Week 1: Electric Force and Electric Fields
Learning Outcomes: At the end of this week students should be able to:





Explain the concept of electric charges and electric force.
Apply Coulomb’s Law to calculate electric force between two or more point charges.
Calculate electric fields as vectors due to point charges.
Define linear, surface and volume charge densities in a uniform distribution of charges.
Calculate electric fields due to uniform charge distributions by treating small charged elements
as point charges and integrating electric field vectors set up by each element.
1.0
Introduction
We begin our study of this course with the branch of Physics that is concerned with electric and
magnetic phenomena. You are surrounded by devices that depend on the Physics of electromagnetism,
which is a combination of electric and magnetic phenomena. This Physics plays a major role in the
operation of electronic devices such as computers, television, radio, telecommunications, and
household lighting to name a few. The electromagnetic force between charged particles is one of the
fundamental forces of nature. We will begin this topic by describing some basic properties of electric
charges and forces between them.
1.1
Electric Charges
There are two types of electric charges; positive (+) and negative (-) charge. The nature of the charges is
such that like charges repel and unlike charges attract each other. Ordinary objects often carry equal
number of positive and negative charges, and the object is said to be electrically neutral. Objects that
carry excess of one charge over the other have an unbalanced charge. The unit of electric charge is the
Coulomb (C). Charge is quantized as a multiple of the electron (-) or proton (+) charge. The charges and
masses of the electron, proton and neutron are given in Table 1.1.
Table 1.1: Charge and Mass of the Electron, Proton and Neutron
Particle
Charge (C)
Mass (kg)
Electron (e)
- 1.602 × 10-19
9.109 × 10-31
Proton (p)
+ 1.602 × 10-19
1.673 × 10-27
Neutron (n)
0
1.675 × 10-27
Notice that the electron and proton are identical in the magnitude of their charge but vastly different in
mass. On the other hand the proton and neutron are similar in mass but vastly different in charge.
1.2
Coulomb’s Law
If two charges are brought near each other, they each exert an electrostatic force on the other. The
direction of the force vectors depends on the sign of the charges. If the particles have the same sign of
charge, they repel each other. If the particles have opposite signs of charge, they attract each other.
Figure 1.1 shows the direction lines of forces on charges of the same and opposite signs.
1

Figure 1.1: Two point charges separated by distance r exert a force on each other. The force F12

exerted by q1 on q2 is equal in magnitude and opposite in direction to the force F21 exerted by q2 on q1.
(Serway & Jewett, 2014 pg. 696)
The magnitude of the electric force between two point charges is given by Coulomb’s law:
F
1
q1 q2
4 o
r2
where the constant,  o is the permittivity of free space = 8.85 × 10-12 C2/N m2.
Example 1:
The electron and proton of a hydrogen atom are separated by a distance of approximately 5.3 × 10-11 m.
Find the magnitude of the electric force between the two particles.
Solution:
This is a simple substitution problem. The magnitude of electric force will be evaluated using Coulomb’s
law.
F
1
e e
4 o
r2
1.6  1019   8.2  108 N
1

4 8.85  10 12  5.3  10 12 2
2
Example 2:
Consider three point charges located at the corners of a right angle triangle as shown in figure 1.2,
where q1 = q3 = 5.00 µC, q2 = -2.00 µC, and a = 0.100 m. Find the resultant force on q3.
Solution:
To solve this problem we need to consider the net force on q3. We determine the force vectors acting

on q3 due to q1 and q2 using Coulombs law. The figure 1.2 shows the force exerted by q1 on q3 is F13 .

The force exerted by q2 on q3 is F23 . The force directions are due to the fact that like charges repel and


unlike charges attract. Hence, the force F23 is directed to the negative x direction, and the force F13
makes an angle of 45o with the x axis.
2


Figure 1.2: (Example 2) The force exerted by q1 on q3 is F13 . The force exerted by q2 on q3 is F23 .
(Serway & Jewett, 2014 pg. 696)


We find the magnitudes of F23 and F13 :
F23 
F13 
1
q2 q3
4 o
a2
1
q1 q3
4 o

2.00 106 5.00 106   8.99 N
1
4 8.85  10 12 
0.1002

1
5.00  10 6 5.00  10 6 

 11.2 N
2
2
4 8.85  10 12 
2 0.100
2a
 

To determine the resultant force, F3 we add the force components:
F13x  11.2 N  cos 45o  7.94 N
F13y  11.2 N  sin 45o  7.94 N
F3 x  F13x  F23x  7.94   8.99   1.04 N
F3 y  F13 y  F23 y  7.94  0  7.94 N



The resultant force acting on q3 in unit-vector form: F3   1.04 ˆi  7.94 ˆj N
Example 3:
Two identical small charged spheres each have a mass of 0.03 kg hang in equilibrium as shown in figure
1.3. The length L of each string is 0.15 m, and the angle θ is 5.00o. Find the magnitude of the charge on
each sphere.
Solution:
The two spheres repel each other and move outwards from the center as shown in figure 1.3 (a). The
repulsive electric force is in equilibrium with the gravitational force. A close inspection of the figure 1.3
(b) shows that the sphere on the left is in equilibrium under the application of the tension force T from
the string, the electric force F from the other sphere, and the weight force mg.
3
Figure 1.3: (Example 3) (a) Two identical spheres each carrying a charge q suspended in equilibrium. (b)
Diagram of the forces acting on the left sphere in (a). (Serway & Jewett, 2014 pg. 698)
Applying the principles of bodies in equilibrium:
 F  0, F  T sin  , and
 F  0, mg  T cos 
x
e
y
Dividing the two equations gives
Fe
T sin 
 Fe  mg tan   0.039.8 tan 5o  0.0257 N

mg T cos 
We solve Coulomb’s law for charge q on each sphere, where a  L sin   0.15 sin 5o  0.0131 m
Fe 
1
q
2
4 o 2a 2

q  16 o Fe a 2  16 8.85  10 12 0.0257 0.0131
2
q  4.43  10 8 C
DIY 1.1:
Three point charges lie along the x axis as shown in figure 1.4. The positive charge q1 = 15.0 µC is at x =
2.00 m, the positive charge q2 = 6.00 µC is at the origin, and the net force acting on q3 is zero. What is
the x-coordinate of q3?
Solution:
x = 0.775 m
4
Figure 1.4: (DIY 1.1) Three point charges placed on the x axis. The sum of forces F13 and F23 acting on
charge q3 due to q1 and q2 respectively must add for the net force on q3 to be zero. (Serway & Jewett,
2014 pg. 697)
1.3
Electric Field
The electric field is a region in space where an electric force acts on a charged object. Consider a test
charge qo placed at a point P near a source charge Q as shown in figure 1.5. The source charge produces
an electric field E at point P that exerts a force on the test charge. If the force exerted on the test charge
is Fe , we define the electric field at a point in space as the electric force Fe acting on a positive test
charge qo placed at that point divided by the test charge:

 Fe
E
qo
Unit for electric field is N/C.
Figure 1.5: A small positive test charge placed at point P near an object carrying a much larger positive
charge Q. (Serway & Jewett, 2014 pg. 699)
According to Coulomb’s law, the force exerted by a charge q on the test charge, qo is:

Fe 
q qo
rˆ
4 o r 2
1
where r̂ is a unit vector indicating the direction of the force on the charges. Thus the electric field at a
point P at the position of the test charge by definition is:


Fe
1 q
Ee 

rˆ
qo 4 o r 2
5
This shows that the source charge at a point sets up an electric field in the direction of the force acting
on a positive test charge placed at that point. Thus the direction of the electric field is away from the
source if the charge is positive and directed towards the source if it is negative as shown in figure 1.6.
(b)
(a)
Figure 1.6: At a point P the electric field E (a) is directed radially outward near a positive source charge q
and (b) is directed towards the negative source charge q. (Serway & Jewett, 2014 pg. 700)
We note here that the electric field is a vector quantity. Thus the total electric field from a group of
source charges is a vector sum of the fields from each charge.
Example 4:
Charges q1 and q2 are located on the x axis at distances a and b respectively from the origin as shown in
figure 1.7. Find the components of the net electric field at the point P, which is at position (0, y).
y
Figure 1.7: (Example 4) The total electric field E at P equals the vector sum E1 + E2, where E1 is the field
due to charge q1 and E2 is the field due to charge q2. (Serway & Jewett, 2014 pg. 702)
Solution:
We solve this problem by evaluating the electric fields produced by each charge at point P. The vector
sum of the fields provides the total electric field.
The magnitude of the electric field at P due to charge q1:
6
E1 
1
q1
4 o r1
2

q1
1
where r1  a 2  y 2
2
4 o a  y 2
2
The magnitude of the electric field at P due to charge q2:
E2 
1
q2
4 o r2
2

q2
1
4 o b  y
2
where r2  b 2  y 2
2
2
We write the electric fields in unit vector form as:
1  q1
ˆi  q1 sin  ˆj
 2
cos


4 o  a  y 2
a2  y2


q

1  q2
 2
E2 
cos  ˆi  2 2 2 sin  ˆj
2

4 o  b  y
b y


E1 
Thus the components of the net electric field vector are:
q

1  q1
 2
cos   2 2 2 cos  
2

4 o  a  y
b y

q2

1  q1
 2

Ey 
sin


sin

2
2
2

4 o  a  y
b y

Ex 
1.4
Electric Field of a Continuous Charge Distribution
We have only considered electric fields due to point charges, i.e., charges that have zero size. Often we
have continuous distribution of charges around objects as a collection of discrete charges. The charge
can be described as continuously distributed along some line, over some surface, or throughout some
volume. To realize the behavior of charge distributions we use the concept of charge density in the
following way:
 If a charge Q is uniformly distributed throughout a volume V, the volume charge density ρ is
defined by:


where ρ has the units (C/m3)
If a charge Q is uniformly distributed on a surface of area A, the surface charge density ς is
defined by:


Q
V
Q
A
where ς has the units (C/m2)
If a charge Q is uniformly distributed along a line of length l, the linear charge density λ is
defined by:

Q
l
where λ has the units (C/m)
Thus if a charge is nonuniformly distributed over a volume, surface or line, the amounts of charge dq in
a small volume, surface or length element are:
dq   dV
dq   dA
dq   dl
The process of evaluating electric fields due to continuous charge distribution uses the concept
developed with point charges. First we divide the charge distribution into small elements, each with a
small charge Δq as shown in figure 1.8. We calculate the electric field due to one of these elements as if
they were point charges at the point P. We then calculate the total electric field at P due to the charge
7
distribution by adding the contributions from all the charge elements by the principle of superposition.
The electric field at point P due to one charge element carrying charge Δq is:

E 
q
rˆ
4 o r 2
1
where r is the distance from the charge element to point P and r̂ is the unit vector directed from the
element towards P.
Figure 1.8: The electric field at P due to a continuous charge distribution as a vector sum of field
contributions from all small charge elements Δq. (Serway & Jewett, 2014 pg. 704)
The total electric field is then the vector sum of the field contributions from all elements in the charge
distribution.

E
qi
1
 r rˆ
4
2
o
i
i
i
where i refers to the ith element in the distribution. For a continuous charge distribution the total field
at P is evaluated in the limit as qi  0 :

E
1
lim 
4 o pi 0
i
qi
1
dq
rˆi 
rˆ
2

4 o r 2
ri
where the integration is over the entire charge distribution. In calculating the electric field due to
continuous charge distribution the strategy we use is exactly as we have described above. We apply the
superposition principle by replacing the vector sums by vector integrals to evaluate the total electric
field from individual charges. The charge distribution is divided into infinitesimal pieces and the vector
sum is carried out by integrating over the entire charge distribution. The examples that follow here will
demonstrate the technique.
Example 5
A rod of length l has a uniform positive charge per unit length λ and a total charge Q. Calculate the
electric field at a point P that is along the long axis of the rod and a distance a from one end as shown in
figure 1.9.
8
Figure 1.9: (Example 5) The electric field at point P due to a uniformly charged rod lying along the x axis.
(Serway & Jewett, 2014 pg. 705)
Solution:
We conceptualize the problem by dividing the rod into small elements where each element carries a
small charge dq. Each of the elemental charges contributes a small electric field dE at point P and is
directed in the negative x direction. Notice that since the rod is positively charged all the pieces have
field contributions in the same direction. Thus the sum of their contributions can be handled without
the need to add vectors.
Let us now treat an element on the rod and evaluate the electric field at P (see figure 1.9). Let dx be the
length of the small segment, and dq is the charge on that segment. Because the rod has a uniform
charge distribution with a charge density λ, we see that dq = λ dx. The magnitude of the electric field
due to this segment at point P is:
dE 
dq
1  dx

2
4 o x
4 o x 2
1
Notice we treated the segment as a point charge by applying the same formula used for electric fields
due to point charges. The total electric field is the integral over the entire length of the rod:
l a
E   ke
a
where k e 
1
4 o
dx
x2
is referred to as the coulomb constant.
Evaluating the integral gives:
l a
l a
dx
 1
E  k e   2  k e   
x
 x a
a
Evaluating the limits and then simplifying gives:
k eQ
1
 1
1 

1
E  k e  
   ke  

 la a 
 a l  a  a l  a 
where λ = Q/l
Notice if if point P is very far away from the rod, i.e., a >> l , then we can neglect l in the denominator
since (l + a) ≈ a and E  k eQ a 2 . This is exactly the form expected from a point charge. This shows that
from a distance far away from the charge distribution one cannot distinguish between the object with
charge distribution and a point charge.
9
Example 6
A ring of radius a carries a uniformly distributed charge Q. Calculate the electric field due to the ring at
the point P lying a distance x from its center along the central axis perpendicular to the plane of the ring
as shown in figure 1.10.
Figure 1.10: (Example 6) A uniformly charged ring of radius a. (a) The electric field at point P on the x axis
due to an element of charge dq. (b) The total electric field at P is the vector sum of the elemental
contributions and is directed along the x axis only. (Serway & Jewett, 2014 pg. 706)
Solution:
A closer inspection of figure 1.10a shows the electric field contribution dE at P due to one segment of
charge at the top of the ring. This field vector can be resolved into components dEx, parallel to the axis
of the ring and dE  perpendicular to the axis. The figure 1.10b shows the electric field contributions
from two segments on opposite sides of the ring. Because of symmetry we note that the perpendicular
component of the fields cancel out. This is true for all the pairs of segments around the ring. So we can
ignore the perpendicular component field and focus only on the parallel component which simply adds.
Evaluating the parallel component of the field gives:

dq
dq
dE  k e 2  k e 2
r
a  x2

dq
dE x  dE cos   k e 2
cos 
a  x2
x
x
cos   
r
a2  x2
ke x
dq
x
dE x  k e 2

dq
2
2
2
2
a x a x
a  x 2 3 2
We integrate over the circumference of the ring to obtain the total field a P:
Ex 
ke x
ke x
dq 
Q
a  x   a  x 
2
2 32
2
2 32
Note if x = 0, E= 0 i.e., the electric field at the center of the ring is zero as expected due to symmetry of
the charge distribution. If x >> a, the equation reduces to k eQ x 2 by the far point approximation. So the
ring acts like a point charge for locations far away.
10
Example 7
A disk of radius R has a uniform surface charge density ς. As shown in figure 1.11 calculate the electric
field at a point P that lies along the central perpendicular axis of the disk and a distance x from the
center of the disk.
Figure 1.11: (Example 7) A uniform charged disk of radius R. (Serway & Jewett, 2014 pg. 707)
Solution:
We first divide the disk into a set of concentric rings each with a charge dq. Use the result obtained from
example 6 which gives the electric field at point P due to a single ring. Also note that we have a surface
charge density so:
dq   dA   2 r dr 
where dA is the area of the ring segment which is 2πr dr (see figure 1.11). By symmetry we again note
that the rings only contribute fields that are parallel to the axis of the disk. Thus the electric field at P
due to the entire disk must be parallel and in the positive x-direction. We write the parallel field
component from the ring element as:
dE x 
ke x
r  x 
2
2 32
dq 
ke x
r  x 
2
2 32
2 r dr 
To obtain the total field we integrate the expression over the limits r = 0 to r = R:
R
E x  k e x 
0
2r dr
r  x 
2
2 32
Solving the integration and simplifying gives (see lecture/recording for the integration technique used):
R
 2 


x
E x  k e x 
 2k e 1 
2
2 
2
2 
R x 
 r  x 0

11
Note for large values of x it can be shown by a series expansion that the electric field is equivalent of a
point charge Q. For the field close to the disk along the axis where x << R, the expression in the brackets
reduces to unity to give the near field approximation:
E  2 k e 

2o
Another way of looking at this is by letting the radius of the disk grow so the disk becomes an infinite
plane of charge. The result is the same as above by letting R   .
1.5

Summary
Coulomb’s Law states that the magnitude of electric force exerted by a point charge q1 on a
second point charge q2 is:
F  ke
q1 q2
r2

At a distance r from a point charge q the electric field due to the charge is:

The electric field at some point in space is defined as the electric force that acts on a small
positive test charge placed at that point divided by the magnitude of the test charge:

q
E  k e 2 rˆ
r


Fe
Ee 
qo

The electric field at some point due to a continuous charge distribution is:

dq
E  k e  2 rˆ
r
Self-Assessment Problems
You can now try the self assessment problems to gauge your level of understanding for this week. This
activity is not assessed.
Tutorial Assignment/Online Quiz
You are required to complete the assignment questions and submit online. Take note of the due dates
as these are assessed towards your coursework. You may submit a scanned written work in pdf, or a
word processed document. Make sure the documents submitted are very clear and that it has your
name and ID number. Format your document name as “Wk1_tut_S1101xxxx.pdf” before submitting
online.
References
Serway & Jewett 2014, Physics for Scientists and Engineers, 9th ed., Brooks/Cole Cengage Learning, USA.
12