Transcript Electric Forces and Fields

Electric Forces and Fields
GIANCOLI- CHAPTER 16
Electric Charge
 Electric charge is conserved
 Electric Charge is quantized
 One unit of charge : e= 1.60219 x 10-19 C

C stands for Coulomb, the unit of electric charge

A proton has a charge of +1.60 x 10-19 C

An electron has a charge of -1.60 x 10-19 C
Conductors & Insulators
 Conductors: Materials in which electric charges
move freely

Examples: most metals
 Insulators: Materials in which electric charges do not
move freely

Examples: Plastic, glass, silk, rubber
Charging by contact
 The two objects are
rubbed together and
electrons are
transferred from one to
the other

electrons from the fur are
transferred to the rod
Charging by Induction
 To charge by induction, a charged object is
brought close to (not touching!) a conductor and
then a conducting wire connects the conductor to
the ground and the electrons travel to the ground
Charging by Polarization
 Charging by polarization creates a surface charge
 A charged object is brought close to an insulator and the
electrons and protons realign themselves to create one side
that is more positive and one that is more negative
Coulomb’s Law
 Coulomb’s Law
describes the
mathematical
relationship between
electric force, distance
and electric charge for
two objects
Felectric
 q1q2 
 kC  2 
 r 
kC= 8.99 x 109 Nm2
C2
Electric force= Coulomb’s Constant x (charge 1)(charge 2)
distance2
Coulomb’s Law
 Remember that force is a vector!

For problems involving two charges, the direction is either
“attractive” or “repulsive”

the direction of the force between a positive charge and
negative charge is attractive a
the direction of the force between two negative charges is
repulsive

Example Problem
 Two identical conducting spheres are placed with
their centers 0.30 m apart. One is given a charge of
+12 x 10-9 C and the other is given a charge of -18 x
10-9 C


A. Find the electric force exerted on one sphere by the other
B. The spheres are connected by a conducting wire. After
equilibrium has occurred, find the electric force between the
two spheres.
Use Coulomb’s Law
Felectric
 q1q2 
 kC  2 
 r 
9
9


12
x
10
C

18
x
10
C 



9 Nm 
5


F   8.99 x10

2.2
x
10
N Attractive
2
2 

C 
 0.30m 



2
 What does it mean “after equilibrium has
occurred”?

The charge on each sphere is the same
9
9


3
x
10
C

3
x
10
C 


9 Nm  
7


F   8.99 x10

8.99
x
10
N Repulsive
2
2 

C 
 0.30m 



2
Three charges in a line
 Three charges are arranged in a line as shown below.
The distance between Q1 and Q2 is 0.35 m and the
distance between Q2 and Q3 is 0.45 m.
 What is the net electrostatic force acting on Q3?
Q1
+4 µC
Q2
-5 µC
Q3
+8µC
 Find the force acting on Q3 from Q1
2


Nm
9
6
6
 9 x10

4
x
10
C
8
x
10
C
2 
C 

F13 
 0.45 N Right
2
0.8m



 Find the force acting on Q3 from Q2
2

9 Nm 
6
6
 9 x10

5
x
10
C
8
x
10
C
2 
C 
F23  
 1.78 N Left
2
0.45m



 Net force acting on Q3= 1.33 N Left
Two Dimensional Problem
 Three charges are arranged in a triangular pattern as
shown below. The distance between Q1 and Q2 is
0.35 m and the distance between Q2 and Q3 is 0.45
m.
 Find the net electrostatic force on Q2
Q1
+4 µC
Q2
-5 µC
Q3
+8µC
Solving the Problem
 We already have the numerical values for the force
acting on Q2from Q3…but the direction is different

F32= 1.78 N Right
 Find the force from Q1 on Q2
2


Nm
9
6
6
 9 x10

4
x
10
C
5
x
10
C
2 
C 

F12 
 1.47 N Up
2
0.35m



Finish the Problem
 Use vector addition (Pythagorean theorem) to find
the net force acting on Q2.
Fnet  1.782  1.47 2  2.31 N
1.47 N
1.78 N
 Direction?
 1.47 

  tan 
  39.6 N of E
 1.78 
1
Electric Field
 Electric force, like gravitational force, is a field
force

Remember: Field forces can act through space even when
there is no physical contact between the objects involved
 A charged object has an electric field in the space
around it
Electric Field Lines
 Electric Field Lines point in the direction of the
electric field
 The number and spacing of field lines is proportional
to the electric field strength

The electric field is strong where the field lines are close
together and weaker when they are far apart
Electric Field Lines
 The lines for a positive charge point away from the
charge
 The lines for a negative charge point towards the
charge
Electric Field Lines
 This diagram shows the electric field lines for two
equal and opposite point charges

Notice that the lines begin on the positive charge and end on
the negative charge
Electric Field Lines
 This diagram shows the electric field lines for two
positive point charges

Notice that the same number of lines emerges from each
charge because they are equal in magnitude
Electric Field Lines
 If the charges are unequal, then the number of lines
emerging from them will be different

Notice that the positive charge has twice as many lines
Calculating Electric Field Strength
 The equation for the electric field produced
by a point charge is:


kc q
E 2
r
Kc=9x109 Nm2/C2 ,r is the distance from the charge and q is the
charge producing the field
The unit for E is N/C
 Electric field strength is a vector!!
 If q is positive, then E is directed away
from q
 If q is negative, then E is directed toward
q
Calculating the force from an electric
field
 If a charged object is placed in an electric field, we
can calculate the force acting on it from the electric
field
F  qE
 Remember that F is a vector!!
Sample Problem p. 647 #3
 An electric field of 2.0 x 104 N/C is directed along the
positive x-axis
a. What is the electric force on an electron in this field?
b. What is the electric force on a proton in this field?
Sample Problem p. 647 #3
 E= 2.0 x 104 N/C , q= 1.6 x 10-19 C
 F=qE= 3.2 x 10-15 N for both the electron and the
proton
 What about the direction?
 The electric field is pointing along the positive x axis (to
the right) which means there’s a positive charge to the left
E field
+
For the proton
 Since the electric field is pointing to the right, if you
put a proton in it, the proton will want to move away
towards the right and the direction of the force on it
will be to the right
+
+
F
 Answer: 3.2 x 10-15 N along the positive x axis (to the
right)
For the electron
 Since there’s a positive charge causing the electric
field to point towards the right, an electron would
feel attracted to the positive charge. Therefore, the
force acting on it is toward the left
 Answer: 3.2 x 10-15 N along the negative x axis (to the
left)
+
F
-
Sample Problem
 Find the electric field at a point midway between
two charges of +30 nC and 60 nC separated by a
distance of 30.0 cm
+30 nC
+60 nC
 For the 30 nC charge:


kc q
9 x109 (30 x10 9 )
N
E 2 

12
,
000
r
C
0.15m 2

For the 60 nC charge:


kc q
9 x109 (60 x10 9 )
N
E 2 

24
,
000
r
C
0.15m 2
 Direction of the E-field for both charges is “away”
since they’re both positive
+30 nC
+60 nC
Which one will win?
 At the midway point, the 30nC charge’s field
strength is 12000 N/C toward the 60 nC charge
and the 60 nC charge’s field strength is 24,000
N/C toward the 30 nC charge.
 The 60 nC charge will win. Since the field’s point in
opposite directions, you have to subtract
 Answer: 12,000 N/C toward the 30 nC charge
Sample Problem
 A constant electric field directed along the positive x-
axis has a strength of 2.0 x 103 N/C.


Find the electric force exerted on a proton by the field
Find the acceleration of the proton
Answer 
 F=qE=(1.6x10-19 C)(2.0 x 103 N/C)=
 3.2 x 10-16 N
 Direction?
E field
+
+
F
Answer: 3.2 x 10-16 N along the positive x-axis (to the right)
Answer 
 B. What is the acceleration?
 Ask Newton!
 F=ma
 a = F/m= 3.2 x 10-16 N/1.6x10-27 kg
 a= 2 x 1011 m/s2 along the positive x axis