Transcript File

Static Electricity / Electrostatics
• Where have you experienced this?
– Shocked a friend? (or yourself)
– Clothes in the dryer stick together
– Stroke a cat and see the hair standing up on
end.
– Rub a rubber balloon in somebody’s hair
– Bolt of lightning
Electrons are migrants.
They are attracted to the
protons in the nucleus,
but they can be
persuaded to become
attracted to protons in a
different nucleus and join
another electron shell if
energy is added.
The First Law of Electrostatics
Like charges repel; unlike charges attract.
Neg
Pos
Neg
Pos
Neg Pos
http://www.colorado.edu/physics/2000/waves_particles/wavpart2.html
Here’s the fun part!
• Charged objects attract neutral objects!
The electrons in the neutral
object are attracted (or
repelled) by the electrons
(or protons) in the charged
object. They move
accordingly, and the object
becomes polarized.
The Quantity of Charge
The quantity of charge (q) can be defined in
terms of the number of electrons, but the
Coulomb (C) works better
The Coulomb: 1 C = 6.25 x 1018 electrons
Which means that the charge on a single electron is:
1 electron: e- = -1.6 x 10-19 C
Units of Charge
The coulomb (selected for use with electric
currents) is actually a very large unit for static
electricity. Thus, we often encounter a need to
use the metric prefixes.
1 mC = 1 x 10-6 C
1 nC = 1 x 10-9 C
1 pC = 1 x 10-12 C
Ex: Determine the quantity and type of
charge on an object which has 3.62 x 1012
more protons than electrons.
3.62 x 1012
x
1C
6.25 x 1018 e
= +5.8 x 10^-7 Coulombs (rounded)
Example 1. If 16 million electrons are
removed from a neutral sphere, what is
the charge on the sphere in coulombs?
1 electron: e- = -1.6 x 10-19 C
-19

-1.6
x
10
C
6 q  (16 x 10 e ) 

1e


+ +
+ + +
+ + + +
+ + +
+ +
q = -2.56 x 10-12 C
Since electrons are removed, the charge
remaining on the sphere will be positive.
Final charge on sphere:
q = +2.56 pC
Coulomb’s Law
The force of attraction or repulsion between two
point charges is directly proportional to the product
of the two charges and inversely proportional to the
square of the distance between them.
- q
F
F
r
q’
q
q’
-
-
+
F
qq '
F 2
r
Coulomb’s Law
F12
Q1
F21
r
Q2
F  F12  F21
Force (N)
Q1Q 2
Fk
r2
Charge (Q)
Coulombs (C)
1 C = 6.2421 x 1018 e
e = 1.602 x 10-19 C
Distance (m)
Coulomb’s constant (k)
k = 8.988 x 109 Nm2/C2
Coulomb’s Law
The force is along the line connecting the charges,
and is attractive if the charges are opposite, and
repulsive if they are the same.
Example 2. A –5 mC charge is placed 2 mm from a
+3 mC charge. Find the force between the two
charges.
-5 mC
Draw and label
givens on figure:
q
+3 mC
q’
F
-
+
r
2 mm
9 Nm2
C2
kqq ' (9 x 10
F 2 
r
)(-5 x 10 C)(3 x 10 C
-6
-3
(2 x 10 m)
F = 3.38 x 104 N;
-6
2
Attraction
Note: Signs are used ONLY to determine force direction.
Problem-Solving Strategies
1. Read, draw, and label a sketch showing all
given information in appropriate SI units.
2. Do not confuse sign of charge with sign of
forces. Attraction/Repulsion determines the
direction (or sign) of the force.
3. Resultant force is found by considering force
due to each charge independently. Review
practice on vectors, if necessary.
4. For forces in equilibrium: SFx = 0 = SFy = 0.
Example 3. A –6 mC charge is placed 4 cm from a +9
mC charge. What is the resultant force on a –5 mC
charge located midway between the first charges?
1 nC = 1 x 10-9 C
1. Draw and label.
2. Draw forces.
3. Find resultant;
right is positive.
-6 mC
q1
-
r1
q3
2 cm
kq1q3 (9 x 109 )(6 x 10-6 )(5 x 10-6 )
F1  2 
;
2
r1
(0.02 m)
-
F1
F2 +9 mC
q2
+
r2
2 cm
F1 = 675 N
F2 = 1013 N
Example 3. (Cont.) Note that direction (sign) of forces are
found from attraction- repulsion, not from + or – of
charge.
+
F1 = 675 N
F2 = 1013 N
-6 mC
q1
-
r1
q3
-
2 cm
F1
F2 +9 mC
q2
+
r2
2 cm
The resultant force is sum of each independent force:
FR = F1 + F2 = 675 N + 1013 N;
FR = +1690 N
Example 4. Three charges, q1 = +8 mC, q2 = +6 mC
and q3 = -4 mC are arranged as shown below. Find
the resultant force on the –4 mC charge due to the
others.
+6 mC 3 cm q
3
+
q2
-4 mC
4 cm
q1
Draw free-body diagram.
F2
q3
- -4 mC
5 cm
+
53.1o
+8 mC
53.1o
F1
Note the directions of forces F1and F2 on q3
based on attraction/repulsion from q1 and q2.
Example 4 (Cont.) Next we find the forces F1 and F2
from Coulomb’s law. Take data from the figure and
use SI units.
+6 mC 3 cm q
3
- -4 mC
q2 + F2
4 cm F1
q1
+
5 cm
53.1o
+8 mC
Thus, we need to find resultant of two forces:
F1 = 115 N, 53.1o S of W
F2 = 240 N, West
Example 4 (Cont.) We find components
of each force F1 and F2 (review
vectors). F
F1x = -(115 N) Cos 53.1o
= - 69.2 N
53.1o
F1y = -(115 N) Sin
- 92.1 N
Now look at force F2:
2
240 N F1x
=
F2x = -240 N; F2y = 0
F1y
53.1o
q3
- -4 mC
F1= 115 N
Rx = SFx ;
Ry = SFy
Rx = – 69.2 N – 240 N = -309 N
Rx= -309 N
Ry = -92.1 N – 0 = -92.1 N
Ry= -92.1 N
Example 4 (Cont.) Next find resultant R from
components Fx and Fy. (review vectors).
Rx= -309 N
Ry= -92.1 N
Rx = -309 N q
f
R
We now find resultant R,q:
- -4 mC
3
Ry = -69.2 N
R = ( (309)2 + (92.1)2 )1/2 = 322.4
Thus, the magnitude of
the electric force is:
R = 322.4 N
Example 4 (Cont.) The resultant force
is 317 N. We now need to determine
the angle or direction of this force.
q
-309 N
f R
-69.2
-92.1 N
N
-92.1 N
The reference angle is: f = 73.40S of W
Or, the polar angle q is: q = 1800 + 73.40 = 253.40
Resultant Force: R = 322 N, q = 253.40
Summary of Formulas:
Like Charges Repel; Unlike Charges Attract.
1 mC = 1 x 10-6 C
1 pC = 1 x 10-12 C
1 nC = 1 x 10-9 C
1 electron: e- = -1.6 x 10-19 C
The Electric Field
Electric field of a point charge
q

+Q
r
Qq
Fk
r2
Qq
k
2
F
r
E 
q
q
Q
Ek
r2
Electric Charge and Electric Field 16
A proton is released in a uniform electric field, and it
experiences an electric force of 3.75 × 10-14 N toward
the south. What are the magnitude and direction of the
electric field?
E
+
F
-14 N
F
3.75
x
10
E


1
9
q 1.602 x 10
C
south
2.34 x 105 N/C
Electric Charge and Electric Field 16
What are the magnitude and direction of the electric
field at a point midway between a +7.0 mC and a -8.0 mC
charge 8.0 cm apart? Assume no other charges are
nearby.
d
E1
E  E1  E 2

+
q1
kq1
kq 2
E2
E

d 2 d 2
2
2
   


-
q2
4k
4 8.988 x109 N  m/C
- 6 N  8.0 x10- 6 N

E
q1  q 2  
7.0
x10
d2
0.08 m 2

E  8.4 x107 N/C
Electric Charge and Electric Field 16
A proton (m = 1.67 x 10-27 kg) is suspended at rest in a
uniform electric field E. Take into account gravity at the
Earth's surface, and determine E.
FE
+
q
FG
FE  FG
E
qE  mg
mg
E
q

1.67 x 10- 27 kg 9.80 m/s 2
E
1.602 x 10-19 C

E  1.02 x 10- 7 N/C
Electric Charge and Electric Field 16
The Electric Field
The electric field can be represented by field lines. These
lines start on a positive charge and end on a negative
charge.
Electric Charge and Electric Field 16
The Electric Field
The number of field lines starting (ending) on a positive
(negative) charge is proportional to the magnitude of the
charge.
E
The electric field is stronger where the field lines are closer
together.
E
A
B
C
Electric Charge and Electric Field 16
The Electric Field
Electric field of two charges:
Electric Charge and Electric Field 16
The Electric Field
The electric field between two closely spaced, oppositely
charged parallel plates is constant.
Electric Charge and Electric Field 16
A +2 nC charge is placed at a distance r
from a –8 mC charge. If the charge
experiences a force of 4000 N, what is
the electric field intensity E at point P?
+2 nC
+q +. P
4000 N
E E
r
- --- -Q
- –8 mC
-First, we note that the direction of
E is toward –Q (down).
Electric Field
E = 2 x 1012 N/C
Downward
Note: The field E would be the same for any charge
placed at point P. It is a property of that space.
A constant E field of 40,000 N/C is maintained between the
two parallel plates. What are the magnitude and direction
of the force on an electron that passes horizontally
between the plates.c
+ + + + + + + + +
The E-field is downward,
and the force on e- is up.
e- -
F
E  ; F  qE
q
Fe -.
E
e- -
- - - - - - - - -
F  qE  (1.6 x 10 C)(4 x 10
-19
F = 6.40 x 10-15 N, Upward
4 N
C
)
The E-Field at a distance r
from a single charge Q
Consider a test charge +q placed
at P a distance r from Q.
The outward force on +q is:
kQq
F 2
r
FE
+q +.. P
P
r
kQ
++ E  2
+
+
+
r
+Q +
++
The electric field E is therefore:
F kQq r
E 
q
q
2
kQ
E 2
r
What is the electric field intensity E at point P, a distance
of 3 m from a negative charge of –8 nC?
.
E=?
r
3m
-Q
-8 nC
First, find the magnitude:
P
9 Nm2
C2
-9
)(8 x 10 C)
kQ (9 x 10
E 2 
2
r
(3 m)
E = 8.00 N/C
The direction is the same as the force on a positive
charge if it were placed at the point P: toward –Q.
E = 8.00 N, toward -Q
The Resultant Electric Field.
The resultant field E in the vicinity of a number
of point charges is equal to the vector sum of the
fields due to each charge taken individually.
Consider E for each charge.
Vector Sum:
E = E1 + E2 + E3
Magnitudes are from:
kQ
E 2
r
q1 ER
E2
A
E1
q3 -
E3
+
q2
Directions are based
on positive test charge.
Example. Find the resultant field at point A due to
the –3 nC charge and the +6 nC charge arranged as
shown.
-3 nC
q1 3m
E1
5m
+6 nC

E2 A 4 m
E1 
9 Nm2
C2
(9 x 10
E for each q is shown
with direction given.
+
q2
-9
)(3 x 10 C)
(3 m)
kq1
kq2
E1  2 ; E2  2
r1
r2
2
E2 
9 Nm2
C2
(9 x 10
)(6 x 10-9C)
(4 m)2
Signs of the charges are used only to find direction of E
Example . (Cont.)Find the resultant field at point A.
The magnitudes are:
-3 nC
q1 -
3 cm E 5 cm
1
E1 
+6 nC

E2 A 4 cm
+
q2
E1 = 3 N/C , North
E2 
9 Nm2
C2
(9 x 10
(3 m)2
Next, we find vector resultant ER
E1
ER  E  R ; tan f 
E2
2
1
9 Nm2
C2
(9 x 10
E2 = 3.38 N, West
2
2
)(3 x 10-9C)
)(6 x 10-9C)
(4 m)2
ER
f
E2
E1
Example . (Cont.)Find the resultant field at
point A using vector mathematics.
ER
E1 = 3 N, West
f
E1
E2
E2 = 3.38 N, North
Find vector resultant ER
3.38
N
3.00N
E  (3.00 N)  (3.38 N)  4.52 N; tan f 
3.00
N
3.38N
f = 41.60 N of W; or q = 138.40
2
2
Resultant Field: ER = 4.25 N; 138.40
Electric Potential &
Electric Potential Energy
Review: Work and Energy
Work is defined as the product of displacement d
and a parallel applied force F.
Work = Fd; Units: 1 J = 1 N m
Potential Energy U is defined as the ability to do work
by virtue of position or condition. (Joules)
Kinetic Energy K is defined as the ability to do work
by virtue of motion (velocity). (Also in joules)
Electrical Work and Energy
An external force F moves +q from
A to B against the field force qE.
Work = Fd = (qE)d
At level B, the potential energy U is:
U = qEd (Electrical)
The E-field does negative work;
External force does positive work.
B ++++
Fe
+q + d
qE
E
A - - - -
The external force F against the E-field increases the
potential energy. If released the field gives work back.
Electric Potential Energy and Potential Difference
m
g
h
F = mg
W = mgh
DPE = -W
Electric Potential 17
Electric Potential Energy and Potential Difference
m
a
q
E
g
h
d
F = qE

b
F = mg
W = mgh
DPE = -W
W = Fd
W = qEd
DPE = W
Electric Potential 17
Electric Potential Energy and Potential Difference

q
a
x
E
b

PEa
Va 
q
PEb
Vb 
q
DV  Vb - Va
PEb - PEa
DV 
q
W
DV  ba
q
Wba  qVb - Va 
Wba  qDV
Electric Potential 17
Electric Potential
Electric potential is another property
of space allowing us to predict the
P.E. of any charge q at a point.
Electric
Potential:
U
V ;
q
U  qV
The units are: joules per coulomb (J/C)
U
P. V 
q
r
+ ++
+
++Q++
Potential
For example, if the potential is 400 J/C at point P,
a –2 nC charge at that point would have P.E. :
U = qV = (-2 x 10-9C)(400 J/C);
U = -800 nJ
Problem 17-04
An electron acquires 7.45 x 10-17 J of kinetic energy when it
is accelerated by an electric field from plate A to plate B.
What is the potential difference and which plate is at the
higher potential?
A
B
W  ΔKE
W 7.45 x 10-17 J

V
1.60 x 10-19 C
q
V  466 V
E
Plate B has a positive charge
and is at a higher potential.
Electric Potential 17
Electric Potential Energy and Potential Difference
q
x =1.00 cm
q = 1.60 x 10-19 C
E = 2000 N/C m = 9.1 x 10-31 kg
m
x
E
Find the speed of the
charge at the lower plate.
ΔKE  W
mv 2
 qEx
2
Find the potential through
which the charge moves (ΔV).
W  qDV
qEx  qDV
DV  Ex
DV  2000 N/C 0.01 m
DV  20V

2qEx
v
m

2 1.6 x 10-19 C 2000 N/C0.01 m 
v
9.1 x 10- 31 kg
v  2.65 x 106 m/s
Electric Potential 17
Electric Potential Energy and Potential Difference
q
x
x =1.00 cm
q = 1.60 x 10-19 C
E = 2000 N/C m = 9.1 x 10-31 kg
m
E
Find the acceleration of the charge.
F 3.2 x 10 -16 N
a 
m 9.1 x 10 - 31 kg
Find the force on the charge
as it moves.
F  qE
a  3.5 x 1014 m/s 2
F  1.6 x 10-19 C2000 N/C
F  3.2 x 10-16 N
Electric Potential 17
Electric Potential Energy and Potential Difference
Work is required to move two point charge closer together.
F
r
F
This work is converted to potential energy.
This electric potential energy of two point charges:
Q1Q 2
PE  k
r
Electric Potential 17
Problem 17-20
An electron starts from rest 32.5 cm from a fixed point
charge with Q = -0.125 mC How fast will the electron be
moving when it is 50m away?
PEinitial  KEfinal
Q1Q 2 mv 2
k

r
2

2kQ1Q 2
 v
mr


2 8.99 x 109 N  m 2 /C2 1.6 x 10-19 C 1.25 x 10- 7 C
v
9.1 x 10- 31 kg0.325 m 

v  3.49 x 107 m/s
Electric Potential 17
Relation between Electric Potential and Electric Field
V
V
E
d
V  Ed
E
Let E = 100 N/C
d = 10 cm
V  100 N/C0.10 m
10 V
7.5 V
5d
V
2.5 V
0V
V  10 V
Using potentials instead of fields can make solving problems
much easier – potential is a scalar quantity, whereas the
field is a vector.
Electric Potential 17
Equipotential Lines
Electric Potential 17
The Electron Volt, a Unit of Energy
One electron volt (eV) is the energy gained by an electron
moving through a potential difference of one volt.
1 eV  1.6 x 10-19 J
Electric Potential 17
Electric Potential Due to Point Charges
The electric potential due to a point charge
-r
+r
k - Q 
V
r
kQ
Vr
V
-V
-r
k  Q 
V
r
kQ
V
r
+r
Electric Potential 17
Problem 17-18
(a) What is the electric potential a distance of 2.5 x 10-15 m
away from a proton?
-19 C
1.60
x
10
Q
 8.99 x 109 Nm 2 /C2
Vk
r
2.5 x 10-15 m


 5.8 x 105 V
(b) What is the electric potential energy of a system that
consists of two protons 2.5 x 10-15 m apart?

Q1Q 2
 8.99 x 109 Nm 2 /C2
PE  k
r
PE  9.2 x 10
-14


2
19
1.60 x 10
C
2.5 x 10 -15 m
J
Electric Potential 17
Potential For Multiple Charges
The Electric Potential V in the vicinity of a number
of charges is equal to the algebraic sum of the
potentials due to each charge.
Q1 - r1
r3
Q3 -
A
r2
+
Q2
kQ1 kQ2 kQ3
VA 


r1
r2
r3
kQ
V 
r
Potential is + or – based on sign of the charges Q.
Example 5: Two charges Q1= +3 nC and Q2
= -5 nC are separated by 8 cm. Calculate the
electric potential at point A.

B
kQ1 kQ2
VA 

r1
r2

2 cm
9 x 10
(3 x 10 C)
kQ1
C2

 450 V
r1
(0.06 m)
kQ2

r2

9 x 10
9 Nm 2
9 Nm 2
C
2

-9
(-5 x 10-9C)
(0.02 m)
VA = 450 V – 2250 V;
 -2250 V
Q1 + +3 nC
6 cm
A 
2 cm
VA = -1800 V
Q2 = -5 nC
Capacitance
A capacitor consists of two conductors that are close
but not touching. A capacitor has the ability to store
electric charge and stores energy in an electric field.
-Q
+Q
Capacitance
C
Q
V
Coulomb
1
 1 Farad
Volt
V
Battery
Electric Potential 17
Problem 17-36
The charge on a capacitor increases by 18 mC when the
voltage across it increases from 97 V to 121 V. What is the
capacitance of the capacitor?
Q 2 - Q1
18 μC
C

V2 - V1
24 V
 7.5 x 10- 7 F
Electric Potential 17
Storage of Electric Energy
A charged capacitor stores electric energy; the energy
stored is equal to the work done to charge the capacitor.
-Q
+Q
C
QV
Energy 
2
CV 2

2
Q2

2C
V
Battery
Q  CV
2
Q
V2 
C2
Energy density of an electric field
energy 1
energy density 
 εoE2
volume 2
Electric Potential 17