Download Electric Fields and Forces

Electric Forces,
Fields, Potential and Energy
Fundamental Charge: The charge
on one electron.
e = 1.6 x 10
-19
C
Unit of charge is a Coulomb (C)
Two types of charge:
Positive Charge: A shortage of electrons.
Negative Charge: An excess of electrons.
Conservation of charge – The net charge of a
closed system remains constant.
Nucleus
-
-
n + n
+ +
n
+
n
n
+
+ n
-
-
-
-
Negative
Neutral Atom
Atom
Positive
Atom
Number
Numberof
ofelectrons
electrons><=Number
Numberof
ofprotons
protons
Number
of
electrons
Number
of
protons
-19
-2e
=
-3.2
x
10
CC
+2e = +3.2 x 10-19
Electric Forces
Like Charges - Repel
F
+
+
Unlike Charges - Attract
-
F
F
+
F
Coulomb’s Law – Gives the electric force
between two point charges.
q1q2
F k 2
r
Inverse Square
Law
k = Coulomb’s Constant = 9.0x109 Nm2/C2
q1 = charge on mass 1
q2 = charge on mass 2
r = the distance between the two charges
The electric force is much stronger than the
gravitational force.
Example 1
Two charges are separated by a distance r and have a force
F on each other.
qq
F k
F
1 2
2
r
q2
q1
F
r
If r is doubled then F is :
¼ of F
If q1 is doubled then F is :
2F
If q1 and q2 are doubled and r is halved then F is : 16F
Example 2
Two 40 gram masses each with a charge of 3μC are
placed 50cm apart. Compare the gravitational force
between the two masses to the electric force between the
two masses. (Ignore the force of the earth on the two
masses)
3μC
40g
3μC
40g
50cm
m1m2
Fg  G 2
r
 6.67 10
11
(.04)(.04)
2
(0.5)
 4.27 10
q1q2
FE  k 2
r
6
6
(
3

10
)(
3

10
)
9
 9.0 10
(0.5) 2
13
N
 0.324 N
The electric force is much greater than the
gravitational force
Example 3
Three charged objects are placed as shown. Find the net
force on the object with the charge of -4μC.
F k
- 5μC
45º
20cm
202  202  28cm
q1q2
r2
(5 106 )(4 106 )
F1  9 10
 4.5N
2
(0.20)
9
(5 106 )(4 106 )
F2  9 10
 2.30 N
2
(0.28)
9
F1 45º
- 4μC
5μC
20cm
F2
F1 and F2 must be added together as vectors.
F1
2.3cos45≈1.6
45º
F2
2.3sin45≈1.6
F1 = < - 4.5 , 0.0 >
+ F2 = < 1.6 , - 1.6 >
Fnet = < - 2.9 , - 1.6 >
- 1.6
- 2.9
29º
θ
3.31
Fnet  2.9 2  1.6 2  3.31N
  1.6 

  tan 

29

  2.9 
1
3.31N at 209º
Example 4
Two 8 gram, equally charged balls are suspended on earth
as shown in the diagram below. Find the charge on each
ball.
20º
L = 30cm
FE
q
10º 10º
L = 30cm
30sin10º
q
r
r =2(30sin10º)=10.4cm
2
q1q2
q
FE  k 2  k 2
r
r
FE
Draw a force diagram for one charge and treat as an
equilibrium problem.

T
FE
q
T sin 80  .08
.08
T
 .081N

sin 80
Tsin80º
80º
Tcos80º
Fg = .08N
FE  T cos 80
q2

k

(.
081
)
cos
80
.104 2
.014
2
q 
(.104) 2
k
q  1.3 10 7 C
Electric Fields: A property of space around a charge that
causes forces on other charges. The force per unit charge felt
by a positive test charge placed in the field
+ FE
Use a positive test charge
FE
_
FE
+
+
FE
FE
+
+
FE
+
+
Use a positive test charge
The
gravitational
Force
= (The
mass)xx(The
(TheElectric
Gravitational
The
Electric Force
= (The
charge)
Field) Field)
FgE  mg
qE
q1eqm2
M
FFg EGk 22
rr
M e FEFg
q
Eg  kG 2  
r r2 q m
The
units
forfor
a gravitational
field
N/kg
The
units
a electric field
areare
N/C
Things to Know
• Electric field lines go out of a positive
charge and into a negative charge.
• Positive charges experience a force with
the field.
• Negative charges experience a force
against the field.
• The Electric Field inside a conductor is
zero.
Example 3
What is the electric field at a point 25cm away from a point
charge of 2µC?
q
Ek 2
r
6
2 10
E  9 10
(.25) 2
9
5 N
E  2.88  10
C
Away from the charge
2µC
25cm
E
Example 4
A 5 gram mass with a charge of -3µC is released from rest in a
constant field of 20000N/C directed in the positive x direction.
Find its position and velocity after 4.0 seconds.
Negative charges experience a force against the field
FE = qE
 F  ma
FE  ma
qE  ma
qE
m
a
 12 2
m
s
5g
-3µC
m
v  v0  at  (12)( 4)  48
s
1 2
x  x0  v0t  at
2
1
x  (12)( 4) 2  96m
2
Example 6
What is the electric field at the origin?
E3
EE21
kq
kq1 1
EE32 1 222
rr
 666
10
10
(9(910
109 9)()(121..5010
) ))
EE32 1
222
5
.
0
34.5
NN
720
EE32 1844
735
CC
To the right (+)
Up (+)
To the right (+)
The electric field due to each charge
must be calculated individually and
then added together as vectors.
Now add the vectors together:
E3=844N/C
E1 = < 735, 0
>
E2 = < 720, 0
>
E3 = <
E
E1=735N/C
θ
1455N/C
E2=720N/C
844N/C
0 , 844 >
E = < 1455, 844 >
E  14552  844 2
N
E  1682
C
1  844 
  tan 

 1455 
  30

The electric field due to more than one point charge.
F
+
F
Two charged parallel plates
The Electric field is constant between the two plates.
The Electric Field inside a
conductor is zero
E = 0N/C
An electron has a velocity of 6x105m/s as it enters a constant electric
field of 10000N/C as shown in the diagram below. How far will the
electron travel in the x direction before striking one of the plates?
2.82 x 10-3m
_
v
4cm
FNow
First
mawe
have
projectile motion
1a simple
2 acceleration
the
force
and
of the1
2
y  y0 electron
v0t  must
at
x

x

v
t

at
problem!
0
0
be
calculated.
FE  ma
2
2
5 shown
9
 ma
The qE
Electron
path
as
1 will follow15a parabolic
2 x  (6  10 )( 4.7  10 )
.02 
(1.8 10 19
)t
qE2 (1.6 10 )(10000)
315 m
x  2.82
a

 1
.810
10 m 2

31
s
t  4m
.7 10 9 s9.1110
Let’s make a parallel comparison of Gravitational Potential
Energy (Ug) to Electric Potential Energy (UE).
U g  mgh
Units:
U E  qEd
Units:
N
kg   m  Nm  J
kg
N
C   m  Nm  J
C
The Electric Potential Energy at a point is the work
done to move a charge from infinity to that point.
Now we will do an example:
An electron is released from rest in an electric field of
2000N/C. How fast will the electron be moving after traveling
30cm?
E0  W  E f
v=?
_
v = 0m/s
_
30cm
UE  K
1 2
qEd  mv
2
2qEd
v
m
2(1.6 1019 )( 2000)(.3)
v
9.111031
m
v  1.45 10
s
7
A new quantity is defined called the Electric Potential
Difference. It is the work done per unit charge to move a
small positive test charge between two points.
Electric Potential Difference 
UE
V
q
qEd
V
q
V  Ed
Electric Potential Energy
Ch arg e
The Electric Potential Difference can also
called the Electric Potential, the Potential,
or the voltage.
Units:
J
 volt  V
C
Remember Energy and voltage are scalars, so
you don’t have to deal with vectors (direction)
The electric potential energy can now be written in
terms of the electric potential.
U E  qEd
V  Ed
U E  qV
U E  qV
Let’s do an example using this new concept:
The potential difference between two charge plates is 500V.
Find the velocity of a proton if it is accelerated from rest from
one plate to the other.
E W  E
0
High
Low
Potential
+
Potential
-
+
-
+
+
-
+
-
+
-
500V
Positive charges move from high to low potential
Negative charges move from low to high potential
f
UE  K
1 2
qV  mv
2
2qV
v
m
2(1.6 10 19 )(500)
v
1.67 10  27
5 m
v  3.110
s
Example:
E0  W  E f
U E  K1  K 2
q1q2 1 2 1 2
k
 mv  mv
r
2
2
q2
k
 mv 2
r
2
kq
v2 
rm
Two 40 gram masses each
with a charge of -6µC are
20cm apart. If the two
charges are released, how
fast will they be moving when
they are a very, very long way
apart. (infinity)
kq2
(9  109 )(6  10 6 ) 2
m
v

 6.36
rm
(.2)(. 04)
s
Let’s use the electric potential energy between two charges
to derive an equation for the electric potential (voltage) due
to a single point charge.
The electric potential energy is given by:
q2P
r
q1
q1q2
UE  k
r
Let’s remove charge q2 and consider the electric
potential (voltage) at point P.
kq1q2
UE
q1
r
V

k
q
q2
r
To find the potential due to more than one point charge simply
add up all the individual potentials:
i
q
V  k
r
i
Example 1: The electron in the Bohr model of the atom can exist at only certain
orbits. The smallest has a radius of .0529nm, and the next level has a
radius of .212nm.
a) What is the potential difference between the two levels?
b) Which level has a higher potential?
q
V k
r
e
V1  k
r1
r1
r2
19
1.6 10
V1  (9 10 )
 27.2V
9
.0529 10
19
1
.
6

10
V2  (9 109 )
 6.79V
9
.212 10
+e
9
r1 is at a higher potential.
potential diff  V  27.2  6.79  20.4V
Example 2
What is the electric potential at the center of the square?
45º
45º
r
r
r
2k
V V 
qq2
kr rr
 0.10 
 6 6

10

5

2
r

9 10  10  
9.01
 
10 
V V 9 910

071
.
.
071


m
r  .071
65 J
10
3410
V V 16..27
C
2
r
r
V  k
i
2
qi
r
Vtotal  1.27 106  (1.27 106 )  6.34 105  6.34 105
J
Vtotal  1.27 10
C
6
Example 3: A proton is moved from the negative plate to the positive plate of a
parallel-plate arrangement. The plates are 1.5cm apart, and the electric field is
uniform with a magnitude of 1500N/C.
a) How much work would be required to move a proton from the negative to the
positive plate?
b) What is the potential difference between the plates?
c) If the proton is released from rest at the positive plate, what speed will it have
just before it hits the negative plate?
W  U E  qEd
N
W  (1.6 10 C )(1500 )(.015m)
C
W  3.6 10 18 J
19
Example 3: A proton is moved from the negative plate to the positive plate of a
parallel-plate arrangement. The plates are 1.5cm apart, and the electric field is
uniform with a magnitude of 1500N/C.
b) What is the potential difference between the plates?
V  Ed
N
V  (1500 )(.015m)
C
J
V  22.5
C
Example 3: A proton is moved from the negative plate to the positive plate of a
parallel-plate arrangement. The plates are 1.5cm apart, and the electric field is
uniform with a magnitude of 1500N/C.
c) If the proton is released from rest at the positive plate, what speed will it have
just before it hits the negative plate?
Use conservation of energy
E0  W  E f
UE  K
1 2
qV  mv
2
2qV
v
m
2(1.6 10 19 )( 22.5)
v
1.67 10  27
m
v  6.57 10 4
s
Example 4:
Compute the energy necessary to bring together the charges in the
configuration shown below:
Calculate the electric potential
energy between each pair of
charges and add them together.
qq12q1qq332
UU13
2312kk
rr
66 6
666


(
4

10
)(

4

10
(
4

10
)(
4

10
))
99
9
UU13
(910
10 ))
2312(9
.2
UU13
72JJ
231200..72
U total  .72 J  (.72 J )  (.72 J )
U total  .72 J