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Chapter 5.1

Q1 The charge is positive. A positive charge brought near the electroscope will attractelectrons towards the ball of the electroscope making the leaves even more positive and

hence the leaves will diverge further. If the unknown charge is negative, then electronswill be pushed towards the leaves, neutralizing some of the positive charge there makingthe leaves diverge less. However, if the body that is brought nearby is neutral, thencharge separation will take place in the body with the result that more positive chargemust now exist on the ball of the electroscope. This means that negative charge has leftthe ball for the leaves, neutralizing positive charge there and the leaves will diverge less.So if the leaves diverge less, we cannot be sure if the body was charged or neutral.

Q2 The flame ionized air molecules creating electrons and positive ions. The electrons orthe ions would be attracted to the charged electroscope neutralizing its charge.

Q3 The net charge available is 6 C! and this would be shared equally by the three

identical spheres. Hence each would get 2 C! .

Q4 (a)9 6 6

1 2

2 2 2

8.99 10 2.0 10 4.0 1028.8 29 N

(5.0 10 )

kQQF

r

! !

!

" " " " "= = = #

"

. (b) The force would

be becomes 4 times as small.

Q5 The middle charge is attracted to the left by the charge on the left with a force of9 6 6

21 2 2 2

8.99 10 2.0 10 4.0 1045 N

(4.0 10 )

kqQF

r

! !

!

" " " " "= = =

"

. It is attracted to the right by the

charge on the right with a force of9 6 6

22 2 2 2

8.99 10 2.0 10 3.0 10 135 N(2.0 10 )

kqQFr

! !

!

" " " " "= = =

"

.

The net force is thus 135 45 90 N! = directed towards the right.

Q6 (a) Suppose we call the distance (in cm) from the left charge x . Then we need

9 6 6 9 6 6

2 2

2 2

8.99 10 2.0 10 4.0 10 8.99 10 2.0 10 3.0 10

(6 )

4.0 3.0

(6 )

x x

x x

! ! ! !

" " " " " " " " " "=

!

=

!

This means that2 2

2 2

2

4.0(6 ) 3.0

4(36 12 ) 3

48 144 0

x x

x x x

x x

! =

! + =

! + =

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The solution is 3.22 cmx = . (b) If we displace the middle charge a bit to the left of the

equilibrium position the force from the left charge will be bigger and so the charge willmove towards the left i.e. away from the equilibrium position. The same is true if it isdisplac3d a bit to the right of the equilibrium position. Hence the position is unstable.

Q7 The forces are as shown. The distance between the charge Q and the charge 2Q is5.0 cm .

F1

F2

The magnitudes are9 6 6

1 2 2 2

(2 ) 8.99 10 3.0 10 6.0 1064.7 N

(5.0 10 )

kQ QF

d

! !

!

" " " " "= = =

"

and

9 6 2

2 2 2 2

8.99 10 (3.0 10 )89.9 N

(3.0 10 )

kQQF

d

!

!

" " "= = =

"

. We need to find the components of1

F :

1

464.7 cos 64.7 51.76 N

5x

F != = " = and1

364.7 sin 64.7 38.82 N

5yF != = " = . The

components of the net force are: 51.76 Nx

F = ! and 38.82 89.9 51.08 NyF = ! = ! . The

net force has magnitude 2 251.76 51.08 72.7 73 NF = + = ! and direction

51.08180 arctan 224.6 225

51.76+ = !

! ! ! .

Q8 The directions of the forces are as shown.

F1F2

F3

The magnitudes are2 9 6 2

1 3 2 2 2

8.99 10 (5.0 10 )15.60 N

(12 10 )

kQF F

d

!

!

" " "= = = =

"

and

2 9 6 2

2 2 22

8.99 10 (5.0 10 )7.80 N

2 (12 10 )( 2)

kQF

d

!

!

" " "= = =

" "

. The direction of the force is clearly

along the direction of2

F and has magnitude2 2

7.80 15.60 15.60 29.9 N+ + = .

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Q9 (a) A diagram is the following in which the angle ! of each string to the vertical is

given by5

sin 3.3785

! != " =! .

F F

!

We have that

cosT mg! = and2

2sin

kQT F

d! = = so that dividing side by side gives

2 2 6 2

9

2 9

tan 100 10 9.8 0.1 tan 3.37tan 8.0 10 C

8.99 10

kQ mgd Q

mgd k

!!

"

"# # # #

= $ = = = #

#

!

. (b)

This corresponds to

9

10

19

8.0 105.0 10

1.6 10

!

!

"= "

" electronic charges.

Q10 Assume for the sake of discussion that the charge on the sphere of the van de Graafgenerator is negative. This charge causes a small separation of the charge inside theplastic sphere: electrons are made to move slightly away leaving positive charge to facethe negative charge on the generator. As the charge on the generator increases the forceof attraction increases as well bringing the small plastic sphere closer and closer to thegenerator. Once it touches, the plastic sphere will become negatively charged and so willbe immediately repelled by the negative charge on the generator.

Q11 (a) Since the molar mass of water is 18 g per mole, a mass of 60 kg corresponds to

360 103333

18

!= moles i.e. 23 273333 6.02 10 2 10! ! = ! molecules of water. A molecule of

water contains 10 electrons (2 from hydrogen and 8 from oxygen) and so we have28

2 10! electrons in each person. (b) The electric force is therefore9 28 19 2

26 271 21 2 2

9 10 (2.0 10 1.6 10 )9 10 10 N

(10)

kQQF

r

!

" " " " "= = = " # , an enormous force. (c)

Assumptions include the use of Coulombs law for objects that are not point charges,

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assuming the same distance between charges etc. (d) We have neglected the existence ofprotons which gives each person a zero electric charge and hence zero electric force.

Q12 The electroscope will become negatively charged everywhere and the leaves willdiverge.

Q13

+

+ + +

_ _ __ _ _

+

_

_ _

The diagram assumes that the earth has been removed before the rod was also removed.If the rod is removed before the earth is, the electroscope will be left neutral.

Q14 This question is removed.

Q15 As the negatively charged sphere is lowered in the container, positive charge will beinduced in the inside surface of the container and an equal amount of negative charge onthe outside surface. The amount of positive charge on the inside is equal to the negative

charge on the sphere. When the sphere touches, its charge will be neutralized, leaving thesphere with zero net charge. The law of conservation of charge holds since the negativecharge on the sphere now finds itself on the outside surface of the container.