Scalar field: Temperatures
73
77
82
84
83
72
71
75
77
68 64
80
73
82
88
55
66
80 88
75
88
90
83
92
91
The shown temperatures are samples of the field
Vector field: Winds
73
77
72
71
82
84
83
88
75
68 64
80
73
57 56 55
66
88
80
75
90
83
92
91
77
3. Electric field
Coulomb’s low and electric field

r
Q
q - test charge
P


Qq
Q
F  k 2 rˆ  qk 2 rˆ  qE
r
r

r
rˆ 
r
Charge Q creates an electric (electrostatic) field E.
Charge q is used to find this electric field E.
Definition


F  qE


F
E
q

E - electricfield
Units
If the electric force on a test charge q located at
point P is F, then the electric field at point P is F/q.
Because the force is always proportional to q, the
electric field is independent of the test charge!


F 
E     N /C
q

(Action at a distance?)
Example: A negative charge, placed in the electric field between two
charged plates, experiences an electric force as shown below.
What is the direction of the electric field?
+
A. Left B. Right C. Upward D. Downward
• The negative charge is attracted by
the positive plate and is repelled from
the negative plate
• The electric field is directed from the
positive to the negative charge!
q

F

E
-
-
Example: Between the red and the blue charges, which of them
experiences the greater electric field due to the green charge?
+1
d
+2
+1
d
Both charges feel the same electric field due to the green
charge because they are at the same point in space!
+1
Example (Electron in a uniform electric field): Describe the motion of an
electron that enters a region with a uniform electric field having initial
velocity perpendicular to the direction of the field
v0
E
electron
parabola
F = –|qe|E
Once the electric field is known, finding the force on a given charge is simple…


 F  qe E
a 
m
m
Constant acceleration in the –y direction. Identical to projectile motion!
Two most important questions:
1) How can one find force, F on the electric charge, q, exerted by field E?
2) How can electrostatic field E be created?
Answers:
1)


F  qE
2) Field E is due to other charges
2a) Field due to a single charge:

Q
E  k 2 rˆ
r
2b) Field due to a number of charges:

Q1
Q2
E  k 2 rˆ1  k 2 rˆ2  ...
r1
r2
Principle of superposition has been used in 2b)
Principle of superposition

 



Fnet  F1  F2  ...  qE1  qE2  ...  qEnet
Q2
q1
Q3
Q1
Q6
Q4
Q7
Q5
Q8
These charges
create electric field

 
Enet  E1  E2  ...
test charge


Q1 q1
Q2 q1
F1  k
rˆ  k
rˆ  ...  q1 E
2
2
r1
r2

Q1
Q2
E  k 2 rˆ  k 2 rˆ  ...
r1
r2

E.
Charges Q1, Q2… create electric field
This field is independent from the
test charge q1.

If we will replace the charge q1 with
 another charge q2, then the force F2 on the
new charge will be different then F1, but the electric field is independent from q.



F1
F2
E

q1
q2
Definition of electric field
Example (Net electric field):
Which of the three vectors
best represents the direction
of the net electric field at the
location of charge Q?
Example: Calculate the electric field at
the center of a square 52.5 cm on a side
if one corner is occupied by a charge
+45μC and the other three are occupied
by charges -27μC.
Q2  27.0C
Q2
E1
q1 < 0
A
Enet
E1
q2 > 0
Q
B
E2
d
C
E2
Q2
Q1  45.0 C
Q2
Q1  Q2
Q1
E  E1  E 2  k 2
k 2
k

2
d /2
d /2
d /2
6
9
2
2 45  27  10 C
 9  10 Nm / C

2
2
52.5  10 m / 2

 4.7  106 N / C

Electric field lines
Definition:
• Electric field lines indicate the direction of the force due to the given field
on a positive charge, i.e. electric force on a positive charge is tangent to
these lines
• Number of these lines is proportional to the magnitude of the charge
Properties:
• Electric field lines start on positive charges or came from infinity, they end
on negative charges or end at infinity
• Density of these lines is proportional to the magnitude of the field
+
-
-
+Q
-Q
-2Q
+
+
+
+
-
Electric Field Lines Around Electric Charges
A single positive charge
(an electric monopole)
A positive charge and a
negative of equal magnitude
(an electric dipole)
Two equal positive
charges
Example:
The electric field lines in a certain region of space are as shown below.
Compare the magnitude of the electric field at points 1, 2 and 3.
A. E1 = E2 > E3
1
2
3
B. E1 > E2 > E3
C. E1 > E2 ; E3 = 0
The magnitude of the electric field is proportional to the
local density of lines. Being on the same line or being
between the lines is totally irrelevant.
Electric field in conductors
• The electric field inside a conductor in equilibrium is always zero.
If E  0  F  0  a  0
 motion of charges (conductor, charges can move)
 non equilibrium
• The electric field right outside a conductor in equilibrium is
perpendicular to the surface of the conductor.
We cannot have a force parallel to
the surface (would produce motion),
but perpendicular to it is OK.
E=0
Example: A 4.7μC and a -3.5μC
charge are placed 18.5 cm apart.
Where can a third charge be placed
so that it experiences no net force?
Q1  4.7C
Q2  3.5C
Q
–
d
x
To experience no net force, the third charge Q must be closer to the smaller
magnitude charge (the negative charge). The third charge cannot be between
the charges, because it would experience a force from each charge in the same
direction, and so the net force could not be zero. And the third charge must be
on the line joining the other two charges, so that the two forces on the third
charge are along the same line. Equate the magnitudes of the two forces on the
third charge, and solve for x > 0.
F1  F2
xd

 k
Q1 Q
d  x
Q2
Q1 
Q2

2
k
Q2 Q
x
 18.5 cm 
2

 xd

Q2
Q1 
Q2
3.5  106 C
6
6
4.7  10 C  3.5  10 C


 116 cm