Lightning
Becomes very
“negative”
Becomes very
“positive”
T . Norah Ali Almoneef
1
Arbitrary numbers of protons (+)
and electrons (-) on a comb and in
hair
(A) before and
(B) after combing. Combing
transfers electrons from the hair to
the comb by friction, resulting in a
negative charge on the comb and a
positive charge on the hair.
• If a positively charged rod is
brought near a trickle of water, the
water moves towards it. What
happens if we use a negatively
charged rod?
T . Norah Ali Almoneef
2
2
Electric Charge
• Types:
– Positive
• Glass rubbed with silk
• Missing electrons
– Negative
• Rubber/Plastic rubbed with fur
• Extra electrons
• Arbitrary choice
– convention attributed to ?
• Units: amount of charge is measured in
[Coulombs]
• Empirical Observations:
– Like charges repel
– Unlike charges attract
T . Norah Ali Almoneef
3
•
In the process of rubbing two solid
objects together, electrical charges are
NOT created. Instead, both objects
contain both positive and negative
charges. During the rubbing process,
the negative charge is transferred from
one object to the other and this leaves
one object with an excess of positive
charge and the other with an excess of
negative charge. The quantity of
excess charge on each object is
exactly the same.
T . Norah Ali Almoneef
4
Charge Properties
CONSERVATION OF ELECTRIC CHARGE
– Charge is not created or destroyed, only transferred.
• , however, it can be transferred from one object to another. The net amount
of electric charge produced in any process is zero.
• Quantization
– The smallest unit of charge is that on an electron or proton. (e = 1.6 x 10-19 C)
• It is impossible to have less charge than this
• It is possible to have integer multiples of this charge
– A coulomb is the charge resulting from the transfer of 6.24 x 1018 of the
charge carried by an electron.
– The magnitude of an electrical charge (q) is dependent upon how many
electrons (n) have been moved to it or away from it.
Mathematically,
Q  Ne
T . Norah Ali Almoneef
5
• Electric Charge and Electrical Forces:
• Electrons have a negative electrical charge.
• Protons have a positive electrical charge.
• These charges interact to create an electrical force.
– Like charges produce repulsive forces – so they repel each other (e.g.
electron and electron or proton and proton repel each other).
– Unlike charges produce attractive forces – so they attract each other
(e.g. electron and proton attract each other).
It is impossible to have less charge than this
It is possible to have integer multiples of this charge
How many electrons constitute 1 mC?
qtotal
# electrons 
1.6 x1019
T . Norah Ali Almoneef
Q  Ne
6
Electric Force - Coulomb’s Law
• Consider two electric charges: q1 and q2
• The electric force F between these two charges separated
by a distance r is given by Coulomb’s Law
kq1q2
F 
r2
• The constant k is called Coulomb’s constant
and is given by k 
109 Nm 2 /C 2
9
• Coulomb law
– The electrical force between two charged bodies
is directly proportional to the charge on each
body and inversely proportional to the square of
the distance between them.
T . Norah Ali Almoneef
7
• The coulomb constant is also written as
2
1
12 C
k
where  0  8.85 10
4  0
Nm2
• 0 is the “electric permittivity of vacuum”
– A fundamental constant of nature
1
q1q 2
F
40 r 2
According to the superposition principle the resultant force on a point charge q
equals the vector sum of the forces exerted by the other point charges Qi that are
present:
• The force between two charges gets stronger as the
charges move closer together.
• The force also gets stronger if the amount of charge
becomes larger.
T . Norah Ali Almoneef
8
the resultant force on any one of them
equals the vector sum of the forces
exerted by the various individual
charges. For example, if four charges
are present, then the resultant force
exerted by particles 2, 3, and 4 on
particle 1 is
q1 q2
Fe  ke
r2
or
q1q2
F12  ke 2 rˆ12
r
T . Norah Ali Almoneef
9
example
• A positive charge of 6.0 x 10 -6C is 0.030m from a second positive
charge of 3.0 x 10 -6C. Calculate the force between the charges.
Fe  ke
q1 q2
r2
= (8.99 x 109 N m2/C2 ) (6.0 x 10 -6C) (3.0 x 10 -6C)
( 0.030m )2
= (8.99 x 109 N m2/C2 ) (18.0 x 10 -12C)
(9.0 x 10 -4 m2)
= + 1.8 x 10 -8 N
T . Norah Ali Almoneef
10
Three point charges, q1 = - 4 nC, q2 = 5 nC, and q3 = 3 nC, are placed as in the Fig.
T . Norah Ali Almoneef
11
Coulomb's Law
• The force between charges is directly
proportional to the magnitude, or amount,
of each charge.
• Doubling one charge doubles the force.
• Doubling both charges quadruples the force.
• The force between charges is inversely
proportional to the square of the distance
between them.
• Doubling the distance reduces the force by a
factor of 22 = (4), decreasing the force to
one-fourth its original value (1/4).
• .
T . Norah Ali Almoneef
12
• Compare electrical to gravitational force in a hydrogen atom
Felectrical  9 109
1.6 10 
0.53 10 
Fgravity  6.67 10
19 2
10 2
 8.2 10 8 N
 27
31
1
.
67

10

9
.
11

10
11
0.53 10 
10 2
 3.6110  47 N
•Electron and proton attract each other 1040 times stronger electrically than
gravitationally
T . Norah Ali Almoneef
13
An object, A, with +8.25 x 10-6 C charge, has two other charges nearby. Object B, -3.5 x 10-6
C, is 0.030 m to the right. Object C, +2.50 * 10-6 C, is 0.050 m below. What is the net force
and the angle on A?V

X

Y
T . Norah Ali Almoneef
14
Example
• What is the magnitude of the electric force of attraction between
an iron nucleus (q=+26e) and its innermost electron if the distance
between them is 1.5 x 10-12 m
.The magnitude of the Coulomb force is
F = kq1q2/r2
= (9.0 X 109 N · m2/C2)(26)(1.60 X 10–19 C)(1.60 X10–19 C)/(1.5 X10–12 m)2
=
2.7 X 10–3 N.
Example
• We have a charge of 5 C at X = 2 m.
• What will the force on a charge of -4 C be at X = 0 m.
• F = kq1 q2 / r2
• , F = 5 C x (4C) x 9x109 / 4 m2
• F = 45 x109 N
T . Norah Ali Almoneef
15
example
q1=+q
q22q
Q3=+3q
Given that q = -13 µC and d = 13 cm, find the direction and magnitude of
the net electrostatic force exerted on the point charge q1 in Figure.
F12(force on q1 from q2)= k (q1)(q2)/(d)2
= (9x109 N m2/C2) (13x10 -6C)(26x10-6C)/(0.13m)2
= 180 N
towards q2 (attractive)
F13 (force on q1 from q3)= k(q2)(q3)/(2d)2
= (9E9 N m2/C2) (13x10-6C)(-39x10-6C)/(0.26m)2
= +67.4 N
away from q3 (repulsive)
Net force = (180 – 67.4)=112 N Towards q2
T . Norah Ali Almoneef
16
example
Two point charges lie on the x axis. A charge of -9.5 µC is at the origin, and a charge of +
2.5 µC is at x = 10.0 cm.
(a) What is the net electric field at x = -2.0 cm?
Electric field from charge –9.5x10-6C is
E1 = (9x10+9 Nm2/C29.5x10-6C)/(0-(-0.02m))2
E1 = 2.14E8 N/C
field points towards charge –9.5x10-6 (positive x-direction)
Electric field at x=-0.02 m from charge 2.5x-6 C is
E2 = (9x10+9 Nm2/C22.5x10-6C)/(0.10m-(-0.02m))2
E2 = 1.56E6 N/C
field points away from charge 2.5x10-6 (negative x-direction)
Net electric field E = 2.14E8 i N/C  1.56E6 i N/C
= [2.12e+08] i N/C
T . Norah Ali Almoneef
17
example
Three point charges, q1 = -1.2 x 10-8 C, q2 = -2.6 x 10-8 C and q3 = +3.4 x 10-8 C,
are held at the positions shown in the figure, where a = 0.16 m
T . Norah Ali Almoneef
18
example
Two electrostatic point charges of +20.0 μC and –30.0 μC
exert attractive forces on each other of –145 N .What is
the distance between the two charges?
r
kq
1
q
2
F
9
6
6
9 x10 x 20 x10 x30 x10
r
145
r  0.193m
T . Norah Ali Almoneef
19
Three point charges lie along the x axis as shown in Figure. The positive charge q1 !
15.0 *C is at x ! 2.00 m,the positive charge q2 ! 6.00 *C is at the origin, and the
resultant force acting on q3 is zero. What is the x coordinate of q3?
Solution Because q3 is negative and q1 and q2 are positive,
For the resultant force on q3 to be zero, F23 must be equal in magnitude and opposite
in direction to F13. Setting the magnitudes of the two forces equal,
T . Norah Ali Almoneef
20
 Consider two charges located on the x axisThe charges are described by
q1 = 0.15 mC
q2 = 0.35 mC
x1 = 0.0 m
x2 = 0.40 m
x1
x2
Where do we need to put a third charge for that charge to be at an equilibrium
point?
At the equilibrium point, the forces from the two charges will cancel.
third charge to be at an equilibrium point when
F1  F2
kq1q3
kq2 q3

2
x
(0.4  x) 2
q1
q2

2
x
(0.4  x) 2
x = 0.12m
or
= 0.72m
X
T . Norah Ali Almoneef
21
Example
Suppose two charges having equal but opposite charge are separated by
6.4 × 10-8 m. If the magnitude of the electric force between the charges
is 5.62 ×10–14 N, what is the value of q
T . Norah Ali Almoneef
22
example
Charged spheres A and B are fixed in position, as shown, and have charges
of +7.9 x 10-6 C and -2.3 x 10-6 C, respectively. Calculate the net force on
sphere C, whose charge is +5.8 x 10-6 C.
6
6
7
.
9

10
2
.
3

10
Fc  9 109  5.8 106 [

]
2 2
2 2
(25 10 ) (15 10 )
F  1.26 N ..to the Left
T . Norah Ali Almoneef
23
F
T . Norah Ali Almoneef

1.1  7.9
2
2
24
Example:
What is the force between two charges of 1 C separated by 1 meter?
kq1q2
F
2
r
9 x109 x1x1
9

9
x
10

F= 12
Example:
What is the electric force between a +5 mC charge and a –3 mC
charge separated by 3 cm?
F = (9 x 109 Nm2/C2)(.005C)(.003C) / (.03 m)2
= 150,000,000 N
T . Norah Ali Almoneef
25
25
How far apart would two objects, each with a charge of 1Coulomb,
have to be so that they only exerted a 1 Newton electric force on
one another?
T . Norah Ali Almoneef
26
Object A has a charge of +2 μC, and object B has a
charge of +6 μC. Which statement is true about the electric
forces on the objects?
•
FAB = –3FBA
•
FAB = –FBA
•
3FAB = –FBA
•
FAB = 3FBA
•
FAB = FBA
•
3FAB = FBA
From Newton's third law, the
electric force exerted by object B
on object A is equal in magnitude
to the force exerted by object A on
object B and in the opposite
direction.
T . Norah Ali Almoneef
27
Where do I have to place the + charge in order for the force
to balance, in the figure at right?
•
•
•
•
Force is attractive toward both negative charges, hence could
balance.
Need a coordinate system, so choose total distance as L, and
position of + charge from q charge as x.
Force is sum of the two force vectors, and has to be zero, so
2qQ
qQ
F  F1  F2  k

k
0
2
2
( L  x)
x
A lot of things cancel, including Q, so our answer does not depend
on knowing the + charge value. We end up with
2
1

( L  x) 2 x 2
•
( L  x) 2
Lx

2

 2
x2
x
L
Solving for x, x 
, L so slightly less than half-way
 0.412
1 2
between.A
T . Norah Ali Almoneef
2q
L
x
q
28
A charged particle, with charge Q, produces an
electric field in the region of space around it
A small test charge, qo, placed in the field, will
experience a force


F q E
0


F
Electric Field
E 
q0
•Mathematically,
•Use this for the magnitude of the field
•The electric field is a vector quantity The direction of the field is
defined to be the direction of the electric force that would be
exerted on a small positive test charge placed at that point
•For a point charge
T . Norah Ali Almoneef
29
• The force (Fe) acting on a “test” charge (qo) placed in an electric
field (E) is
Fe = qoE
• Note the similarity of the electric force law to Newton’s 2nd Law
(F=ma)
• Formal definition of electric field:
– the electric force per unit charge that acts on a test charge
at a point in space or
E = Fe/qo
•The electric field exists whether or not there is a test
charge present
•The Superposition Principle can be applied to the
electric field if a group of charges is present
T . Norah Ali Almoneef
30
Electric Field
• The electric force on a positive test charge
q0 at a distance r from a single charge q:

qq0
Fe  k 2 rˆ
r
• The electric field at a distance r from a single
charge q:

 Fe
q
E
 k 2 rˆ
q0
r
T . Norah Ali Almoneef


Fe  q0 E
31
Example 9
A positive charge is released from rest in a region of electric
field. The charge moves:
a) towards a region of smaller electric potential
b) along a path of constant electric potential
c) towards a region of greater electric potential
A positive charge placed in an electric field will experience a force
given by
F qE
Therefore
F qE
Since q is positive, the force F points in the direction opposite to
increasing potential or in the direction of decreasing potential
T . Norah Ali Almoneef
32
The electric field at a given point P
is the sum of the electric fields due to
every point charge
P
+
+
+
+
-
A charge distribution
T . Norah Ali Almoneef
N
E   Ei
i 1
N
qi
  k 2 rˆi
ri
i 1
33
The electric force on a charge q is
F  qE
which, together with Newton’s 2nd Law,
can be used to calculate the
motion of an electric charge,
of mass m
F  ma
Newton’s 2nd Law for an electric charge can be written as
q
E a
m
If E is constant, both in direction and
magnitude, so to is the acceleration of
the charge.
Note that the acceleration depends on the charge to mass ratio.
T . Norah Ali Almoneef
34
Electric field & line
Point charge
•The lines radiate equally in all directions
•For a positive source charge, the lines will
radiate outward
For a negative source charge, the lines will
point inward
T . Norah Ali Almoneef
35
– A map of the electrical field can be made by bringing a positive test charge into
an electrical field.
• No two field lines can cross.
• You can draw vector arrows to indicate the direction of the
electrical field.
This is represented by drawing lines of force or electrical field
lines,
• These lines are closer together when the field is stronger
and farther apart when it is weaker.
•Field lines must begin on positive charges (or from infinity) and end on
negative charges (or at infinity). The test charge is positive by convention
•The number of lines drawn leaving a positive charge or approaching a
negative charge is proportional to the magnitude of the charge
•the magnitude of the electric force will increase proportionally with an
increase in charge and/or and increase in the electric field magnitude
•
The line must be perpendicular to the surface of the charge
T . Norah Ali Almoneef
36
Shark
Fish to detect object
T . Norah Ali Almoneef
37
Electric Field lines
T . Norah Ali Almoneef
38
example
4
4
mg   r 3 g  (1000)  (0.6 10 6 ) 3 (9.8)  8.87 10 15
3
3
qE  mg
q  8.87 10 15 / 462  1.9 10 17
N  120e
T . Norah Ali Almoneef
39
example
Which of the following statements about electric field
lines associated with electric charges is false?
•
1) Electric field lines can be either straight or curved.
•
2) Electric field lines can form closed loops.
•
3 )Electric field lines begin on positive charges and end on negative
charges.
•
4 ) Electric field lines can never intersect with one another.
T . Norah Ali Almoneef
40
You are sitting a certain distance from a
point charge, and you measure an electric
field of E0. If the charge is doubled and
your distance from the charge is also
doubled, what is the electric field strength
now?
(1) 4 E0
(2) 2 E0
(3) E0
(4) 1/2 E0
(5) 1/4 E0
Remember that the electric field is: E = kQ/r2. Doubling
the charge puts a factor of 2 in the numerator, but doubling
the distance puts a factor of 4 in the denominator, because
it is distance squared!! Overall, that gives us a factor of
1/2.
T . Norah Ali Almoneef
41
Between the red and the blue
charge, which of them
experiences the greater electric
field due to the green charge?
1)
+1
2)
+2
3) the same for both
+2
+1
+1
d
+1
d
Both charges feel the same electric field due
to the green charge because they are at the
same point in space!
T . Norah Ali Almoneef
Q
Ek 2
r
42
Between the red and the
blue charge, which of
them experiences the
greater electric force due
to the green charge?
1)
+1
2)
+2
3) the same for both
+2
+1
+1
d
+1
d
The electric field is the same for both charges, but the
force on a given charge also depends on the
F  qE
magnitude of that specific charge.
T . Norah Ali Almoneef
43
example
Find electric field at point P in the figure
kq
9 109  8 10 9
E 2 
 18 N / C
r
4
o
1.6
  tan
 306.87
1.2
1
T . Norah Ali Almoneef
44
Example
In figure shown, locate the point at which
the electric field is zero? Assume a = 50cm
E1 = E2
d = 30cm
Example
Find the electric field at point p
in figure .due to the charges
shown.
T . Norah Ali Almoneef
45
Example
In figure shown, locate the point at which
the electric field is zero? Assume a = 50cm
E1 = E2
d = 30cm
Example
Find the electric field at point p
in figure .due to the charges
shown.
T . Norah Ali Almoneef
46
Example
In figure shown, locate the point at which
the electric field is zero? Assume a = 50cm
E1 = E2
d = 30cm
Example
Find the electric field at point p in figure .due
to the charges shown.
Ex = E1 - E2 = -36´104N/C
Ey = E3 = 28.8´104N/C
Ep = [(36´104)2+(28.8´104)2 ] = 46.1N/C
 = 141o
T . Norah Ali Almoneef
47
Example
Three identical charges (q = –5.0 mC)
lie along a circle of radius 2.0 m at angles
of 30°, 150°, and 270°, as shown. What is
the resultant electric field at the center of
the circle?
E x  E 1 cos 30 0  E 2 cos 30 0  0
q


k
E1 E2
r2
E y  (E 1 y  E 2 y )  E 3
E y  (E 1 sin 30 0  E 2 sin 30 0 )  k

Etot  0
E2
E1
30°
E3
q
1 q 1 q
q

(
k

k
)

k
0
2
2
2
2
2 r 2 r
r
r
T . Norah Ali Almoneef
48
Example A +100 mC point charge is separated from a
-50 mC charge by a distance of 0.50 m as shown below. (A) First calculate the
electric field at midway between the two charges. (B) Find the force on an
electron that is placed at this point and then calculate the acceleration when
it is released.
Q1
Q2

E
_
+
In part A we found that E = 2.1x107 N/C and is directed to the right.


F
E 
q
F  q E  eE

F   1.6  1019 C  2.1107 N

C
 3.4 1012 N
( to left )


F  ma
F
a 
m
3.4  10 12 N

 3.7 10 18 m 2
31
s
9.110 kg
T . Norah Ali Almoneef
( to left )
49
16.3 Electric Field due to arrangements of charges
the Field on Electric Dipole

kq
E 
y
2
(r  a)

kq
 y
E 
2
(r  a)

The total field at P is

1
1
E  E  kq{

}
2
2 y
(r  a) (r  a)

4kar
E [ 2
]y
2 2
(r  a )
When r is much greater than a we can neglect
a in the denominator then
1
E
4kqa
r
3

y
E
r
3
T . Norah Ali Almoneef
50
Note
• A set of two (equal and opposite) charges separated by a distance
•the dipole electric field reduces as 1/r3, instead of 1/r of a single charge.
• although we only calculate the fields along z-axis, it turns out that this
also applies to all direction.
• p is the basic property of an electric dipole, but not q or d. Only the
product qd is important.
kq
E 
x;
2
R  a 
 kq
E 
x
2
R  a 


1
1
E  E   E   kq

x
2
2 
R  a  
 R  a 
4kqa
R  a  E  
x
3
R
T . Norah Ali Almoneef
E
p
E
51
Example
Find the electric field due to electric dipole shown in figure along x-axis at point p which
is a distance r from the origin. then assume r>>a
Solution
When x>>a then
T . Norah Ali Almoneef
52
A test charge of +3 μC is at a point P where an
external electric field is directed to the right and has a
magnitude of 4 × 106 N/C. If the test charge is replaced with
another test charge of –3 μC, the external electric field at P
•
A )is unaffected
•
B )reverses direction
•
C )changes in a way that cannot be determined
There is no effect on the electric field if we assume that
the source charge producing the field is not disturbed by
our actions. Remember that the electric field is created
by source charge(s) (unseen in this case), not the test
charge(s).
T . Norah Ali Almoneef
53
A Styrofoam ball covered with a conducting paint has a mass of
5.0 × 10-3 kg and has a charge of 4.0 μC. What electric field
directed upward will produce an electric force on the ball that will
balance the weight of the ball?
(a) 8.2 × 102 N/C
(b) 1.2 × 104 N/C
(c) 2.0 × 10-2 N/C
(d) 5.1 × 106 N/C
The magnitude of the upward electrical force must equal the
weight of the ball. That is:
T . Norah Ali Almoneef
54
•An electric dipole consists of two equal and
opposite charges
•The high density of lines between the charges
indicates the strong electric field in this region
Two equal but like point charges
•At a great distance from the charges, the field
would be approximately that of a
single charge of 2q
•The bulging out of the field lines between the
charges indicates the repulsion between the
charges
•The low field lines between the charges indicates
a weak field in this region
T . Norah Ali Almoneef
55
In a uniform electric field in empty space, a 4 C
charge is placed and it feels an electrical force of
12 N. If this charge is removed and a 6 C charge
is placed at that point instead, what force will it
feel?
1) 12 N
2) 8 N
3) 24 N
4) no force
5) 18 N
Since the 4 C charge feels a force, there must
be an electric field present, with magnitude:
E = F / q = 12 N / 4 C = 3 N/C
Once the 4 C charge is replaced with a 6 C
Q
charge, this new charge will feel a force of:
F = q E = (6 C)(3 N/C) = 18 N
T . Norah Ali Almoneef
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T . Norah Ali Almoneef
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example
Determine the point (other than infinity) at which the total electric field is zero
we will call the position of the negative charge x = 0,
which means the positive charge is at x = 1 m. We will call the position where electric
field is zero x. The distance from this point to
the negative charge is just x, and the distance to the positive charge is 1 + x. Now write
down the electric field due to each charge:
E
E
pos
neg


k ( 2.5 x10 6 )
x
2
k (6 x10  6 )
(1 x)
k ( 2.5 x10  6 )
2
x
2.5 (1 x)
x  1.82,
2

2
k (6 x10  6 )
(1 x)
6
2
x
2
 0.39
We wrote down the distance x the distance to
the left of the negative charge. A negative value
of x is then in
the wrong direction, in between the two
charges, which we already ruled out. The
positive root, x = 1.82, means a distance 1.82m
to the left of the negative charge. This is what
we want.
T . Norah Ali Almoneef
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A charge q1 = 7.0 µC is located at the origin, and a second charge
q2 =5.0 µC is located on the x axis, 0.30 m from the origin .Find the
electric field at the point P,
which has coordinates (0, 0.40) m.
T . Norah Ali Almoneef
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T . Norah Ali Almoneef
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TT ..Norah
Norah Ali
Ali Almoneef
Almoneef
61
Electric fields of Concentric spherical shells
E
If the smaller shell has a radius R :
Surface area of the sphere is :
E

A  4

KQ
R
2
(Out side)
2
R
4 K Q
A
T . Norah Ali Almoneef
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T . Norah Ali Almoneef
63
Planar Symmetry
For two oppositely charged plates
placed near each other, E field
outer side of the plates is zero
while inner side the E-field
=E = 2π KQ / A
 Two conducting plates with charge density 1
 All charges on the two faces of the plates
 For two oppositely charged plates placed near each other, E field outer
side of the plates is zero while inner side the E-field= 2 1/0
T . Norah Ali Almoneef
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E field of a single uniformly charged plate
• Charged infinite plane: E = 2k (: Q/A)
• E = 2π KQ / A
For two oppositely charged plates placed near each other, E field outer
side of the plates is zero while inner side the E-field
• E =4π KQ / A

Einside 
Q
0

Eoutside  0
A
T . Norah Ali Almoneef
65
• The electrostatic force is a conservative (=“path independent”)
force
• It is possible to define an electrical potential energy function with
this force
• Work done by a conservative force is equal to the negative of the
change in potential energy
• There is a uniform field between the two plates
• As the positive charge moves from A to B, work
is done
• WAB=F d=q E d
• ΔPE =-W AB=-q E d
– only for a uniform field
T . Norah Ali Almoneef
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Potential Difference (=“Voltage Drop”)
The electric potential V at a given point is the electric potential
energy U of a small test charge q0 situated at that point divided by
the charge itself:
The electric potential difference between any two points i and f in an
electric field.
If we set
at infinity as our reference potential energy,
• The potential difference between points A and B is defined as the
change in the potential energy (final value minus initial value) of
a charge q moved from A to B divided by the size of the charge
• ΔV = VB – VA = ΔPE /q
• Potential difference is not the same as potential energy
SI Unit of Electric Potential: joule/coulomb=volt (V)
T . Norah Ali Almoneef
67
•
It is equal to the difference in potential energy per unit charge
between the two points.
• the negative work done by the electric field on a unite charge as
that particle moves in from point i to point f.
• Another way to relate the energy and the potential difference:
ΔPE = q ΔV
• Both electric potential energy and potential difference are scalar
quantities
• A special case occurs when there is a uniform electric field
• VB – VA= -Ed
• Gives more information about units: N/C = V/m
•The electric potential energy U and the electric potential V are not the
same. The electric potential energy is associated with a test charge, while
electric potential is the property of the electric field and does not depend on
the test charge.
• A larger charge would involve a larger amount of PEe, but the ratio of that
energy to the charge is the same as it would be if a smaller charge was in
that same place.
T . Norah Ali Almoneef
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Energy and Charge Movements
• A positive charge gains electrical potential energy when it is
moved in a direction opposite the electric field
• If a charge is released in the electric field, it experiences a force
and accelerates, gaining kinetic energy
– As it gains kinetic energy, it loses an equal amount of electrical potential
energy
• A negative charge loses electrical potential energy when it moves
in the direction opposite the electric field
• When the electric field is directed downward,
point B is at a lower potential than point A
• A positive test charge that moves from A to B
loses electric potential energy
• It will gain the same amount of kinetic energy
as it loses potential energy
T . Norah Ali Almoneef
69
•
Electric field always points from higher electric potential to
lower electric potential.
•
A positive charge accelerates from a region of higher
electric potential energy (or higher potential) toward a
region of lower electric potential energy (or lower potential).
– it moves in the direction of the field, Its electrical potential
energy decreases, Its kinetic energy increases
•
A negative charge accelerates from a region of lower
potential toward a region of higher potential.
– It moves opposite to the direction of the field
– Its electrical potential energy decreases
– Its kinetic energy increases
T . Norah Ali Almoneef
70
General Considerations
• If a charged particle moves perpendicular to electric field lines, no
work is done.
if d  E
• If the work done by the electric field is zero, then the electric
potential must be constant
We
V  
 0  V is constant
q
• Thus equipotential surfaces and lines must always be perpendicular
to the electric field lines.
T . Norah Ali Almoneef
71
Electrical Potential Energy in a Uniform Electric Field
• If a charge is released in a uniform electric field at a constant
velocity there is a change in the electrical potential energy
associated with the charge’s new position in the field.
•  PEe = -qE d
The unit: Joules
• The negative sign indicates that the electrical potential energy will
increase if the charge is negative and decrease if the charge is
positive.
• The  V in a uniform field varies with the displacement from a reference
point.
•  V = E d
• The displacement is moved in the direction of the field.
• Any displacement perpendicular to the field does not change the electrical
potential energy.
T . Norah Ali Almoneef
72
POTENTIAL ENERGY IN A
UNIFORM FIELD
The Electric Field points in the direction of
a positive test charge.
+ Charge
- Charge
Loses PEe
Gains PEe
Opposite E Gains PEe
Loses PEe
Along E
T . Norah Ali Almoneef
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T . Norah Ali Almoneef
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74
T . Norah Ali Almoneef
75
75
Example:
An electron accelerates from rest through an electric potential of 2 V.
What is the final speed of the electron?
Answer: The electron acquires 2 eV kinetic energy. We have
1
J 

19
me v 2  [2 eV] 1.6  10 19

3.2

10
J ,

2
eV


and since the mass of the electron is me = 9.1 × 10–31 kg, the speed is
v 
2  3.2  10 19 J
9.1  10 – 31 kg
 8.4 10 5 m/s .
T . Norah Ali Almoneef
76
example
An electron (mass m = 9.11×10-31kg) is accelerated in the uniform
field E (E = 1.33×104 N/C) between two parallel charged plates.
The separation of the plates is 1.25 cm. The electron is
accelerated from rest near the negative plate and passes through
a tiny hole in the positive plate, as seen in the figure. With what
speed does it leave the hole?
F = qE = ma
Vf 2 = vi2 + 2a(d)
a = qE/m
Vf
GIVEN:
m = 9.11×10-31kg
• Vf
2=
2ad
2=
2ad = 2(qE/m)d
E = 1.33×104 N/C
= 2(qE/m)d
= 2 (1.9 x 10
= 8.3 x 10 6 m/s
-19C)
d = 1.25 cm
(1.33×104 N/C) (1.25m) / 9.11×10-31kg
T . Norah Ali Almoneef
77
“Electric field lines always point in the direction of decreasing electric potential”
The unit: V m-1 Or NC-1
T . Norah Ali Almoneef
78
78
e-
T . Norah Ali Almoneef
79
eeee-
But greater
a more the
usefulmagnitude
concept isofthethe
The
electric the
potential
energy
of electric
each
charge,
greater
is the
charge energy
potential
T . Norah Ali Almoneef
80
Clicker Question
• In the figure, a proton moves from
point i to point f in a uniform
electric field directed as shown.
Does the electric field do positive,
negative or no work on the proton?
A: positive
B: negative
C: no work is done on the proton
T . Norah Ali Almoneef
81
Electric Potential Energy with a Pair of Charges
• A single point charge produces a non-uniform electric field.
• PEe = kq1q2/r
• The reference point for PEe is assumed to be at infinity.
• The ground is usually the reference point for PEg.
• The PEe is positive for like charges and negative for unlike charges.
Work and Electrical Potential Energy
• In order to bring two like charges near each other work must be
done. (W = Fd)
• In order to separate two opposite charge, work must be done.
• Remember: whenever work gets done, energy changes form.
The potential energy will change to kinetic energy.
T . Norah Ali Almoneef
82
• Since the electrical potential energy can change depending on the amount of
charge you are moving, it is helpful to describe the electrical potential energy
per unit charge.
• Electric potential: the electrical potential energy associated with a charged
particle divided by the charge of the particle.
• A larger charge would involve a larger amount of PEe, but the ratio of that
energy to the charge is the same as it would be if a smaller charge was in that
same place.
T . Norah Ali Almoneef
83
16.4Electrical Potential Energy of Two Charges
• V1 is the electric potential due to q1 at some
point P1
• The work required to bring q2 from infinity to
P1 without acceleration is q2E1d=q2V1
• This work is equal to the potential energy of
the two particle system
PE  q 2V1  k e
q1q 2
r
• If the charges have the same sign, PE is positive
– Positive work must be done to force the two charges near one
another
– The like charges would repel
• If the charges have opposite signs, PE is negative
– The force would be attractive
– Work must be done to hold back the unlike charges from
accelerating as they are brought close together
T . Norah Ali Almoneef
84
Potential Difference Near a Point Charge
• An electric potential exists at some point in an electric field
regardless of whether there is a charge at that point.
• The electric potential at a point depends on only two quantities:
the charge responsible for the electric potential and the
distance (r)from this charge to the point in question.
•  V = kcq/r
• Voltage is a way of using numbers (quantitative) to describe an
electric field.
• Electric fields are measured in volts over a distance. This means
the larger the E the larger the V.
• When an E is attracting or repelling an object, we instead could
say that the object is being driven by the voltage.
T . Norah Ali Almoneef
85
Example:
Calculate the electric potential, V, at 10 cm from
a -60mC charge.
Equation:
Answer:
V = -5.4x 106V
Calculate the electric potential, V, at the midpoint between
a 250 mC charge and a -450 mC separated by a distance of
60 cm.
Equation:
Answer:
T . Norah Ali Almoneef
V = -6.0x106V
86
Example
If you want to move in a region of electric field without
changing your electric potential energy. You would move
a) Parallel to the electric field
b) Perpendicular to the electric field
T . Norah Ali Almoneef
87
Electric Potential
General Points for either positive or negative charges
The Potential increases if you move in the direction
opposite to the electric field
and
The Potential decreases if you move in the same direction
as the electric field
Electric Potential is a scalar field
it is defined everywhere
it doesn’t depend on a charge being there
but it does not have any direction
T . Norah Ali Almoneef
88
Example
Points A, B, and C lie in a
uniform electric field.
A
E
B
C
What is the potential difference between points A and B?
ΔVAB = VB - VA
a) ΔVAB > 0
b) ΔVAB = 0
c) ΔVAB < 0
The electric field, E, points in the direction of decreasing
potential
Since points A and B are in the same relative horizontal
location in the electric field there is on potential difference
between them
T . Norah Ali Almoneef
89
Example
A
Points A, B, and C lie in a
uniform electric field.
E
B
C
Point C is at a higher potential than point A.
True
False
As stated previously the electric field points in the direction of
decreasing potential
Since point C is further to the right in the electric field and the
electric field is pointing to the right, point C is at a lower
potential
The statement is therefore false
T . Norah Ali Almoneef
90
Example
Points A, B, and C lie in a
uniform electric field.
A
E
B
C
Compare the potential differences between points A and C
and points B and C.
a) VAC > VBC
b) VAC = VBC
c) VAC < VBC
In Example 4 we showed that the the potential at points A and B
were the same
Therefore the potential difference between A and C and the
potential difference between points B and C are the same
Also remember that potential and potential energy are scalars
and directions do not come into play
T . Norah Ali Almoneef
91
Example
A
Points A, B, and C lie in a
uniform electric field.
E
B
C
If a negative charge is moved from point A to point B, its electric
potential energy
a) Increases.
b) decreases.
c) doesn’t change.
The potential energy of a charge at a location in an electric field is
given by the product of the charge and the potential at the
location
As shown in Example 4, the potential at points A and B are the
same
Therefore the electric potential energy also doesn’t change
T . Norah Ali Almoneef
92
•
•
•
•
•
•
•
•
•
•
The Joule
The joule is a measure of work accomplished on an object.
It is also a measure of potential energy or how much work an object can do.
In the English system the unit of work and energy is the ft x lb.
F = m x a For a falling object a = g, so
F=mxg
Energy is force x distance.
E=Fxd
For a falling object d=h (h=height)
E=Fxh
PE= m x g x h
The Volt
The commonly encountered unit joules/coulomb is called the volt, abbreviated V, after the
Italian physicist Alessandro Volta (1745 - 1827)
1V=
1J
1C
•
With this definition of the volt, we can express the units of the electric field as
[F ] N J/m V
[E ] 
 

[q ] C
C
m
•
For the remainder of our studies, we will use the unit V/m for the electric field.
T . Norah Ali Almoneef
93
The Electron Volt
There is an additional unit that is used for energy in addition to
that of joules
A particle having the charge of e (1.6 x 10-19 C) that is moved
through a potential difference of 1 Volt has an increase in
energy that is given by
• The electron volt (eV) is defined as the energy
that an electron (or proton) gains when
accelerated through a potential difference of 1 V
• 1 V=1 J/C  1 eV = 1.6 x 10-19 J
W  qV  1.6  1019 joules
 1 eV
T . Norah Ali Almoneef
94
Potential energy
Calculate the work done bringing a 250 mC charge and a -450
mC from infinity to a distance of 60 cm apart.
Equation:
= 9x109(250x10-6)(-450x10-6)
60x10-2
PE = -1.7x103J
T . Norah Ali Almoneef
95
Example
• A proton is placed between two parallel
conducting plates in a vacuum as shown.
The potential difference between the two
plates is 450 V. The proton is released from
rest close to the positive plate.
• What is the kinetic energy of the proton
when it reaches the negative plate?
+
-
The potential difference between the two plates is 450 V. = V(+)-V()
The change in potential energy of the proton is U, and V = U / q
(by definition of V), so
U = q V = e[V()V(+)] = 450 eV
T . Norah Ali Almoneef
96
Example
Suppose an electron is released from rest in a uniform electric field
whose magnitude is 5.90 x 103 V/m. (a) Through what potential
difference will it have passed after moving 1.00 cm? (b) How fast will
the electron be moving after it has traveled 1.00 cm?
(a) |V| = Ed = (5.90 x 103 V/m)(0.0100 m) = 59.0 V
(b) q |V| = mv2/2  v = 4.55x106 m/s
Example
An ion accelerated through a potential difference of 115 V experiences an
increase in kinetic energy of 7.37 x 10 –17 J. Calculate the charge on the
ion.
qV= 7.37x10-17 J , V=115 V
q = 6.41x10-19 C
T . Norah Ali Almoneef
97
Example
A particle has a mass of 1.8x10-5kg and a charge of +3.0x10-5C. It is released from
point A and accelerates horizontally until it reaches point B. The only force acting
on the particle is the electric force, and the electric potential at A is 25V greater than
at C. (a) What is the speed of the particle at point B? (b) If the same particle had a
negative charge and were released from point B, what would be its speed at A?
1
2
mvB2  EPEB  12 mvA2  EPEA
1
2
mvB2  12 mvA2  EPEA  EPEB
1
2
mvB2  12 mvA2  qo VA  VB 
T . Norah Ali Almoneef
98
Example
The work done by the electric force as the test charge (+2.0x10-6C)
moves from A to B is +5.0x10-5J.
(a) Find the difference in EPE between these points.
(b) Determine the potential difference between these points.
WAB  EPE A  EPE B
5
EPE B  EPE A  WAB  5.0 10 J
VB  V A 
 WAB  5.0 10 5 J
VB  VA 

 25 V
-6
qo
2.0 10 C
(a)
EPE B EPE A
 WAB


qo
qo
qo
WAB  EPE A  EPE B
EPE B  EPE A  WAB  5.0 105 J
(b)
 WAB
 5.0 10 5 J
VB  VA 

 25 V
qo
2.0 10-6 C
T . Norah Ali Almoneef
99
Example 3: A proton is moved from the negative plate to the positive plate of a parallelplate 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  Fx cos
1
FE  qE
W  qEx
N
W  (1.6 10 C )(1500 )(.015m)
C
W  3.6 10 18 J
19
T . Norah Ali Almoneef
100
Example:
E0  W  E f
U E  K1  K 2
q1q2 1 2 1 2
k
 mv  mv
r
2
2
2
q
k
 mv 2
r
2
kq
2
v 
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 106 ) 2
m
v

 2.0
rm
(.2)(. 4)
s
T . Norah Ali Almoneef
101
:
Problem
An electron is released from rest in an electric field of 2000N/C.
How fast will the electron be moving after traveling 30cm?
 UE  K
1
 qEd  mv 2
2
v=?
_
v = 0m/s
_
30cm
v 

2qEd
m
2(1.6 10 19 )( 2000)(.3)
v
9.1110 31
m
v  1.45  10
s 102
7
T . Norah Ali Almoneef
F=qE
Ignore triangle on sketch.
• F = qE
 q=+0.045·106 C > 0, Force parallel to E
 Charge accelerates to left
 Work done on charge = qEd = Change in Kinetic Energy





W = (0.045·106C)(1200 V/m)(0.05m) = 54. ·103V ·C =0.054J
(Kf-Ki) = W
(1/2) mvf2 – 0 = 0.054 J
vf2 = 2 (0.054 J) / ( 3.5 ·103kg)= 30.8 m2/s2
vf =5.6 m/s
T . Norah Ali Almoneef
103
Example:
An electron accelerates from rest through an electric
potential of 2 V. What is the final speed of the electron?
The electron acquires 2 eV kinetic energy. We have
1
J 

19
me v 2  [2 eV] 1.6  10 19

3.2

10
J ,

2
eV 

and since the mass of the electron is me = 9.1 × 10–31
kg, the speed is
v 
2  3.2  10 19 J
9.1  10 – 31 kg
 8.4 10 5 m/s .
T . Norah Ali Almoneef
104
An a- particle, mass = 6.7x10-27 kg, initially at rest travels 25 cm
through a uniform electric field of 250 N/C. Calculate the
potential difference across the 25 cm path
Equation:
Answer: 62.5 V
V =Ed
What is the a-particle’s speed after 25 cm of travel?
Equation:
qEd = ½ mv2
T . Norah Ali Almoneef
Answer:7.7x104m/s
105
How much energy (work) is necessary to bring three point charges from
infinity to the vertices of the right triangle shown below.
Equation:
-4.0mC
PE = kq1q2
r
Answer:
5.0 cm
4.0 cm
2.0 mC
PE = -2.16J
PE = 1.8 J
PE = -1.8 J
PE = -2.16J
3.0 mC
3.0 cm
T . Norah Ali Almoneef
106
example
Calculate the work done bringing a 250 mC charge and a -450 mC from infinity to
a distance of 60 cm apart.
qq
PE  K 1 2 = 9x109(250x10-6)(-450x10-6)
r
60x10-2
PE = -1.7x103J
example :
The potential difference between to charge plates is 500V. Find the velocity of a
proton if it is accelerated from rest from one plate to the other.
High
Potential
Low
Potential
+
-
++
-
 U E  K
qV 
v 
500
V
Positive charges move from high to low potential
Negative charges move from low to high potential
T . Norah Ali Almoneef
1
mv 2
2
2qV
m
2(1.6  10 19 )(500 )
v 
1.67  10 27
m
v  3.1  10 5
s
107
example
Which of the following statements is false?
A. The total work required to assemble a collection of discrete charges is the
electrostatic potential energy of the system.
B. The potential energy of a pair of positively charged bodies is positive.
C. The potential energy of a pair of oppositely charged bodies is positive.
D. The potential energy of a pair of oppositely charged bodies is negative.
E. The potential energy of a pair of negatively charged bodies is negative.
example
The figure depicts a uniform
electric field. Along which
direction is the increase in the
electric potential a maximum?
T . Norah Ali Almoneef
108
Potential Difference in a Uniform field
C
+Q
E
WAC  WAB  WBC
W AB  F|| d  QEd ||
WBC  F|| d  0
 QEd ||
U AC  QEd ||
d||
V AC   Ed||
+Q
+Q
A
B
T . Norah Ali Almoneef
109
example
The potential at a point due to a unit positive point charge is
found to be V. If the distance between the charge and the point
is tripled, the potential becomes
A. V/3.
B. 3V.
C. V/9.
D. 9V.
E. 1/V 2 .
example
The figure shows two plates A and B. Plate A has a potential of 0 V and plate B a
potential of 100 V. The dotted lines represent equipotential lines of 25, 50, and 75
V. A positive test charge of 1.6 × 10–19 C at point x is transferred to point z. The
electric potential energy gained or lost by the test charge is
A.
B.
C.
D.
E.
8 × 10–18 J, gained.
8 × 10–18 J, lost.
24 × 10–18 J, gained.
24 × 10–8 J, lost.
40 × 10–8 J, gained.
T . Norah Ali Almoneef
110
example
Two parallel metal plates 5.0 cm apart have a potential difference
between them of 75 V. The electric force on a positive charge of 3.2
× 10–19 C at a point midway between the plates is approximately
A.
B.
C.
D.
E.
4.8 × 10–18 N.
2.4 × 10–17 N.
1.6 × 10–18 N.
4.8 × 10–16 N.
9.6 × 10–17 N.
example
When +2.0 C of charge moves at constant speed from a point with zero
potential to a point with potential +6.0 V, the amount of work done is
A. 2 J.
B. 3 J.
C. 6 J.
D. 12 J.
E. 24 J.
T . Norah Ali Almoneef
111
The concept of “potential difference" or “voltage" in
electricity is similar to the concept of "height" in gravity,
or “pressure” in fluids
T . Norah Ali Almoneef
112
A point charge of +3 μC is located at the origin of a coordinate
system and a second point charge of -6 μC is at x = 1.0 m. At what
point on the x-axis is the electrical potential zero?
1.
2.
3.
4.
-0.25 m
+0.25 m
+0.33 m
+0.75 m
A positive charge of 6 nC attracts a negative charge of 5 nC from a distance of 4 mm. What
is the work done? What does the sign mean?
T . Norah Ali Almoneef
113
Example :
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.
q1q2
U12  k
r
6
6
(
4

10
)(
4

10
)
9
U12  (9 10 )
.2
U12  0.72 J
U total  .72 J  (.72 J )  (.72 J )
U total  .72 J
T . Norah Ali Almoneef
114
Two point charges of values +3.4 and +6.6 μC respectively, are
separated by 0.20 m. What is the potential energy of this 2-charge
system?
1. +0.34 J
2. -0.75 J
3. +1.0 J
4. -3.4 J
What will be the electrical potential at a distance of 0.15 m
from a point charge of 6.0 μC?
1.
2.
3.
4.
5.4 x 104 V
3.6 x 105 V
2.4 x 106 V
1.2 x 107 V
T . Norah Ali Almoneef
115
Example
1. The potential difference between the two terminals on a battery
is 9 volts. How much work (energy) is required to transfer 10
coulombs of charge across the terminals?
V = 9.0 V
W= ?
q= 10 c
9.0 = W/10
W = 90 J
Example
2. The work required to transfer 30 coulombs of charge across two
terminals is 50 joules.
What is the potential difference?
V= ?
W= 50 J
q= 30 c
V=50/30
V =1.7 V
TT ..Norah
Norah Ali
Ali Almoneef
Almoneef
116
Problem
Show that the amount of work required to assemble four
identical point charges of magnitude Q at the corners of a
square of side s is 5.41keQ2/s.
U  U1U 2 U 3 U 4
U  0  U 12  U 13  U 23   U 14  U 24  U 34 
keQ 2 keQ 2  1
1
 keQ 2 

U  0


1

1


1

s
s  2 
s 
2 
keQ 2 
keQ 2
2 
U 
4
 5.41


s 
s
2
T . Norah Ali Almoneef
117
Example
Consider a positive and a negative charge, freely moving in a
uniform electric field. True or false?
(a) Positive charge moves to points with lower potential.
(b) Negative charge moves to points with lower potential.
(c) Positive charge moves to a lower potential energy
position.
(d) Negative charge moves to a lower potential energy
position
(a) True
(b) False
(c) True
(d) True
+V
–Q
T . Norah Ali Almoneef
+Q
0
–V
118
PROBLEM
• A proton is released from rest in a uniform electric field with
a magnitude of 8 E 4 V/m. The proton is displaced 0.5 m as a
result. A) Find the potential difference between the
proton’s initial and final positions. B) Find the change in
electrical potential energy of the proton as a result of this
displacement.
E = 8 E 4 V/m q = + 1.6 10 -19 C d = 0.5 m
A) V = -Ed  V = - (8 E 4 V/m)(0.5 m) =
V = - 40000 V
B) V = PEe/q  PEe = V q 
(-40000 V)(+1.6 10 -19 C) = - 6.4 E-15 J
T . Norah Ali Almoneef
119
The Potential due to a Point Charge
Although only changes in potential are physically
relevant, it is often convenient to choose the location
of the zero of the potential. For
a car battery, this is typically the car’s
chassis; for an electrical outlet it
is the ground. For an isolated point
charge, it is convenient to choose
the potential to be zero at infinity
VAB
1 1
 V ( B)  V ( A)  kq   
 rB rA 
T . Norah Ali Almoneef
120
The Potential of a Point Charge
The potential difference between two points A and B from a point
charge
1 1
VAB  V ( B)  V ( A)  kq   
 rB rA 
can be re-written as
1 1
V ( B)  V ( A)  kq   
 rB rA 
When rA = infinity the last term vanishes. We are free to choose V(A)
as we please, e.g., V(A) = 0.
kq
With this choice, the potential of a point charge becomes
T . Norah Ali Almoneef
V (r ) 
r
121
Electric Potential of Many Point Charges
• Electric potential is a SCALAR not a vector.
• Just calculate the potential due to each individual
point charge, and add together! (Make sure you
q4
get the SIGNS correct!)
• Superposition principle applies
r4
• The total electric potential at some
point P due to several point charges is
q5
the algebraic sum of the electric
potentials due to the individual charges
– The algebraic sum is used because
potentials are scalar quantities
qi
V  k
ri
i
q3
r3
P
r5
r2
q2
r1
q1
T . Norah Ali Almoneef
122
Potential from more than one
charge
Principle of superposition
Total Potential is sum of all
individual potentials
V  V1  V2  ...
Each potential is
Thus Total potential is
Q1
V1 
40 r1

1  Q1 Q2
V
 ...
 
40  r1 r2

Which can be written
T . Norah Ali Almoneef
1
1
Q
V

40
r
123
A
Electric Potential of a single charge
Remember that
It can be shown that
E
E 
Q
1
40 r 2
B
so
r
if V = 0 at rA=
+
Potential energy
Arbitrary shape
V  Er
Q 1
V 
40 r
Potential difference
Arbitrary shape
T . Norah Ali Almoneef
This looks a bit like the
formulae for the
potential in a Uniform
Field
V AC   Ed||
124
Potential Energy in 3 charges
Q2
Q1
Energy
when we
bring in Q2
U12  Q2V  Q2
U12 
Q3
1 Q1
40 r12
1
Q
V
r
40
1 Q1Q2
40 r12
Now bring in
U  U12  Q3V3
Q3
1  Q1 Q2 
 U12  Q3
  
40  r13 r23 
1  Q1Q2 Q1Q3 Q2Q3 
U




40  r12
r13
r23 
So finally we find
U  U12  U13  U 23
T . Norah Ali Almoneef
125
Example:
How many electrons should be removed from an initially
uncharged spherical conductor of radius 0.300 m to
produce a potential of 7.5 kV at the surface?
Substituting given values into V =
=
ke q
= 7.50 x 103 V
r
(8.99x10 9 Nm 2 /C 2 )q
(0.300m)
q  2.50  10 7 C
N = 1.56 x 1012 electrons
T . Norah Ali Almoneef
126
Pictures from Serway & Beichner
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 .212m.
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
.0212 10
+e
9
r1 is at a higher potential.
potential diff  V  27.2  6.79  20.4V
T . Norah Ali Almoneef
127
Example: Finding the Electric Potential at
Point P (apply V=keq/r).
5.0  10 6 C
V1  (8.99  10 Nm / C )
 1.12  10 4 V,
4.0m
6
(

2
.
0

10
C)
9
2
2
V2  (8.99  10 Nm / C )
 3.60  10 3 V
(3.0m) 2  (4.0m) 2
9
2
2
Superposition: Vp=V1+V2
Vp=1.12104 V+(-3.60103 V)=7.6103 V
5.0 mC
-2.0 mC
T . Norah Ali Almoneef
128
PROBLEMS
• As a particle moves 10 m along an electric field of strength 75 N/C,
its electrical potential energy decreases by 4.8 x10 – 16 J. What is the
particle’s charge?
• What is the potential difference between the initial and final
locations of the particle in the problem above?
• An electron moves 4.5 m in the direction of an electric field of
strength 325 N/C/ Determine the change in electrical potential
energy.
1) E = 75 N/C PEe = - 4.8 x10 -16 J d = 10 m
q = ? Must be a + q
since PEe was lost
PEe = - qEd  q = (- 4.8 x10 -16)/(-(75)(10)) =
Q = +6.4 E -19 C
2) V = PEe/q  (-4.8 x10 -16 J)/(6.4 E -19 C)
V = - 750 V
3) q = - 1.6 x10 -19 C d = 4.5 m E = 325 N/C
PEe = ?
PEe = - qEd 
-(-1.6 x10 -19 C)(325 N/C)(4.5 m) = 2.3 x10 -16 J
T . Norah Ali Almoneef
129
T . Norah Ali Almoneef
130
Example:
Consider three point charges q1 = q2 = 2.0 mC and q3 = -3 mC which are
placed as shown. Calculate the net force on q1 and q3
T . Norah Ali Almoneef
131
The force on q1 is F1 = F12 + F13.
T . Norah Ali Almoneef
132
Similarly, F3 = F31 + F32
T . Norah Ali Almoneef
133
Example
At locations A and B, find the total electric potential.
VA
8.99 10

VB
9

 


N  m 2 C2  8.0 108 C
8.99 109 N  m 2 C2  8.0 108 C

 240 V
0.20 m
0.60 m
8.99 10

9

 


N  m 2 C 2  8.0 108 C
8.99 109 N  m 2 C 2  8.0 108 C

0V
0.40 m
0.40 m
T . Norah Ali Almoneef
134
Example
A charge +q is at the origin. A charge –2q is at x = 2.00 m on the x
axis. For what finite value(s) of x is (a) the electric field zero ? (b)
the electric potential zero ?
q

2q
0
E  k  2 
2 
( x  2.00) 
x
x2 + 4.00x – 4.00 = 0
(x+4.83)(x0.83)=0
x = - 4.83 m
(other root is not physically valid)
q

2q
  0
V  

 x (2.00  x) 
x = 0.667 m and x= -2.00 m
T . Norah Ali Almoneef
135
• In the drawing on the right, q = 2.0 μC and d = 0.96 m. Find the
total potential at the location P, assuming that the potential is
zero at infinity.
q
q
q
q
V K
K K
K
2d
d
d
d
 q q q q
V  K
   
 2d d d d 
6
2 10
q
9
V  K
 9 10 
2d
0.96  2
V  9.4 10 V
3
T . Norah Ali Almoneef
136
The Potential due to a Point Charge:
T . Norah Ali Almoneef
137
137
T . Norah Ali Almoneef
138
T . Norah Ali Almoneef
139
Example
The three charges in Fig. , with q1 = 8 nC, q2 = 2 nC, and q3 = - 4 nC, are
separated by distances r2 = 3 cm and r3 = 4 cm. How much work is required to
move q1 to infinity?
T . Norah Ali Almoneef
140
T . Norah Ali Almoneef
141
Example
Calculate the electric potential, V, at 10 cm from a
60mC charge.
Equation:
-
Answer: V = -5.4x 106V
Example
Calculate the electric potential, V, at the midpoint between a 250
mC charge and a -450 mC separated by a distance of 60 cm.
Equation:
Answer: V = -6.0x106V
T . Norah Ali Almoneef
142
T . Norah Ali Almoneef
143
Example
Points A, B, and C lie in a
uniform electric field.
A
E
B
C
If a negative charge is moved from point A to point B, its electric
potential energy
a) Increases.
b) decreases.
c) doesn’t change.
The potential energy of a charge at a location in an electric field is given by the
product of the charge and the potential at the location
As shown in Example 4, the potential at points A and B are the
same
Therefore the electric potential energy also doesn’t change
T . Norah Ali Almoneef
144
E. Potential & Potential Energy vs
Electric Field & Coulomb Force
Electric Field is
Coulomb Force
divided by test
charge
Coulomb Force is
thus Electric Field
multiplied by
charge
F
E
Q0
F  QE
U
V 
Q0
U  QV
D Potential is D
Energy divided
by test charge
D Energy is D
Potential
multiplied by
test charge
If we know the potential field this allows us to calculate
changes in potential energy for any charge introduced
T . Norah Ali Almoneef
145
Electric Potential Difference
The electric potential energy depends on the charge
present
We can define and electric potential V which
does not depend on charge by using a “test”
charge
Change in potential is change in
potential energy for a test charge
divided by the unit charge
U
V 
Q0
Remember that for uniform field
U  Q0 Ed
T . Norah Ali Almoneef
U
V 
  Ed
Q0
146
Parallel-Plate Capacitor
•
•
•
•
•
The capacitor consists of two parallel plates
Each have area A
They are separated by a distance d
The plates carry equal and opposite charges
When connected to the battery, charge is pulled off one
plate and transferred to the other plate
• The transfer stops when Vcap = Vbattery
T . Norah Ali Almoneef
147
147
Capacitors
Storing a charge between the plates
• Electrons on the left plate are
attracted toward the positive
terminal of the voltage source
• This leaves an excess of positively
charged holes
• The electrons are pushed toward
the right plate
• Excess electrons leave a negative
charge
+
+
+
T . Norah Ali Almoneef
_
_
-
148
Capacitance of parallel plates
Capacitance – is a measure of the capacitor’s ability to store electric energy
where: q – magnitude of the charge
q
C
V
V – potential/potential difference
C - capacitance
V
The bigger the plates the more surface area
over which the capacitor can store charge
CA
E
+Q
-Q
Moving plates togeth`er Initially E is constant
(no charges moving) thus V=Ed decreases
charges flows from battery to increase
V C  1/d
+
Never Ready
T . Norah Ali Almoneef
149
In SI units,
q – coulomb (C)
V – volt (V)
C – coulomb/volt =
farad (F)
A Farad is very large Often will see µF or pF
• The capacitance of a device depends on the geometric
arrangement of the conductors
• For a parallel-plate capacitor whose plates are separated by air:
(C does not depend on Q or V)
[V = Ed, E=Q/(A 0), V = Qd / (A 0)]
For any geometry, capacitance scales as Area/distance
A
C  o
d
T . Norah Ali Almoneef
150
Electric Field
•
•
A uniform electric field (The field strength is the
same magnitude and direction at all points in the
field)can be produced in the space between two parallel
metal plates.
The plates are connected to a battery.
E
The field strength at any point in this field is:
V
E 
d
E = field strength (Vm-1)
V = potential difference (V)
d = plate separation (m)
T . Norah Ali Almoneef
151
Capacitance
Experiments show that the charge in a capacitor is proportional to the electric
potential difference (voltage) between the plates.
The constant of proportionality C is the capacitance which is a property of the
conductor
Q  V
V  Ed 
C 
Q
V
Q


0
0
A
A
d
Q  CV
d
Q
C 
V
To increase C, one either increases ,
increases A, or decreases d.
Note that if we doubled the voltage, we would not do anything to the
capacitance. Instead, we would double the charge stored on the
capacitor.
However, if we try to overfill the capacitor by placing too much voltage
across it, the positive and negative plates will attract each other so
strongly that they will spark across the gap and destroy the capacitor.
Thus capacitors have a maximum voltage!
T . Norah Ali Almoneef
152
Example
The plates of the capacitor are separated by
a distance of 0.032 m, and the potential difference
between them is VB-VA=-64V. Between the
two equipotential surfaces shown in color, there
is a potential difference of -3.0V. Find the spacing
between the two colored surfaces.
V
 64 V
3
E

 2.0 10 V m
d
0.032 m
V
 3.0 V
3
d 


1
.
5

10
m
3
E
2.0 10 V m
T . Norah Ali Almoneef
153
The two plates of a capacitor hold +5000mC and -5000mC, respectively, when
the potential difference is 200V. What is the capacitance?
Equation:
q 5000 10 6
C 
V
200V
C  25mF
How much charge flows from a 12 V battery when connected to a
20 mF capacitor?
Equation:
q = CV
q =20mF x12V = 240 mC
T . Norah Ali Almoneef
154
Example
• How strong is the electric field between the plates of a 0.80
mF air gap capacitor if they are 2.0 mm apart and each has a
charge of 72 mC? V  Q  72  90 V
C
0. 8
V
90
E

 45000 V / m
d
0.002
Example
Find the capacitance of a 4.0 cm diameter if the plates are separated by 0.25 mm.
8.85 1012 1.26 10 3
C
 4.5 10 11 F
0.25 10 3
 A
C  8.85  1012 F/m  r 
 d 
A  πr 2    0.02 m 2   1.26 103 m 2
8.85 1012 1.26 10 3
C
 4.5 10 11 F
0.25 10 3
T . Norah Ali Almoneef
155
Example
When a potential difference of 150 V is applied to the plates of a parallel-plate
capacitor, the plates carry a surface charge density of 30.0 nC/cm2. What is the
spacing between the plates?
d
0 A
Q 
 V 
d
8.85 1012 C 2
0  V 


30.0 10
9
C cm
2

N  m 2 150 V 
1.00  10
4
2
cm m
2

 4.42 mm
Example
What is the charge on a 250 microfarad capacitor if it has been charged to 12 V?
q = CV = (250E-6 F)(12 V) = 3E-3 Coulombs
T . Norah Ali Almoneef
156
Show Demo Model, calculate its capacitance , and show
how to charge it up with a battery.
Circular parallel plate capacitor
r
r = 10 cm
r
A = r2 = (.1)2
A = .03 m 2
s
S = 1 mm = .001 m
C
0A
S
C  (10 11 )
.03 Coulomb
.001 Volt
Farad
C  3  10 10 F
C  300 pF
T . Norah Ali Almoneef
p = pico = 10-12
157
Example: Thundercloud
• Suppose a thundercloud with horizontal dimensions of 2.0 km by 3.0
km hovers over a flat area, at an altitude of 500 m and carries a
charge of 160 C.
• Question 1:
– What is the potential difference between
the cloud and the ground?
• Question 2:
– Knowing that lightning strikes require
electric field strengths of approximately
2.5 MV/m, are these conditions sufficient
for a lightning strike?
T . Norah Ali Almoneef
158
Example:
• Question 1
• We can approximate the cloud-ground system as a parallel plate
capacitor whose capacitance is
0 A
(8.85·10-12 F/m)(2000 m)(3000 m)
C

 0.11 mF
d
500 m
• The charge carried by the cloud is 160 C, which means that the “plate
surface” facing the earth has a charge of 80 C
q
80 C
V 
 7.2 10 8 V
C 0.11 mF
++++++++++++
…++++++++++++ …
• 720 million volts
T . Norah Ali Almoneef
159
• Question 2
• We know the potential difference between the cloud and
ground so we can calculate the electric field
V 7.2 108
E 
 1.44 106 V/m
d
500
• E is lower than 2.5 MV/m, so no lightning cloud to ground
– May have lightning to radio tower or tree….
T . Norah Ali Almoneef
160
Problem
A parallel-plate capacitor has an area of 5.00 cm2 and the plates
are separated by a distance of 2.50 mm Calculate the capacitance.
C  0
A
d
C  8.85
12
 5  10  4
2.5  10 3
C  1.8 pF
Problem
What is the capacitance if it has .014 V across it when it has a
charge of 2.13x10-15 C?
C = q/V = (2.13E-15 C)/(.014 V) = 1.52x10-13 F
T . Norah Ali Almoneef
161
Problem
• If a spark jumps across a 1mm gap when you reach for a
doorknob, what was the potential difference between you and the
knob and how much charge was on you?
• Assume your finger and the knob form a parallel plate capacitor
with area 1 cm2 and breakdown electric field = 3MV/m.
V = Q/C
V = Ed
Q = CV=CEd = (A 0/d)(E d) = A 0 E
Q = (0.01m)2 (8.8510-12 F/m)(3.06 V/m)
Q = 2.7 10-15 (C/V)(Vm2/m 2) = 2.7 femto-Coulomb
T . Norah Ali Almoneef
162
Problem
(a) What plate area is required if an air-filled, parallel plate capacitor
with a plate separation of 2.6 mm is to have a capacitance of 12
pF?
(a) C = A 0/d,
A = Cd/ 0 = (12·10 12 F)(2.6 ·10 3 m) / (8.8510 12 F/m)
A = 3.5 ·103 m2 = (6cm)2
Problem
A 0.25 μ F capacitor is connected to a 400 V battery. What is the
charge on the capacitor?
1. 1.2 x 10-12 C
2. 1.0 x 10-4 C
3. 0.040 C
163
4. 0.020 C
T . Norah Ali Almoneef
Problem
The two plates of a capacitor hold +5000mC and -5000mC,
respectively, when the potential difference is 200V. What is the
capacitance?
Equation:
q
C
V
Answer
25 mF
Problem
How much charge flows from a 12 V battery when connected to a
20 mF capacitor?
Equation:
q = CV
Answer:
T . Norah Ali Almoneef
240 mC
164