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Electric Fields
Electric Fields – The force of a charged object on a point charge. It is defined as a force per unit charge.
Very similar to the force of gravity (force per unit mass)
N
Newton
Unit  Coulomb 
C
Therefore, the force on a point charge in an electric field is:
which is similar to Fg = mg
Think of E as being like g
 DEMO 16-4
On the board or on an overhead, use the idea of a small positive test charge to map out the force
field from a positive charge q located at the origin. Use arrows to indicate the direction and magnitude of the
field at many points. (Figure 16-23 will give you a start.) Remind the students that the field is threedimensional, not flat like the board. Once you have the field set up, be sure to discuss the direction of the
Fb
force on a negative test charge.
b
a
Fa
Electric Field Lines:
Electric lines of force - lines drawn so that tangent to a line shows direction of the force
Direction of E field  direction of force on a positive charge
1. Electric field vector E is tangent to electric field lines at each point
2. The density of field lines is proportional to the strength of the field
 When E is large, lines are close together.
 E is small, lines are far apart.
Electric field exists around charged object in space.
 Any charged object in the field experiences force
 Like mass in gravity field.
Q
qo
E
Electric field  vector quantity
1
Electric field lines for four arrangements of charges.
Summary of field lines:
1. Field lines indicate the direction of the field; the field is tangent to the line.
2. The magnitude of the field is proportional to the density of the lines.
3. Field lines start on positive charges and end on negative charges; the number is proportional to
the magnitude of the charge.
How to solve problems with Electric Fields:
1. Draw a diagram; show all charges, with signs, and electric fields and forces with directions
2. Calculate forces using Coulomb’s law
3. Add forces vectors to get result (if there is more than 1 field)
Examples:
1. An electric field has a field intensity of 2.0 x 104 N/C. If the force acting on a test charge is 6.2 N, what
is the magnitude of the test charge?
This is a simple plug and chug problem.
E
F
q
q
F
E

6.2 N
2.0 x 104
N
C

3.1 x 104 C
Once we can find forces, we can use Newton’s laws to calculate all sorts of wondrous things. Like
velocity or acceleration!
2
2. An electron travelling at 2.3 x 105 m/s with a direction as shown in the drawing enters a uniform 280
N/C electric field. (a) Analyzing the electric lines of force, what is the direction of the force acting on
the particle? (b) What is the magnitude of the force? (c) What is the acceleration acting on the electron?
(d) If the electron travels a distance of 3.0 mm in the field, what is the distance it will be deflected from
its original path?
v
E
(a) The direction of the force will be down. The arrows on the lines of force show the direction of a force acting
on a positive test charge. Since the electron has a negative charge, the force will be in the opposite direction.
(b) We know the charge of an electron, so we can figure out the force acting on it.
E
F
q
F  Eq  280


N
1.6 x 1019 C  448 x 1019 N
C

4.5 x 1017 N
(c) We can use the second law to find the acceleration. We’ll need the mass of the electron, but we can look
that up.
F  ma
a
a  0.494 x 1014
m
s2
F
m
 4.5 x 1017

kg  m 
1
2 
s  9.11 x 1031 kg
4.9 x 1013



m
s2
(d) The electron is moving horizontally at a constant velocity. It will be accelerated downward by an
electromagnetic force and also by gravity. Looking at the acceleration from the electric field, we can see that
the acceleration from gravity is way way smaller, so we can ignore gravity - it is totally insignificant. (Hey,
what is a lousy 9.8 m/s2 compared with 1013 m/s2?)
We need to figure out the length of time it will be accelerated. It is moving through a field – this is when it will
be accelerated – a distance of 3.0 mm. We know the horizontal speed, so we can find the time to travel that
distance.
x
x  vt t 
v




1
 0.0030 m 
 0.0013 x 105 s  1.3 x 108 s

m
 2.3 x 105

s 

Armed with the time, we can find the distance it will be displaced.
1
y  at 2
2
1
m
  3.2 x 1013 2  1.3 x 108 s
2
s 


2

2.1 x 103 m or
2.1 mm
3
Shape of Conductor:
Spherical:
sphere  uniform distribution
charge density  charge per unit area is uniform.
Non-spherical -- charge not uniform.
Charge is denser at greatest curvature.
Electric Fields and Objects:
The electric charge on an object, such as a conductive sphere, say for example, always is on the outside of the
object. It is on the outer surface. Why is this so?
Well, it’s very fundamental. The free electrons repel each other. This means that they try to get as far away
from one another as they possibly can. In order to do this, they collect on the outside surface and spread out. If
they were on the inside, they would be closer together, so they don’t do that.
Electrons repel each
other to the outside
Electrons end up on
the outer surface
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