d2 =
By symmetry, the resultant electric force is d 2 mB 1
F res = 2 × 4.608 cos 30° = 7.98 N , pointing ∴d = 0.866 m

towards the mid-point between two positive

charges. 12. When they are released from rest, they move to-
(b) From (a), the electric force F on the +4 μC wards each other due to the electrostatic attraction
charge by the −8 μC charge is also 4.608 N. between them. Once they are in contact, they will
share their charges. After sharing, the charge on
As F ∝ Q 1Q 2 , the electric force F ′ on the +4 μC
each ball Q = −15+5 = −5 nC.
charge by the other +4 μC charge charge is 2

4.608/2 = 2.304 N. After that, they repel each other and return to their
original positions.
By Coulomb’s law, the magnitude of the electric
1 (5 × 10−9 )2
force F = 4πε 0 12
= 2.25 × 10−7 N .

13. (a) Their attractive force is doubled when the

electric force between them equals the
gravitational force. By Newton’s law of
gravitation and Coulomb’s law,

Gm 2 1 q2
= ⇒ q 2 = 4πε0Gm 2
r2 4πε0 r 2
From the above igure, we have: ∴ q = 8.61 × 10−13 C
′
x –component of F res = (F + F ) cos 60° = 3.456 N
8.61 × 10−13
y –component of F res = (F − F ′ ) sin 60° = 1.995 N So, the mass of electron required = 1.60 × 10−19
×
1.995 9.11 × 10−31 = 4.90 × 10−24 kg .
tan θ = 3.456 ⇒ θ = 30°
p
∴ Fres = 3.4562 + 1.9952 = 3.99 N , pointing 30° (b) The constant in the Newton’s law of gravitation
below the x –axis. is G = 6.67 × 10−11 N m2 kg−2 , while that in the
1
Coulomb’s law is k = 4πε 0
= 9 × 109 N m2 C−2 .
10. If an additional charge is placed at the top of the
As G is much smaller than k , the mass of the
circle, the net force acting on the central charge
balls have to be much greater than that of the
becomes zero by symmetry.
electrons if the magnitude of the gravitational
So, the electric force by the charge at the top cancels force equals that of the electric force between
out the resultant force by the other four charges them.
on the circumference, thus they have the same
magnitude.
The required resultant force has a magnitude of
2
1 q
4πε0 r 2 , pointing upwards.
8| Chapter 20 Electrostatics Exercise Active Physics Full Solutions to Textbook Exercises

14. According to Example 20.5, the electric force equals (b) (i) See the diagram below.
9 × 10−8 N.
This force provides the required centripetal force
for the electron to orbit around the proton, so we
have

mv 2 (ii) See the diagram below.

= 9 × 10−8
r
√ ( )
r 9 × 10−8
∴v =
m
√( )( )
0.5 × 10−10 9 × 10−8
=
9 × 10−31
(iii) See the diagram below.
= 2.23 × 106
2.23 × 106 ( )
= 8
× 3 × 108
3 × 10
≈ 7 × 10−3 c

So the linear speed of the electron is 0.7% of the

speed of light in vacuum.
3. B If the initial velocity has a component perpen-
dicular to the ield, it would move out of a plane. So
Exercise 20.4 (p.43)
C is incorrect.

1. (a) See the diagram below. By F = qE and Newton’s second law, a uniform electric field
implies a uniform acceleration to a charge.

4. C As the charge moves along the circular path,

a centripetal force is required. Only ield C can
provide the required force.

5. By FE = qE ,

(b) See the diagram below. F E 3.10 × 10−5

q= = = 4.25 × 10−11 C
E 7.30 × 105

6. As the water droplet is in equilibrium, the weight is

balanced by the electric force which points upwards.
Also, the water droplet carries negative charge, so
the electric ield points downwards.
(c) See the diagram below.
Consider the magnitude of the electric ield, we have

mg = qE
(0.50 × 10−3 )(9.81) = (40.0 × 10−6 )E
∴ E = 123 N C−1

Therefore, the electric ield at that point has a

2. (a) (i) E A = E B = EC magnitude of 123 N C−1 , pointing downwards.
(ii) E A < E B This electric field varies from place to place, and is affected
(iii) E A > E B (= 0) by the weather. In the above calculation, we have ignored
the buoyant force (’floating force’), which actually becomes
considerable if the droplet is very small.
Active Physics Full Solutions to Textbook Exercises Chapter 20 Electrostatics Exercise |9

7. No (c) See the diagram below.

• Electric ield lines do not branch at any point.

• Electric ield lines do not form closed loops.
• Electric ield lines do not have edges.

8. (a) See the diagram below.

2. C By Newton’s second law, F = qE = ma . This

q
implies a = m E . A charged parallel plates produce
a uniform electric ield, which implies that E is a
constant. Therefore, the acceleration of the charged
particles in the parallel plates depends on the
(b) Since the ield line density increases along the charge to mass ratio only, not the charge or the mass
path shown in (a), the electric ield strength alone.
increases. By F = qE , the magnitude of
3. B At the position x = 2, the electric ields due to
the electric force acting on the charge also
the two charges cancel out each other. So it is a
increases.
neutral point, and no ield line passes through it.
On the other hand, due to symmetry, the
Also, charged particle placed at a neutral point stays.
electric force acting on the particle always
points left. 4. It rotates clockwise, and inally comes to a stop after
it becomes horizontal.
9. See the diagram below.
Actually, when the rod rotates to its horizontal position, it
oscillates about the horizontal axis until it comes to a stop.
3
5. (a) The electric ield E = Vd = 252 ×× 10
10
−3 = 80 000 N C−1

(b) As the oil droplet is in equilibrium, it experi-

ences an upwards electric force. So, it carries a
(Accept other reasonable answers.) positive charge.

The line density decreases from left to right.

qE = mg
(1.30 × 10−9 )(9.81)
Exercise 20.5 (p.60) ∴q= ≈ +1.59 × 10−13 C
80 000

1. (a) See the diagram below. (c) It stays at the new position.

6. The electric ields due to A and B alone point left

and right at P respectively, the resultant electric
ield is
( )
1 1 × 10−9 3 × 10−9
E = E A − EB = −
4πε0 0.22 0.72
≈ 170 N C−1

(b) See the diagram below.

∴ The electric ield is 170 N C−1 , pointing left.

7. The electric ield due to the negative charge is

1 Q Q
4πε0 ( 2d )2 = 8πε d 2 (along −y ).
p
0

As the two positive charges are symmetric about

the y -axis, the x -components of their electric ields
cancel out each other.
10 | Chapter 20 Electrostatics Chapter Exercise Active Physics Full Solutions to Textbook Exercises

So the electric ield due topthe two positive charges So by Newton’s second law, we have
1 Q
× cos 45 ◦ = 4πε d 2 (along +y ).
2Q
is 2 × 4πε 0 d2 0 qV0 qV0
p = ma ⇒ a= (pointing upwards)
Q(2 2 − 1) d dm
Thus, the resultant electric ield is E = in
8πε0 d 2
the +y -direction. Consider the horizontal motion.
Time taken to pass through the gap
8. The centripetal force is provided by the electric
t = vℓ0 = 10
1
= 0.1 s.
force:
Consider the vertical motion. Take the upward
v
2 2 u 2 direction as positive. By y = u y t + 21 at 2 ,
1 e me v u e
t
= ⇒ v=
4πε0 r 2 r 4πε0 m e r
d 0 1
= *
u y
t + at 2
2 2
9. (a) See the diagram below. 1 (0.4 × 10−6 )V0
0.03 = (0.1)2
2 (0.06)(1 × 10−4 )
∴ V0 = 90 V

(b) Yes, it is still parabolic. Chapter Exercise

The electric ield provides a constant
downward electric force, therefore the Multiple-choice Questions (p.65)
resultant force acting on the electron still acts
downwards. So it still travels in a parabolic 1. B Both (1) and (2) are incorrect. X carries a
path. positive charge while Y is neutral. The key here is
to regard X , Y and the Earth as a single conductor.
10. By conservation of energy,
Repelled by the charged rod, electrons tend to stay
1 away from the rod and low to the Earth. They do
mv 2 = qV
2 not accumulate in Y .
1
(5 × 10−9 )(1 × 102 )2 = (1.25 × 10−9 )V On the side of X close to the rod, there is a shortage
2
V = 20 kV of electrons and thus a positive charge is induced.

2. D As the air near the sphere is ionized by the

V
So, the magnitude of the electric ield E = d = α source, there are many positive and negative
20 × 103
4 × 10−2
= 5.00 × 105 N C−1 . charges around the sphere. So, the sphere is
discharged, and remains in its original position.
Alternative Solution:
Take the direction to the right as positive. Do not treat the α source as a positive charge. α particles
By v 2 = u 2 + 2as , emitted from the source carry positive charge, but α source itself
does not.
0 = (1 × 102 )2 + 2a(4 × 10−2 )
∴ a = −1.25 × 105 m s−2 3. C If A and B carry like charges, the electric force
becomes 12 F . If they carry unlike charges, the
By Newton’s second law, electric force becomes 0.

qE = ma 4. D The free-body diagram:

(5 × 10−9 )(−1.25 × 105 )
∴E = = 5.00 × 105 N C−1
−1.25 × 10−9

11. Let V0 be the maximum pd. When the particle

projects between the plates, the only force acting
on it is the electric force. (Gravitational force is
neglected.)
Active Physics Full Solutions to Textbook Exercises Chapter 20 Electrostatics Chapter Exercise | 11

FA WA FB WB
The maximum electric force F on A = T − mg = By sine law, we have sin 45° = sin 60° and sin 15° = sin 60° .
10 − (0.05)(9.81) = 9.51 N. By Coulomb’s law, sin 45°
Since F A = FB , we have WB = sin
3 × 10−6 )(5 × 10−6 ) 15° · W A = 2.73W A .
9.51 = ( 4πε d 2
. So d ≈ 11.9 cm.
0 So mass of B is 2.73m .
5. B In arrangements (1) and (2), the electric forces In equilibrium (i.e. zero net force), all the forces acting on an
on X by the charges on opposite vertices cancel out. object must form a closed loop when they are arranged tip-to-tail,
6. Statement (1) must be correct. By as long as the forces are drawn in scale.
B
¯ Coulomb’s ¯
qP ¯ q A q ¯
law, the net force acting on P = 4πε0 ¯ R 2 − r B2 ¯ = 0.
Statement (2) may be incorrect. If P is positively
charged, it still experiences no net force.

7. C As no net electric force is acting on P at Y ,

E Y = 0. So, the electric ield at distance R away
from charge A cancels that at distance r away from
charge B . Thus, by comparing the electric ield at X
due to charge B and that at Z due to charge A , we
know that E Z > E X .
1
10. A Let k = 4πε and n be the number of protons in
8. C The attractive electric forces FE between the two 0

1 Q
2 the nucleus.
spheres is 4πε 0 L2
. To keep their separation, both
spheres must have the same acceleration. The centripetal force is provided by the electric
2 2
force F = ke
r2
, therefore ke
r2
= mr ω2 and thus
So, the net force acting on them must be the same. 1
1 Q
2 r ∝ ω2/3 .
Thus, F = 2FE = 2πε 0 L2
, pointing outwards along the
1
line joining the two particles. When ω doubles, r is reduced to p
3 of its original
4
value.

11. A Statement (3) is incorrect. Only when the

point charge is positive, the electric force is in the
direction of the ield line.

12. A Statement (1) is correct but statement (2) is

incorrect. By E = Vd0 , the electric ield would in-
9. B The free-body diagram of A and B : crease when the voltage increases or the separation
decreases. And stronger electric ield increases the
de lection of the foil.
Statement (3) is incorrect. The electric ield
between the plates is uniform.

13. A The electric force acting on it always has the

same magnitude and points left.

14. B Both statements (1) and (2) are incorrect.

Assume P carries positive charge and ind out the

As they are in equilibrium, the forces acting on each charge of other objects. Observe that P and R carry
of them form a closed triangle. opposite charge, so as Q and S .

15. D Since the three charges stay at rest, charges Q 1

and Q 2 carry the same sign and have a sign opposite
to that of Q 1 .
Consider charge Q 1 , the force on it by Q 2 must
be balanced by the force on it by Q 3 . By F ∝ r12 ,
doubling the distance means reducing the force to a
quarter. Therefore, FQ 2 = 4FQ 3 and thus |Q 3 | = 4|Q 2 |.
12 | Chapter 20 Electrostatics Chapter Exercise Active Physics Full Solutions to Textbook Exercises

Structured Questions (p.67)

18. (a) See the diagram below.

16. (a) (i) When C comes close, opposite charge is
induced on B and like charge is induced in
A. (1A)

So, A is repelled by C and de lects. (1A)

(ii) Sphere A falls back to its initial position.

(1A)

(b) (i) See the diagram below.

Correct arrows: 1A
(b) The electric ield strength at A is greater than
that at B (1A)

because the ield lines at A is denser than that

at B . (1A)

19. (a) The resultant electric ield lines in the

As ball A is in equilibrium, all the forces space between the charges are in the same
acting on it forms a closed loop. (1A) direction. (1A)

The electric force is along AB , the weight is (b) It lies on the right of both charges. (1A)

along OB , and the tension is along O A . (1A) (c) See the diagram below.
Note that triangle O AB is an isosceles
triangle. So the magnitude of tension
equals that of the weight, and the tension
is mg . (1A)

(ii) Its swings back a little, but still de lects at

some angle. (1A)

1 1 × 10−9
1 Q
By E = 4πε 0 r2
, EB = 4πε0 0.82 ≈ 14 N C−1 ,
17. (a) As unlike charges attract, the electric force
points downward. (1M)
acting on A points right and that on B points p
AX = 12 + 0.82 ≈ 1.28 m
left. (1A)
1 Q
When a leftward electric ield is applied on the Vertical component of E A = 4πε0 r 2 sin θ =
−9
1 4 × 10 0.8 −1
spheres, the forces produced would cancel out 4πε0 1.282 · 1.28 ≈ 14 N C (1M)

1 Q
the attractive forces. (1A) Horizontal component of E A = 4πε0 r 2 cos θ =
1 4 × 10−9 1 −1
So, the uniform electric ield points left. (1A) 4πε0 1.282 · 1.28 ≈ 17 N C (1M)

(b) Electric force between the spheres = The vertical component roughly cancels out E B .
−9 −9
1 (2.0 × 10 )(2.0 × 10 ) −5 Therefore the electric ield strength is about
4πε0 −2 2
= 2.25 × 10 N (1M)
(4 × 10 )
−5 17 N C−1 , (1A)
× 10
By F = qE , E = Fq = 2.25
2.0 × 10−9
= 11 300 N C−1 . (1M+1A)
pointing towards the right. (1A)

(c) No, they cannot. (1A)

(d) See the diagram below.
Alternative Solution:
(a) The electric ield on the position of sphere A
produced by sphere B points right. (1A)

The uniform electric ield has to cancel out the

electric ield produced by sphere B . (1A)

So, the uniform applied electric ield should points

left. (1A)

If you consider the position of sphere B , you will get the

same answer.
(b) The electric ield on the position of sphere More ield lines around A : 1A
1 2.0 × 10−9
A produced by sphere B = 4πε −2 2 = Correct directions of ield lines: 1A
0 (4 × 10 )
11 300 N C−1 (2M+1A) Neutral point: 1A
Active Physics Full Solutions to Textbook Exercises Chapter 20 Electrostatics Chapter Exercise | 13

20. (a) Both of them are positive. (1A) (c) No, the de lection angle cannot be too large.
(b) At x = 8 cm, E = 0. (1A)
(1A)

1 Q
By E = 4πε , 1 QA 1
− 4πε
QB
=0 (1M)
If the de lection angle was too large, the sphere
0 r2 4πε0 82 0 22
∴ Q A : QB = 16:1 . (1A)
would touch the metal plate and share charges
with the plate. Therefore the result found
(c) The electric force points towards the right and
would be wrong. (1A)
gets weaker when the test charge is moving
from A to B . (1A) 22. (a) See the diagram below.
It then drops to zero at x = 8 cm. (1A)

After that, it points towards the left and get

stronger. (1A)

21. (a) (i) Draw a best- it lines for the data points.

Correct path inside the plates: 1A

Correct path outside the plates: 1A
( )
(b) ∆KE = 12 m v 22 − v 12 = 1.09 × 10−17 J . (1M+1A)

(c) By ∆U = q∆V and E = − Vd0 ,

∆U = qE d (1M)

−17
1.09 × 10
∴E = ( )( ) (1M)
1.60 × 10−19 2 × 10−2
= 3.41 × 103 N C−1 (1A)
0.28−0
The slope of the graph = 70−0 =
4 × 10−3 V−1 (1A)
(d) By F = ma ,
−3 −1
(Accept answers between 3.7 × 10 V
and 4.2 × 10−3 V−1 ) qE = ma ⇒ a =
qE
(1M)

(ii) By the free-body diagram: m

Consider the vertical motion.

By s = ut + 21 at 2 ,
( )
1 qE 2
s = ut + t
2 m
( )( )
1 1.60 × 10−19 3.41 × 103 2
2 = 0+ t (1M)
2 9.11 × 10−31
∴t = 8.17 × 10−8 s (1A)

Fe 23. (a) The mixture is charged by friction. (1A)

We have tan θ = mg (1M)

By F = qE and E = Vd0 , we have Two kinds of plastic carry opposite charges.

(1A)

qE q
tan θ = = V0 (1M) (b) The negatively charged plastic in the mixture
mg mg d
is attracted by the positive electrode and falls
further away.
∴ The slope represents q (1A)
mg d
. (1A)

q
The positively charged plastic is attracted
(b) As slope = mg d ,
( )( ) by the earthed drum and then sticks on the
q = 4 × 10−3 0.01 × 10−3 (9.81)(0.08) =
drum. (1A)
+3.14 × 10−8 C . (1A)
14 | Chapter 20 Electrostatics Chapter Exercise Active Physics Full Solutions to Textbook Exercises

(c) No (1A) (ii) See the diagram below.

When the mixture falls on the drum, all the
charged metals are already discharged. (1A)

So, the mixture behaves similarly when

passing the electrode and hence it cannot be Correct charge distribution: 1A
separated. (1A) (iii) The positive charge in the sphere is
24. (a) (i) Horizontal plates (1A)
attracted by the rod, (1A)

and the repulsion between the negative

(ii) Vertical plates (1A)

3
charges is smaller than the attraction. (1A)
(b) (i) By geometry, tan θ ≈ 31.5 .
vx ( 3
)( ) (c) (i) It means connecting the sphere to the
As tan θ = v y , v y = 31.5 3 × 107 =
earth/ground. (1A)
2.86 × 106 m s−1 . (1M+1A)
(ii) They move down to the ground (1A)
(ii) Time t spent in horizontal plates
due to the repulsion by the rod. (1A)

0.03 (iii) See the diagram below.

t= = 1 × 10−9 s (1M)
3 × 107
0
>
u + at ⇒ a = vt , we have
As v = 

2.86 × 106 Correct charge distribution: 1A

a= −9
= 2.86 × 1015 m s−2 (1M)
1 × 10 (d) Electrostatic spraying (1A)

By F = ma and F = qE , we have E = ma (Accept other reasonable answers.)

q

( )( )
9.11 × 10−31 2.86 × 1015
∴E =
1.60 × 10−19
= 16.3 × 103 N C−1 (1M)

So, the magnitudes of the electric ield

in the horizontal and vertical plates are
16.3 kN C−1 and 0 respectively. (1A)

Correct igure: 1A
25. (a) Concentrated solution is formed in interval
The droplets sprayed are charged. (1A)
2. (1A)
Due to the repulsion between like charges, the
Under the electric ield, sodium ions shift to the
droplets repel and hence spray evenly over a
right while chloride ions shift to the left. (1A)
large area. (1A)
So, the ions in intervals 1 and 3 would enter
Other applications involving electrostatic charging
interval 2 while that in interval 2 cannot
include electroscope, Van de Graff generator, ink jet printer,
escape. (1A)
photocopier, etc.
So ions accumulate in interval 2 to form
concentrated solution. 27. (a) See the diagram below.
(b) The solution formed in interval 2 would be
desalted. (1A)

The solutions formed in intervals 1 and 3

would consist a large amount of sodium and
chloride ions respectively. (1A)

26. (a) During rubbing, electrons (1A)

are transferred from the cloth to the rod. (1A)

(b) (i) The electrons move to the right side of the Correct shape: 1A
sphere. (1A)
Lines perpendicular to and touching both the plate
It is because they are repelled by the and the sphere: 1A
negative charge on the rod. (1A) Correct direction: 1A
Active Physics Full Solutions to Textbook Exercises Chapter 20 Electrostatics Chapter Exercise | 15

F
(b) (i) See the diagram below. (b) (i) tan θ = W (1A)

V
(ii) By F = qE and E = , d (1M)

F qE qV
tan θ = = =
W mg mg d
mg d tan θ
∴q= (1M)

( V )
0.07 × 10−3 (9.81)(0.1)(tan 2°)
=
4000
From the free-body diagram,
= 6.00 × 10−10 C (1A)

{
W = T cos 20°
(1M) (c) Fix the output voltage of the EHT and the
F = T sin 20°
plates separation. (1A)

∴ F = W tan 20° (1M)

Move the point of support of the thread to
= 3.6 × 10−6 N (1A) place the ball at different positions between
the plates. (1A)

(ii) E = Fq (1M)
OR: Fix the point of support of the thread and move
3.6 × 10−6
= 1.2 × 10−9
= 3.0 × 103 (1A)
the polystyrene tile such that the ball is at different
The unit for the electric ield strength is position between the plates. (1A)

N C−1 . (1A)

1 q1 q2
If the electric ield between the plates are uni-
(c) By F = 4πε0 r 2 , (1M)
form, the angle θ should remain unchanged.
( )2
1.2 × 10−9
(1A)

3.6 × 10−6 = (1M)

4πε0 r 2
Shoot-the-stars Questions (p.71)
∴r = 6 × 10−2 m (1A)

(d) See the diagram below. 1. A Treat A and B as one body. As there is no

net electric force acting on it, the string at the

top is vertical. Then consider ball B only. The
electric force acting on it points right, so the string
connecting A and B de lects to the right.

2. B Initially, it moves in the tangential direction

of the ield line, but not along the ield line. When
it moves forwards, the electric force acting on it
Correct ield pattern: 1A
rotates clockwise, so its path bends towards the
The right half of this electric ield pattern
ield line.
matches that in (a). (1A)
In order for the test charge to cross the ield line, a
28. (a) See the diagram below. sideway force has to be provided. As the direction of
the electric force is tangential, the test charge never
cross the ield line.

3. (a) The net electric force points towards right. (1A)

As the sphere on the right is nearer to the rod, more

charges is induced on it, so it attracts the ball more.
(b) As the ball is attracted to the sphere on the
T : tension
right, it acquires the induced charge on the
F : electric force
sphere (1A)
W : weight
and then repels. (1A)
Any two forces correct: 1A
It acquires opposite charges when touching the
All forces correct: 1A
other sphere and the process repeats. (1A)
Correct electric ield direction: 1A
Hence, the ball shuttles between the spheres.
16 | Chapter 20 Electrostatics Chapter Exercise Active Physics Full Solutions to Textbook Exercises

(c) As the ball shuttles between the spheres, it 4. (a) Horizontal displacement d = 0.1 × sin 20° =
carries charges from one sphere to another. 0.0342 m
(1A) ∴ Work done = qE d =
When the repulsive force produced by the (0.15 × 10−6 )(2.38 × 104 )(0.0342) = 1.22 × 10−4 J

accumulated charges is larger than the (b) Vertical dispacement h = 0.1 − 0.1 × cos 20° =
attractive force, the ball slows down and then 0.006 03 m
stops.
∴ Increase in grav. PE = mg h =
(1A)

(d) The ball shuttles for a while and then stops. (0.001)(9.81)(0.00603) = 5.92 × 10−5 J
(1A)
(c) No
The bead will move past the equilibrium
position due to inertia. This leads to energy
dissipation during the process, and thus the
increases in grav. PE is smaller than the work
done by the electric ield.