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Chapter 16
Electric Forces , Fields And Potentials
Electric forces:
Coulomb’s Law
The electric force:
•is inversely proportional to the square of the separation r
between the particles and directed along the line joining them;
• is proportional to the product of the charges q1 and q2 on the
two particles;
• is attractive if the charges are of opposite sign and repulsive if the
charges have the same sign;
• is a conservative force.
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Electric forces:
Coulomb’s law
The Coulomb constant ke in SI units has the value
Ke = 9x109 N.m2/C2
This constant is also written in the form
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Coulomb constant
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Example:
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?
(a) FAB = -3FBA
(b) FAB = -FBA
(c) 3FAB = -FBA
(d) FAB = 3FBA
(e) FAB = FBA
(f) 3FAB = FBA
(b) FAB = -FBA
Example:
The electron and proton of a hydrogen atom are
separated (on the average) by a distance of approximately
5.3 x 1011 m. Find the magnitudes of the electric force and
the gravitational force
between the two particles.
(1.6 x10 19 ) 2
Fe  9 x10 x
 8.2 x10 8 N
11 2
(5.3x10 )
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The gravitational force between charged atomic particles is
negligible when compared with the electric force.
Example:
Consider three point charges located at the corners of a
right triangle as shown in Figure, where q1 = q3 = 5.0 µC,
q2 = -2.0 µC, and a = 0.10 m. Find the resultant force
exerted on q3.
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example:
Three point charges lie along the x axis as shown in Figure. The positive
charge q1 =15.0 µC is at x = 2.00 m, the positive charge
q2
=6.00 µC is at the origin, and the resultant force acting on q3 is zero.
What is the x coordinate of q3?
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The Electric Field
the electric field vector E at a point in space is defined as the
electric force Fe acting on a positive test charge q0 placed at that
point divided by the test charge:
The Electric Field
At any point P, the total electric field due to a group of source
charges equals the vector sum of the electric fields of all the
charges.
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Example:
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 x 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
(b) reverses direction
Electric field diagrams
•The electric field vector E is tangent to the electric field line at each point. The
line has a direction, indicated by an arrowhead, that is the same as that of the
electric field vector.
• The number of lines per unit area through a surface perpendicular to the lines
is proportional to the magnitude of the electric field in that region. Thus, the field
lines are close together where the electric field is strong and far apart where the
field is weak.
•The lines must begin on a positive charge and terminate on a negative charge.
In the case of an excess of one type of charge, some lines will begin or end
infinitely far away.
• The number of lines drawn leaving a positive charge or approaching a negative
charge is proportional to the magnitude of the charge.
• No two field lines can cross
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Electric field diagrams
Electric field diagrams
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Electric field diagrams
Electric Field
.B
Rank the magnitudes E of
the electric field at points A,
B, and C shown in the figure.
A) EC>EB>EA
B) EB>EC>EA
C) EA>EC>EB
D) EB>EA>EC
E) EA>EB>EC
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.C
.A
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Example:
Example:
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.
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Significance of the electric field
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The Electric Field
Significance of the electric field
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Problem #2: Calculate the electric field at point P below.
Charge A = 6C
Charge B = 2C
Distance from A to P = .4 meters
Distance from P to B = .2 meters
A
.4
P
.2
6C
B
2C
EB= (9x109)(2)
.22
EA= (9x109)(6)
.42
E=4.5x1011 N/C
E= 3.375x1011 N/C
Since the arrows point
opposite directions,
subtract the two
magnitudes to find the net
electric field
A positive test charge placed
at point P would move to the
right, away from charge A
A positive test charge placed
at point P would move to the
left, away from charge B
3.375x1011
4.5x1011
4.5 x 1011
3.375 x 10111.125 x 1011 N/C
The resulting direction
would be to the left
since that electric field
was stronger
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Electric Field of a Dipole


Start with
1 q
1 q

4 0 r2 4 0 r2

1
q
1
q

4 0 ( z  d / 2) 2 4 0 ( z  d / 2) 2

q
d
d
[(1  )  2  (1  )  2 ]
4 0 z 2
2z
2z
If d << z, then,
[(1 

E  E  E 
So
d 2
d
2d
2d
2d
)  (1  ) 2 ]  [(1 
 ...)  (1 
 ...)] 
2z
2z
2 z (1!)
2 z (1!)
z
E
q 2d
1 qd

4 0 z 2 z
2 0 z 3
p  qd
1 p
E
2 0 z 3
E ~ 1/z3
 E =>0 as d =>0
 Valid for “far field”

Spherical Conductors

Because it is conducting, charge on a metal sphere will go everywhere over the
surface.

You can easily see why, because each of the charges pushes on the others so that
they all move apart as far as they can go. Because of the symmetry of the situation,
they spread themselves out uniformly.

There is a theorem that applies to this case, called the shell theorem, that states
that the sphere will act as if all of the charge were concentrated at the center.
Note, forces are equal and opposite
These two situations are the same
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Electric charge
Condense charge E = (kQ/r2) r̂
outside spherically symmetric charge distribution
Properties of Electric fields

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Concentric spherical shells:
4kQ 4kQ
E

Field just outside sphere
A
4R 2
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Uniformly charge planes
E
2kQ
xˆ
A
Uniformly charge planes
E
4kQ
xˆ
A
Example 16.3/384
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The electrical potential
B
U  U B  U A    qE.ds qEl
A
V  VB  V A  El
l
Oppositely charged plates
V 
4kQ
l
A
l
The potential due to a point charge:
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Example:
A charge q1 = 2.00 µC is located at the origin, and a charge
q2 = -6.00 µC is located at (0, 3.00) m, as shown in Figure.
(A) Find the total electric potential due to these charges at
the point P, whose coordinates are (4.00, 0) m. Figure a
(B) Find the change in potential energy of the system of two
charges plus a charge q3 = 3.00 µC as the latter charge moves from
infinity to point P. Figure b
A)
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B)
Capacitance:
The capacitance C of a capacitor is defined as the ratio of the
magnitude of the charge on either conductor to the magnitude of
the potential difference between the conductors:
C 
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Q
V
Farad (F)
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The parallel plate capacitor:
C
A
4kd
Example:
A parallel-plate capacitor with air between the plates has an
area A = 2.00 x 10-4 m2 and a plate separation d = 1.00 mm.
Find its capacitance.
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