Transcript Document

We define the electric field by the force it exerts on a test charge q0:
E=
F0
q0
This is your second starting equation. By convention the direction of the electric field is the
direction of the force exerted on a POSITIVE test charge. The absence of absolute value signs
around q0 means you must include the sign of q0 in your work.
Any time you know the electric field, you can use this equation to calculate the force on a
charged particle in that electric field. F = qE
The units of electric field are Newtons/Coulomb.
 F0 
N
 E  =   =
q0  C
Later you will learn that the units of electric field can also be
expressed as volts/meter:
N V
E = =
C m
The electric field exists independent of whether there is a
charged particle around to “feel” it.
Electric Field Lines
The charge on the right is twice the magnitude of the charge on the left (and opposite
in sign), so there are twice as many field lines, and they point towards the charge
rather than away from it.
Remember: the electric field direction is the direction a + charge would
feel a force.
+
A + charge would be repelled by another + charge.
Therefore the direction of the electric field is away from positive (and towards negative).
Electric Field Lines
Like charges (++)
Opposite charges (+ -)
Electric Field Lines: a graphic concept used to draw
pictures as an aid to develop intuition about its behavior.
The text shows a few examples. Here are the drawing rules.
•
•
•
•
•
E-field lines begin on + charges and end on - charges. (or infinity).
They enter or leave charge symmetrically.
The number of lines entering or leaving a charge is proportional to the charge
The density of lines indicates the strength of E at that point.
No two field lines can cross.
5
QUICK QUIZ 1Rank the magnitudes of the electric field at
points A, B, and C in the figure below, largest
magnitude first.
QUICK QUIZ 2Figure shows the electric field lines for two point charges
separated by a small distance. (a) Determine the ratio q1/q2.
(b) What are the signs of q1 and q2?
Electric Field Lines
A parallel-plate capacitor consists of two
conducting plates with equal and opposite
charges. Here is the electric field:
The Electric Field Due to a Point Charge
q1q 2
F =k 2 ,
12
r12
Coulomb's law says
... which tells us the electric field due to a point charge q is
E q =k
q
r
2
, away from +
q
E=k 2
r
…or just…
We define r̂ as a unit vector from the source point to the field point:
source point
+
r̂
The equation for the electric field of a point charge then becomes:
q
E=k 2 rˆ
r
field point
Electric Field: Example 1
Find the electric force on a proton placed in an
electric field of 2.0 X 104 N/C
E=
F
q
F = qE
F = (1.602 X 10-19 C)(2.0 X 104 N/C)
F = 3.2 X 10-15N
The Electric Field
Example 21-6: Electric field of a single point charge.
Calculate the magnitude and direction of the electric field at a point P which
is 30 cm to the right of a point charge Q = -3.0 x 10-6 C.
E=k
q
r2
E = (9.0 X 109 N-m2/C2)(3.0 X 10-6 C)
(0.30 m)2
E = 3.05 X 105 N/C
towards the charge
(b) due to a positive charge Q, each 30 cm from P.
Solution: Substitution gives E = 3.0 x 105 N/C. The
field points away from the positive charge and
towards the negative one
Example
• A charge
is located at the origin of the x axis. A
second charge
is also on the x axis 4 m from
the origin in the positive x direction
(a) Calculate the electric field at the midpoint P of the line joining the
two charges.
(b) At what point on that line is the resultant field zero?
(a) Since q1 is positive and q2 is negative, at any point
between them, both electric fields produced by them
are the same direction which is toward to q2. Thus,
11
The resultant electric field E at P is
(b) It is clear that the resultant E can not be zero at any point between q1 and
q2 because both E1 and E2 are in the same direction. Similarly E can
not be zero to the right of q2 because the magnitude of q2 is greater then q1
and the distance r is smaller for q2 than q1. Thus, can only be zero to the
left of q1 at some point to be found. Let the distance from
to q1 be x.
Apparently, we need to take x which is positive.
12
The Electric Field
Example : E at a point between two charges.
Two point charges are separated by a distance of 10.0 cm. One has a
charge of -25 μC and the other +50 μC. (a) Determine the direction and
magnitude of the electric field at a point P between the two charges that is
2.0 cm from the negative charge. (b) If an electron (mass = 9.11 x 10-31 kg)
is placed at rest at P and then released, what will be its initial acceleration
(direction and magnitude)?
Solution: a. The electric fields add in magnitude, as both are directed towards the
negative charge. E = 6.3 x 108 N/C.
b. The acceleration is the force (charge times field) divided by the mass, and will
be opposite to the direction of the field (due to the negative charge of the
electron). Substitution gives a = 1.1 x 1020 m/s2
The Electric Field
E:
above two point charges.
Calculate the total electric
field (a) at point A and (b) at
point B in the figure due to
both charges, Q1 and Q2.
Solution: The geometry is shown in the figure. For each point, the process is:
calculate the magnitude of the electric field due to each charge; calculate the x and
y components of each field; add the components; recombine to give the total field.
a. E = 4.5 x 106 N/C, 76° above the x axis.
b. E = 3.6 x 106 N/C, along the x axis.
23-7 Motion of a Charged Particle in a Uniform Electric
Field
Consider a particle with charge q and mass m, moving in a region of
space where the electric field E is constant.
As always, the force on a charge q is F = q E = m a.
Constant acceleration,
a = F/m = (q E /m)
and will move in a parabola.
- - - - - - - - - - - - •If E is uniform (that is, constant in
magnitude and direction), then the
acceleration is constant.
• If the particle has a positive charge, then its
acceleration is in the direction of the electric
field.
•If the particle has a negative charge, then its
acceleration is in the direction opposite the
electric field.
-
F
+ + + + + + + + + + + + +
E
A positive point charge q of mass m is released from rest in a
uniform electric field E directed along the x axis. Describe its
motion.
we can apply the equations of kinematics in one dimension
Taking xi = 0 and vxi = o
The kinetic energy of the charge after it has moved a distance x = xf-xi, is
We can also obtain this result from the work–kinetic energy theorem because the
work done by the electric force is Fex = qEx and W = ∆K
An electron enters the region of a uniform electric
field with vo=3.00x106 m/s and E= 200 N/C. The
horizontal length of the plates is l = 0.100 m.
(a) Find the acceleration of the electron while it is
in the electric field.
(b) Find the time it takes the electron to travel through the field.
(c) What is the vertical displacement y of the electron while it is in the field?
If the separation between the plates is less than this, the electron
will strike the positive plate.
Example - Motion of a charged particle in an
Electric Field
Determine the final velocity and kinetic energy of an electron
released from rest in the presence of a uniform electric field of
300 N/C in the x direction after a period of 0.5 ms.
E
-e
F



N ˆ

19
ˆ
F  qE  eEi
F  1.60  10 C  300 i
C


17
F  4.80 10 N iˆ

 F  4.80 1017 N iˆ


13 m  ˆ
a 
a   5.3 10
i
2 
m
9.111031 kg
s 

E
  
v  vo  at
-e
F


13 m ˆ 
6

v  0   5.3 10
i
0
.
5

10
s

2
s 


7 m ˆ
v  2.6  10
i
s
1
K  mv 2
2


1

31
7 m 
K  9.1110 kg   2.6 10

2
s

K  3.16  10
16
J
2
Example
• Electron accelerated by electric field. An electron (mass m = 9.1x10-31kg)
is accelerated in the uniform field E (E=2.0x104N/C) between two parallel
charged plates. The separation of the plates is 1.5cm. The electron is
accelerated from rest near the negative plate and passes through a tiny hole
in the positive plate. (a) With what speed does it leave the hole? (b) Show
that the gravitational force can be ignored. Assume the hole is so small that it
does not affect the uniform field between the plates.
The magnitude of the force on the electron is F=qE and is directed to the right. The
equation to solve this problem is
F  qE  ma
F qE

a
m
m
The magnitude of the electron’s acceleration is
Between the plates the field E is uniform, thus the electron undergoes a uniform acceleration
eE 1.6  10 C  2.0  10
a

m
 9.110 kg 
19
e
31
4
N /C
  3.5 10
15
m s2
Example
Since the travel distance is 1.5x10-2m, using one of the kinetic eq. of motions,
v2  v02  2ax v  2ax  2  3.5 1015 1.5 102  1.0 107 m s
Since there is no electric field outside the conductor, the electron continues moving with this speed
after passing through the hole.
• (b) Show that the gravitational force can be ignored. Assume the
hole is so small that it does not affect the uniform field between
the plates.
The magnitude of the electric force on the electron is



Fe  qE  eE  1.6  1019 C 2.0  104 N / C  3.2  1015 N
The magnitude of the gravitational force on the electron is


FG  mg  9.8 m s 2  9.1 1031 kg  8.9  1030 N
Thus the gravitational force on the electron is negligible compared to the electromagnetic force.
The electron and proton of a hydrogen atom are separated (on the average) by a
distance of approximately 5.3 x10-11 m. Find the magnitudes of the electric force
and the gravitational force between the two particles.
Using Newton’s law of gravitation
Thus, the gravitational force between charged atomic
particles is negligible when compared with the electric
force.