1

1 ELECTRIC CHARGES
AND FIELDS

Learning Outcomes:

At the end of this unit, the learner will be able to:

∙ Identify electrical terms, parameters, units and electrical quantities relative to

electrical charges and fields. Distinguish different types of electrical conductor
materials and properties and analyze resistance of materials considering some
factors affecting them.

∙ Analyze and calculate forces between charges, and strengths of electric fields.

Pre-Test:

Instructions: Answer the each question/problem quietly and encircle the letter of the correct
answer.

1. What is resinous electricity?

I. Positive; II. Charge; III. Negative; IV. Electric fields
(a) I & III (b) II only (c) III only (d) I, III & IV

2. The algebraic sum of the electric charge in any closed system remains constant.
(a) Law of conservation of charge (b) Coulomb’s Law
(c) Law of conservation of energy (d) Ampere’s Law

3. A charge A of +250 statC is placed on a line between two charges B of +50.0 statC and C of - 300
statC. The charge A is 5.0 cm from B and 10.0 cm from charge C. What is the force on A? (a)
150 dynes
(b) 2,150 dynes
(c) 1,250 dynes
(d) 6,550 dynes
2
Content

Electrostatics - deals with the phenomena common to electric charges (electricity) at rest.
Electrification - the process of rubbing two materials together and then separating them to acquire the
ability to attract light objects.

First Law of Electrostatics:

- Objects that are similarly charged repel each other; bodies with unlike charges attract
each other.

(a) Like charges repel

(b) Unlike charges attract

Figure 1.1 (a) Like charges repel; (b) Unlike charges attract

Positive and Negative Charges:

Vitreous electricity – positive (Du Fay)

Resinous electricity – negative

Law of Conservation of charge

- The algebraic sum of the electric charge in any closed system remains constant.

Law of Conservation of charge

- The algebraic sum of the electric charge in any closed system remains constant.
3

Displacement of Charges
- The positively charged-rubber rod causes the adjacent side of the uncharged object to
 become negatively charged, while the opposite side becomes negatively charged.

Figure 1.2 A charged body brought near an isolated conductor causes charges in the
conductor to separate. This results in an attraction of the conductor
by the charge.

Charging by Induction
- A charged object will induce a charge on a nearby conductor. In this example, a negatively
charged rod pushes some of the negatively charged electrons to the far side of a nearby
copper sphere because like charges repel each other. The positive charges that remain on
the near side of the sphere are attracted to the rod. If the sphere is grounded so that the
electrons can escape altogether, the charge on the sphere will remain if the rod is removed.

Figure 1.3 A body charge by induction

4

Electron Theory and Atomic Structure

- All matters are composed of atoms, each consisting of a nucleus, a small, tightly packed,
positively charged mass, and a number of larger, lighter, negatively charged particles called
electrons, which revolve about the nucleus at tremendous speeds.
 Figure 1.4 The atomic structure where each atom consists of a positively charged
nucleus surrounded by electrons.

Hydrogen Helium Lithium (Research)

Figure 1.5 The three simplest atoms, hydrogen, helium, and lithium, are represented
diagrammatically.

Mass of electron at rest = 9.1083 x10– 31 kg.

Mass of proton or a neutron is about 1836 times that of electron, the mass of the atom is almost entirely
concentrated in the nucleus.

Conductors and Insulators

- Free electrons – are electrons that are loosely bounded by the nucleus. The number and
freedom of motion of these electrons determine the properties of the material as a
conductor of electricity.

- Insulator or poor conductor – is a substance that contains few free electrons.

Table 1.1 Electrical Conductors

Good Conductors Poor Conductors Insulators (very poor conductors)

Silver tap water Glass

Copper Moist earth Mica

Other metals Moist Wood Paraffin

Carbon, Graphite Dry Wood Hard rubber

Certain solutions Leather Amber

 5

The Leaf Electroscope

- Is a common device for studying electrostatic phenomena, used to detect the presence of an
electric charge.

Force Between Point Charges

- The quantity of electricity, or charge Q, possessed by a body is simply the aggregate of the
amount by which the negative charge exceeds or is less than the positive charges in the
body.

Coulomb’s Law of Electrostatics

- The force F between two point charges Q and Q’ varies directly with each charge, inversely
with the square of the distance s between the charges, and is a function of the nature of the
medium surrounding the charges.

QQ
'
F=k
Eq. 1.1
2
s
QQ
F=k
' -
2 NOTE The direction of force is always along ; : '.
the line joining Q and Q
s

System of units in Electrostatics

- Two families of units are useful in the areas of electrostatics: the mks system and the system
of cgs electrostatic units (esu).

A. Electrostatic units (esu): F is expressed in dyne (dyn)

K = 1 dyn-cm2/charge2 (for empty space or vacuum)
Q and Q’ are charges in statcoulomb (statC)
s – in centimeters (cm)

B. mks units: F is expressed in Newton (N)

1
k ; is the permittivityof the medium (F / m)
=ε=
πε
4
2
12
C
ε =− ⇒
 For empty space or vacuum Air x F m
⋅
( ) : 8.85 10 /
Nm
0
2
Q and Q’ are charges in Coulomb (C)
s – in meters (m)

1 C = 3 x 10 9 statC
1 N = 105 dynes
6

______________________________________________________________________________________________________________
_ Illustrative Problem 1.1

A charge A of +250 statC is placed on a line between two charges B of +50.0 statC and C of -
300 statC. The charge A is 5.0 cm from B and 10.0 cm from charge C. What is the force on
A?

Solution:
Given: Electrostatic Units

Figure

B = +50statC A = +250statC C=
-300statC
s2
s1
5.0 cm 10.0 cm

Applying Equation 1.1: (Note: esu system)

Force on A due to B

( )( )( )
AB
QQ Fk 250 50
⋅ ⎢ ⎤
⎣⎡
AB 500
=⋅ 1 dynes
=⋅
22
s 5
⎥= ⎦
() 1

Force on A due to B = FAB is toward the right (repulsive), since they are
similarly positively charged

Also, applying Equation 1.1:

Force on A due to C

( )( )( )
 QQ F kA C 250 300
⋅ ⎢⎡ ⎤
⎣ −
AC
⎥=−
=⋅ 22
s
=⋅
750 ( )
1 dynes Attractive ( )

2
10 ⎦

Force on A due to C = FAC is toward the right (attractive), since they are
oppositely charged

Forces FAB and FAC are in the same direction to the right, hence the two forces are
aiding. Therefore
F F F dynes A AB AC = / / + / / = 500 + 750 = 1,250

Force on A due to charges B and C is towards the right

Figure:
FCA FA
FBA A A

Resultant force on charge

A due to charges B and C
(a) Forces due to charges B and
C acting on charge A
(b)
7

______________________________________________________________________________________________________________
_ Illustrative Problem 1.2

A negative point charge P of 150 μC is situated in air at the origin of a rectangular

coordinate system. A second negative charge Q of 100 μC is situated on the positive x-axis at
a distance of 300 cm from the origin, and a third positive point charge R of 200μC is
situated on the positive y-axis at a distance of 400 cm from the origin. Find (a) The
magnitude and direction of force on charge P; (b) Find the magnitude and direction of force
on charge R.

Solution:
Given: mks system

Figure
y - axis

R = +200 μC
 S3
S2 = 400 cm

Q = -100 μC
x - axis S1 = 300 cm P

= -150 μC

(a) The magnitude and direction of force on charge P

The force on charge P is due to the charges Q and R:

Applying Equation 1.1:

Force on charge P due to charge Q

8.85 x 1 0

( )( )( )
−−
QQ 66
⋅ 150 10 100 10
F kP Q -12 xx
⎢⎡
⎢⎣ −−
⎥
⎥ ⎦⎤
= ⋅−
PQ 1
Repulsive
=⋅2 =⇒ 300 10 x
14.986 ( )
() Newtons N 2
2
s

Force on charge P due to charge Q = FPQ is to the left (repulsive), since they
are similarly negatively charged.

Force on charge P due to charge R

8.85 x 1 0

( )( )( )
QQ ⎢⎡ −−

⎢⎣ −
66
⋅ 150 10 200 10 x x
⎥
⎥ ⎦⎤
F kP R -1 2
= ⋅−
PR
s 400 10 x
=⋅2 2

= − ⇒ 16.860 ( ) 2
2
()
Newtons N Attractive
 Force on charge P due to charge R = FPR is upward (attractive), since they
are oppositely charged
8

Using component method to get the resultant force FR (refer Figures a.1 and a.2):
=
∑ +∑
22
F Fx Fy P
∑ =−+
Where: ( )
Fx
14.986 0 ∑ =+
=−⇒ N to the left
14.986
Fy =⇒
0 16.860
16.860 N upward

FP (14.986) (16.860) 22.557 N

22
=+=
⎜ 16.860
⎝⎛ ⎞
=− θ
tan 1⎟ = ⎠
o
48.368
14.986
Note: The resultant force FP on charge P has a magnitude of 22.557 N, at the
second quadrant directed 48.368o with the horizontal axis.

Figures below: (a.1) Forces FPQ and FPR are indicated below with their directions
given relative to charge P; (a.2) below: Resultant force FP and component forces.
Figure (a.2)

y - axis
+R FPR
FP
Figure (a.1.) R

FPR
y - axis
axis
Q
-P FPQ Θ=48.368o
x-
axis
FPQ -Q x- P
(b) The magnitude and direction of force on charge R

The force on charge R is due to the charges P and Q:

Applying Equation 1.1:

Force on charge R due to charge P:

8.85 x 1 0

( )( )( )
QQ ⎥
⋅ ⎥ ⎦⎤
−−
F kR P -12 66

⎢⎡ 200 10 150 10 x x
⎢⎣ −
= ⋅−
RP
s 400 10 x
=⋅2 2

= − ⇒ 16.860 ( ) 2
2
()
Newtons N Attractive

Force on charge R due to charge P = FRP is downward (attractive), since

they are oppositely charged
9

Force on charge R due to charge Q:

2
2
Where : s s s By PT distance QR
=+ 2
:..
31

( ) ( ) cm
22
=+=
300 400 500

8.85 x 1 0

( )( )( )
QQ ⎥
⋅ ⎥ ⎦⎤
−−
F kR Q -1 2 66

⎢⎡ 200 10 100 10 x x
⎢⎣ −
= ⋅−
RQ 3
Attractive
=⋅2 =−⇒ 500 10 x
7.193 ( )
() Newtons N 2
2
s

Force on charge R due to charge Q = FRQ is attractive, since they are

 oppositely charged. FRQ is directed from charge R towards charge Q.

Figure (b.1) below: Forces FRP and FRQ are indicated below with their directions
given relative to charge R.

Figure (b.1)
y - axis
+R
FRQ

S2 = 400 cm

FRP S 1 -Q x
- axis
-P = 300 cm x’
- axis

Using component method to get the resultant force FR (refer Figure b.1 with respect
to horizontal x’-axis):

=
∑ +∑
22
F Fx Fy R
∑ =⋅+⋅
Fx Cos Cos o o
Where:
7.193 53.13 16.860 90 = ⇒
4.314 N to the right

∑ =−⋅−⋅
Fy Sin Sin o o
( )7.193 53.13 16.860 90

=−⇒
22.614 N downward

FR (4.314) (22.614) 23.021 N

22
=+=
 ⎜ 10
⎝⎛
22.614
⎞
=− θ
o
79.20
4.314
tan 1⎟ = ⎠

Note: The resultant force FR on charge R has a magnitude of 23.021 N, at the

fourth quadrant directed 79.20o with the horizontal axis.

Figure (b.2) below: Resultant force FR and component forces.

Figure (b.2)
y - axis

FRQ_x
Rx - axis

FRQ_y

FRP_y

FR