Soft contact lenses are comfortable to wear because
they attract the proteins in the wearer’s tears,
incorporating the complex molecules right into the
lenses.
They become, in a sense, part of the wearer. Some
types of makeup exploit this same attractive force to
adhere to the skin. What is the nature of this force?
SUMMARY
Electric charges have the following important properties:
• Unlike charges attract one another, and like charges repel one another.
• Charge is conserved.
• Charge is quantized—that is, it exists in discrete packets that are some integral multiple of the
electronic charge.
Conductors are materials in which charges move freely. Insulators are materials in which charges do not
move freely.
The smallest unit of charge known in nature is the charge on an electron or proton, which has an absolute value
of
2
Coulomb’s experiments showed that the electric force between two stationary
charged particles
• is inversely proportional to the square of the separation r between the particles and
directed along the line joining them;
• is proportional to the product of the charges q1 and q2 on the two particles;
• is attractive if the charges are of opposite sign and repulsive if the charges have the
same sign.
From these observations, we can express Coulomb’s law as an equation giving the
magnitude of the electric force (sometimes called the Coulomb force) between
two point charges:
3
Example: Three point charges lie along the x axis as shown
in Figure. The positive charge q1 = 15.0 C is at x = 2.00 m,
the positive charge q2 = 6.00 C is at the origin, and the
resultant force acting on q3 is zero. What is the x coordinate
of q3?
4
Example:
Two identical small charged spheres, each having a mass of
3.0  10-2 kg, hang in equilibrium as shown in Figure. The length of
each string is 0.15 m, and the angle  is 5.0°. Find the magnitude
of the charge on each sphere.
5
Example: Three point charges are located at the corners of an
equilateral triangle, as shown in Figure. Calculate the net
electric force on the 7.00 C charge.
6
Example: Two small beads having positive charges 3q and q are
fixed at the opposite ends of a horizontal insulating rod extending
from the origin to the point x = d. As shown in Figure, a third small
charged bead is free to slide on the rod. At what position is the
third bead in equilibrium? Can it be in stable equilibrium?
7
FIZ 140-21 Physics II
Electric field
Contents
• The Electric Field
• Electric Field Lines
• The Electric Field Due to a Point Charge
• The Electric Field Due to an Electric Dipole
• The Electric Field Due to a continues charge distribution
• Line of Charge
• Charged Disk
• A Dipole in an Electric Field
• Torque
• Potential energy
• Work
9
10
Electric Fields
In this chapter we will introduce the concept of an electric field. As long as
charges are stationary, Coulomb’s law describes adequately the forces among
charges. If the charges are not stationary we must use an alternative approach
by introducing the electric field (symbol E ). In connection with the electric
field, the following topics will be covered:
-Calculating the electric field generated by a point charge.
-Using the principle of superposition to determine the electric field created by
a collection of point charges as well as continuous charge distributions.
-Once the electric field at a point P is known, calculating the electric force on
any charge placed at P.
-Defining the notion of an “electric dipole.” Determining the net force, the net
torque, exerted on an electric dipole by a uniform electric field, as well as the
dipole potential energy.
11
We discussed Coulomb’s law, which gives the force between two point
charges. The law is written in such a way as to imply that q2 acts on q1 at
a distance r instantaneously (“action at a distance”):
q1 q2
F
4 0 r 2
1
Electric interactions propagate in empty space with a large but finite
speed (c = 3108 m/s). In order to take into account correctly the finite
speed at which these interactions propagate, we have to abandon the
“action at a distance” point of view and still be able to explain how q1
knows about the presence of q2 . The solution is to introduce the new
concept of an electric field vector as follows: Point charge q1 does not
exert a force directly on q2. Instead, q1 creates in its vicinity an electric
field that exerts a force on q2 .
charge q1  generates electric field E  E exerts a force F on q2
12
An electric field is said to exist in the region of space around a charged object. When another
charged object enters this electric field, an electric force acts on it.
As an example, consider Figure, which shows a small positive test charge q0 placed near a
second object carrying a much greater positive charge Q. We define the strength (in other
words, the magnitude) of the electric field at the location of the test charge to be the electric
force per unit charge, or to be more specific
13
F
E
q0
Definition of the Electric Field Vector
Consider the positively charged rod shown in the figure.
For every point P in the vicinity of the rod we define
the electric field vector E as follows:
1. We place a positive test charge q0 at point P.
2. We measure the electrostatic force F exerted on q0
by the charged rod.
3. We define the electric field vector E at point P as:
E
F
.
q0
SI Units: N/C
From the definition it follows that E is parallel to F .
Note : We assume that the test charge q0 is small enough
so that its presence at point P does not affect the charge
distribution on the rod and thus alter the electric
field vector E we are trying to determine.
14
E is the field produced by some charge external to the test charge—it is not the field produced by the test
charge itself. Also, note that the existence of an electric field is a property of its source. For example, every
electron comes with its own electric field.
15
The vector E has the SI units of newtons per coulomb
(N/C), and, its direction is the direction of the force a
positive test charge experiences when placed in the
field.
We say that an electric field exists at a point if a test
charge at rest at that point experiences an electric
force. Once the magnitude and direction of the
electric field are known at some point, the electric
force exerted on any charged particle placed at that
point can be calculated from E=F/q0
16
17
E
qo P
q
Electric Field Generated by a Point Charge
Consider the positive charge q shown in
the figure. At point P a distance r from
q we place the test charge q0 . The force
exerted on q0 by q0 is equal to:
q q0
F
4 0 r 2
1
r
F
1 q q0
E 
q0 4 0 q0 r 2
q

4 0 r 2
1
The magnitude of E is a positive number.
q
E
2
4 0 r
1
In terms of direction, E points radially
outward as shown in the figure.
If q were a negative charge the magnitude
of E would remain the same. The direction
of E would point radially inward instead.
18
at any point P, the total electric field due to a group of charges equals the vector
sum of the electric fields of the individual charges.
This superposition principle applied to fields follows directly from the
superposition property of electric forces. Thus, the electric field of a group of
charges can
where ri is the distance from the ith charge qi to the point P (the location of the
test charge) and 𝒓ො 𝑖 is a unit vector directed from qi toward P.
19
Electric Field Generated by a Group of Point Charges. Superposition
The net electric electric field E generated by a group of point charges is equal
to the vector sum of the electric field vectors generated by each charge.
In the example shown in the figure, E  E1  E2  E3 .
Here E1 , E2 , and E3 are the electric field vectors generated by q1 , q1 , and q3 ,
respectively.
Note: E1 , E2 , and E3 must be added as vectors:
Ex  E1x  E2 x  E3 x , E y  E1 y  E2 y  E3 y , Ez  E1z  E2 z  E3 z
20
Example: Electric Field Due to Two Charges
A charge q1 = 7.0 C is located at the origin, and a second charge q2= -5.0
C is located on the x axis, 0.30 m from the origin. Find the electric field at
the point P, which has coordinates (0, 0.40) m.
21
Example: Electric Field Due to Two Charges
A charge q1 = 7.0 C is located at the origin, and a second charge q2= -5.0
C is located on the x axis, 0.30 m from the origin. Find the electric field at
the point P, which has coordinates (0, 0.40) m.
22
Electric Dipole
A system of two equal charges of opposite sign
  q  placed at a distance d apart is known as an "electric
+q/2
+q/2
-q
dipole." For every electric dipole we associate
a vector known as "the electric dipole moment"
(symbol p ) defined as follows:
The magnitude p  qd
The direction of p is along the line that connects
the two charges and points from - q to  q.
Many molecules have a built-in electric dipole
moment. An example is the water molecule (H 2 O).
The bonding between the O atom and the two H
atoms involves the sharing of 10 valence electrons
(8 from O and 1 from each H atom).
The 10 valence electrons have the tendency to remain closer to the O atom.
Thus the O side is more negative than the H side of the H 2 O molecule.
23
Electric Field Generated by an Electric Dipole
24
Electric Field Generated by an Electric Dipole
25
Electric Field Generated by an Electric Dipole
26
Electric Field Generated by an Electric Dipole
We will determine the electric field E generated by the
electric dipole shown in the figure using the principle of
superposition. The positive charge generates at P an electric
1 q
field whose magnitude E(  ) 
. The negative charge
2
4 0 r
1
q
creates an electric field with magnitude E(  ) 
.
2
4 0 r
The net electric field at P is E  E(  )  E(  ) .

1 q
q
1 
q
q
E



 2  2
2
2

4 0  r
r  4 0   z  d / 2 
 z  d / 2  
2
2

d 
d
d

 
E
1    1   
We assume:
2 
4 0 z  2 z 
2z
 2 z  
q  d   d  
qd
1 p
E
1


1

=


 

4 0 z 2 
z 
z   2 0 z 3 2 0 z 3
q
1
27