Download 2_Electric_Field

Electric Fields
The gravitational and electric forces can act through space without
any physical contact between the interacting objects. Just like the
gravitational force the electric force is associated with a field.
The electric field represents a alteration of the space surrounding
an electric charge such that a second charge placed in the field will
experience an electric force.
The electric field vector, E specifies the magnitude and direction at
any point in the electric field.
The magnitude and direction of the electric field vector at some
point in an electric field is defined to be the magnitude and
direction of the electric force exerted on a positive test charge
placed at that point divided by the test charge.
FE
E=
qo
units : N
C
+ q'
FE,q'
E

FE,q' q'
qq
FE,q'
E  q'
+ q'
FE,q'
The electric force on another electric charge, q1, placed in the field
is then given by:
FE,1  q 1E
If the charge q1 is positive, the direction of the force is the same as
the field. If q1 is negative the direction of the force is opposite to
the direction of the field.
From Coulomb’s Law we can derive an expression for the electric
field due to a point charge, q1.
The force on a positive test charge, qo due to q1 is:
Fq q
o
1
k q o q1

2
ro,1
The electric field due to q1 is then:
Fq q
Eq  q
o
o
1
1
k qo q1
2
ro,1


qo
Eq
1
k q1

r12
P
k q1
EP  2
rP,1
rP,1
q1
rQ ,1
Q
k q1
EQ  2
rQ ,1
The electric field points away from a positive source charge and
toward a negative source charge.
The electric field at a point P due to more than one charge is the
vector sum of the individual fields produced by each charge.
k q1 k q 2 k q 3
E total  2  2  2
r1
r2
r3
k qi
E total   2
ri
i
qi
 k 2
i ri
Where ri is the distance from the source charge qi to point P.
Three point charges, q1 = 25mC, q2 = -40mC, and q3 = 60mC lie on a
straight line as shown below.
What is the magnitude and direction of the electric field at point
P, 5.5m from the origin?
+
q
0 1
1m
2m
+
q
4m 2 5m
3m
EP,q 2 
EP,q 3 
P
E P,q
E P,q
q
7m 3
6m
E P,q
2
1
EP,q1 
3
The electric field at point P due to q1 points away from q1.
The electric field at point P due to q2 points toward q2.
The electric field at point P due to q3 points away from q3.
E P  E P,q  E P,q  E P,q
1
2
3
k q1
k q2
k q3
E P  2  ( 2 )  ( 2 )
rP,1
rP,2
rP,3
 q

q
q
1
2
3 
EP  k

(
)

(
2
2 )
r 2
rP,2
rP,3 
 P,1
2 

6
6
6
N
m
25
10
C
40
10
C
60

10
C
EP  8.99  10
 (
)  (
)
2 
2
2
2
C  (5.5m)
(1.5m)
(1.5m)

9
E P  3.92  10 5 N
C
The negative sign indicates that at point P the direction of the
electric field is to the left.
E P  3.92  10 N
C
5
0
q1
1m
2m
3m
q
4m 2 5m
P
6m
q
7m 3
Two point charges, q1 = 25mC and q2 = -40mC lie on a straight line
as shown below. Is there a point where the electric field is zero?
E1 
E1 
E2  +
q
0 1
E2 
1m
2m
E1 
E2 
3m
q
4m 2 5m
6m
For the electric field to be zero the fields due to q1 and q2 must
cancel.
To cancel the fields must be equal in magnitude and opposite in
direction.
There are three possibilities for the location of a point where the
electric field is zero.
andEE22are
areopposite,
opposite qbut
EE11and
q21 > q1
2 >q
E
E
not opposite
and
1 and
2rare
and
r
>
therefore
E
and
and
r
<
r
therefore
E
and
EE22
1
2
1
1
2
1
1. To the right of q2, x > 4m.
can
cancel.
cannot
notequal
be equal
in magnitude
can
be
in magnitude
and
andcancel.
can not cancel.
2. Between q1 and q2, 0 < x < 4m. can
3. To the left of q1, x < 0.
-
+
P
x
0
q1
1m
2m
3m
q
4m 2
x+4
E1,P  E2,P
k q1
k q2
 2
2
r1,P
r2,P
q1
q2
2 
2
x
(x  4)
q1
2 
x
q2
2
(x  4)
q1
q2
x  (x  4)
q1  (x  4)  q 2  x
q1  x  4 q1 
q1  x 
q2  x
q 2  x  4 q1
x( q 1 
q 2 )  4 q1
4 q 1
x
( q1  q 2 )
x
4 25  10 6 C
25  10 6 C  40  10 6 C
0.02
x  0.0013
EP  0
P
x  15.1m
15.1m +
0
q1
1m
2m
3m
q
4m 2
Consider two parallel sheets separated by a distance d and having
equal areas, A. The sheets carry equal but opposite electric
charges. This charge distribution is called a parallel plate
-q
+q
capacitor .
E
A
q
E  A
o
d
Where o is the permittivity of free space:
 o  8.85  10
12
C2
2
N m
At every point between the plates of the capacitor the electric field
is the same and directed from the positive plate toward the
negative plate.
A parallel plate capacitor is composed of plates each with an area of
0.002m2 and separated by a distance of 0.3m. They carry equal and
opposite charges of 0.8mC.
A proton, mp=1.67x10-27kg, qp=+1.6x10-19C is released from rest at
the positive plate.
What will be the velocity of the proton when it reaches the
negative plate?
kinematic data for proton
0.8mC 0.8mC
s = 0.3m
proton
vi  0
E
vf  ?
d=0.3m
a?
t?
A  0.002m 2
Dynamic data for proton
m p  1.67  10 27 kg
q p  1.6  10 19 C
6
q 
0.8  10 C
E   A 
2 
12
2
C
o
8.85  10

0.002m

2 

Nm
7 N
E  4.52  10 C
Fp  q pE  1.6  1019 C4.52  107 N 
C
Fp  7.23  10
12
N
12
Fp
ap  m  7.23  1027 N
1.67  10 kg
p
a p  4.33  1015 m2
s
=0
v 2f  v 2i  2a  s
v f  2a  s  2(4.33  10
15
m )(0.3m)
s2
vf  5.1  10 m
s
7
This velocity is approximately 17% of the speed of light!
How long did it take the proton to cross the capacitor?
vf  vi
t a
m
5.1  10 s  0
t
15 m
4.33  10
2
s
7
8
t  1.18  10 s  11.8ns