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Chapter 25
25-1
25-2
25-3
charges
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electric potential
Potential difference and electric Potential
Potential Difference and electric field
Electric Potential and Potential energy due to point
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1
25-1 Potential difference and electric Potential
dW  dU
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Work and Potential Energy
E  lim
q 0
Electric Field Definition:
F
 F  qE
q
Work Energy Theorem
E
a
b
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Electric Potential Difference
a
b
E
Definition:
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Conventions for the potential “zero point”
Ub  Ua
Wba
Vba  Vb  Va 

q
q
0
Choice 1: Va=0
“Potential”
0
Ub  Ua
Vb  Va 
q
0
Ub
Vb 
q
0
Choice 2: V  0
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25-2 Potential Difference and electric field
When a force is “conservative” ie
gravitational and the electrostatic force
a potential energy can be defined
Change in electric potential energy is negative
of work done by electric force:
∆ V = -∫ E ds = -Ed
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•The change in potential energy is directly related to the change in voltage.
DU = qDV
DV = DU/q
• DU: change in electrical potential energy (J)
• q: charge moved (C)
• DV: potential difference (V)
•All charges will spontaneously go to lower potential energies if they are
allowed to move.
Units of Potential Difference
Vba  Vb  Va 
Ub  Ua
Wba

q
q
 Joules   J 
    Volt  V


Coulomb

 C
Because of this, potential difference is often referred to as “voltage”
In addition, 1 N/C = 1 V/m - we can interpret the electric field as a
measure of the rate of change with position of the electric potential.
So what is an electron Volt (eV)?
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Electron-Volts
• Another unit of energy that is commonly used in atomic and nuclear
physics is the electron-volt
• One electron-volt is defined as the energy a charge-field system
gains or loses when a charge of magnitude e (an electron or a
proton) is moved through a potential difference of 1 volt
1 eV = 1.60 x 10-19 J
• Since all charges try to decrease UE, and DUE = qDV, this
means that spontaneous movement of charges result in
negative DU.
• DV = DU / q
• Positive charges like to DECREASE their potential (DV < 0)
• Negative charges like to INCREASE their potential. (DV > 0)
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B
B
DV  V B V A    E .ds   E .  ds   E .d  E s cos 
A
A
VB – VA = VC - VA
VB = VC
A uniform electric field directed along the
positive x axis. Point B is at a lower electric
potential than point A. Points B and C are at
the same electric potential.
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Example
If a 9 V battery has a charge of 46 C how much chemical
energy does the battery have?
E = V x Q = 9 V x 46C = 414 Joules
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Example
A pair of oppositely charged, parallel plates are separated by 5.33 mm. A
potential difference of 600 V exists between the plates. (a) What is
the magnitude of the electric field strength between the plates? (b)
What is the magnitude of the force on an electron between the
plates?
d  0.00533m
DV  600V
E ?
DV  Ed
600  E (0.0053)
E  113,207.55 N / C
qe   1.6 x10 19 C
Fe
Fe
E

q 1.6 x10 19 C
Fe  1.811014 N
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Example
Calculate the speed of a proton that is accelerated from rest
through a potential difference of 120 V
q p   1.6 x10 19 C
m p   1.67 x10  27 kg
V  120V
v?
W DK
DV 


q
q
v
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2qDV

m
1
2mv2
q
2(1.6 x10 19 )(120)
5

1
.
52

10 m/s
 27
1.67 x10
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25-3 Electric Potential and Potential energy due to
point charges
+Q
kq
E  2 rˆ
r
ds for a point charge
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Recall the convention for the potential “zero
point”
1 1
Vba  Vb  Va  kq   
 rb ra 
V  0
1 1
Vb  Vb  V  kq   
 rb  
kq
V r 
r
Equipotential surfaces are concentric spheres
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Superposition of potentials
V0  V1  V2  V3  ...
+Q1
+Q2
r10
r20
0
+Q3
r30
kQ1 kQ 2 kQ3
V0 


 ...
r10
r20
r30
N
kQi
V0  
i 1 ri0
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E and V for a Point Charge
• The equipotential lines are the
dashed blue lines
• The electric field lines are the brown
lines
• The equipotential lines are
everywhere perpendicular to the field
lines
An equipotential surface is a
surface on which the electric
potential is the same
everywhere.
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Figure 25.4 (Quick Quiz 25.3) Four equipotential
surfaces
Equipotential surfaces (the dashed blue lines are
intersections of these surfaces with the page)
and electric field lines (red- rown lines) for (a) a
uniform electric field produced by an infinite
sheet of charge, (b) a point charge, In all cases,
the equipotential surfaces are perpendicular to
the electric field lines at every point
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Example (25.1)
A 12-V battery connected to two parallel plates. The electric field between the
plates has a magnitude given by the potential difference V divided by the plate
separation d =0.3 cm
Example (25.2)
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Example: (a) In figure a, 12 electrons are equally spaced and fixed
around a circle of radius R. Relative to V=0 at infinity, what are the
electric potential and electric field at the center C of the circle due to
these electrons? (b) If the electrons are moved along the circle until
they are nonuniformly spaced over a 120 are (figure b), what then is
the potential at C?
Solution:
12e
(a) : V   K
R
E0
12e
(b) : V   K
R
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Potential due to a group of point charges
n
Example (25.3)
n
qi
V  Vi 

4 0 i ri
i 1
1
(a) The electric potential at P due to the two
charges q1 and q2 is the algebraic sum of the
potentials due to the individual charges. (b) A third
charge q3 = 3.00 C is brought from infinity to a
position near the other charges.
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Example
An electric dipole consists of two charges q1 = +12nC and q2 = -12nC,
placed 10 cm apart as shown in the figure. Compute the potential at
points a,b, and c.
q1 q2
Va  k  (  )
ra ra
12 x109  12 x109
Va  8.99 x10 (

)
0.06
0.04
Va  899 V
q1 q2
Vb  k  (  )
rb rb
9
9
9
12
x
10

12
x
10
Vb  8.99 x109 (

)
0.04
0.14
Vb  1926.4 V
Vc  0 V
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Example The Total Electric Potential
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 C2  8.0 108 C 8.99 109 N  m 2 C2  8.0 108 C

0V
0.40 m
0.40 m
(a) If two point charges are separated by a distance r12,
the potential energy of the pair of charges is given by
keq1q2/r 12 . (b) If charge q1 is removed, a potential
keq2/r 12 exists at point P due to charge q 2 .
U  V q1
q1 q2
U  ke
r12
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Potential energy due to multiple point
charges
+Q2
+Q1
kq
V r 
r
kq1q 2
U  q2V 
r12
r21
r21
+Q2
+Q1
r23
r13
+Q3
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kq1
V
r12
kq1 kq 2
V

r13
r23
kq1q 2 kq1q 3 kq 2 q 3
U


r12
r13
r23
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Example 1. What is the potential energy if a +2
nC charge moves from  to point A, 8 cm away
from a +6 mC charge?
The P.E. will be positive at point
A, because the field can do +
work if q is released.
Potential Energy:
U
(9 x 10
U = 1.35 mJ
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kQq
U
r
9 Nm 2
C2
A

+2 nC
8 cm
+Q
+6 mC
)(6 x 10-6 C)(+2 x 10-9 C)
(0.08 m)
Positive potential
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Signs for Potential Energy
Consider Points A, B, and C.
A


For +2 nC at A: U = +1.35 mJ
12 cm
8 cm
 C
+Q
Questions:
If +2 nC moves from A to B, does field E do +
or – work? Does P.E. increase or decrease?
+6
mC
Moving
B
4 cm
positive q
+2
nC
The field E does positive work, the P.E. decreases.
If +2 nC moves from A to C (closer to +Q), the field E does negative
work and P.E. increases.
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Example. What is the change in potential energy if
a +2 nC charge moves from A to B?
Potential Energy:
kQq
U
r
A

8 cm
From Ex-1: UA = + 1.35 mJ
UB 
(9 x 19
9 Nm 2
C2

B
12 cm
+Q
+6
mC
)(6 x 10-6C)(+2 x 10-9C)
(0.12 m)
DU = UB – UA = 0.9 mJ – 1.35 mJ
 0.900 mJ
DU = -0.450 mJ
Note that P.E. has decreased as work is done by E.
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Example What is the change in potential energy if a
-2 nC charge moves from A to B?
kQq
U
Potential Energy:
r
From Ex-1: UA = -1.35 mJ
A

8 cm
(Negative due to – charge)
UB 
(9 x 19
9 Nm 2
C2
B
12 cm
+Q
+6
mC
)(6 x 10-6C)(-2 x 10-9C)
(0.12 m)

 0.900 mJ
DU = +0.450 mJ
UB – UA = -0.9 mJ – (-1.35 mJ)
A – charge moved away from a + charge gains P.E.
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Example :Find the potential at a distance of 6 cm
from a –5 nC charge.
P.
q = –4 mC
r
6 cm
- -Q - -
Q = -5 nC
kQ
V

r

9 Nm2
9 x 10
Negative V at Point
P:
C
2

(5 x 10-9C)
(0.06 m)
VP = -750 V
What would be the P.E. of a –4 mC charge placed
at this point P?
U = qV = (-4 x 10-6 mC)(-750 V);
U = 3.00 mJ
Since P.E. is positive, E will do + work if q is released.
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Example : Two charges Q1= +3 nC and Q2 = -5 nC
are separated by 8 cm. Calculate the electric
potential at point A.
kQ1 kQ2
VA 

r1
r2
kQ1
r1
9 x 10


kQ2

r2

9 x 10
9 Nm 2
C
 (3 x 10
2 cm
C)
(0.06 m)
9 Nm 2
C
2

Q1
+
+3 nC
 450 V
6 cm
(5 x 10-9C)
(0.02 m)
VA = 450 V – 2250 V;
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2
-9
B 
 2250 V
A

2 cm
VA = -1800 V
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Q2 = -5 nC
32
Example Calculate the electric potential at point B for
same charges.
kQ1

r1

kQ2

r2
kQ1 kQ2
VB 

r1
r2
9 x 109 Nm
C2

(3 x 10-9C)
(0.02 m)

9 x 10
9 Nm 2
VB = 1350 V – 450 V;
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2
C
2

B 
2 cm
 1350 V
Q1
+
+3 nC
6 cm
(5 x 10-9C)
(0.10 m)
 450 V
A

2 cm
VB = +900 V
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Q2 = -5 nC
33
Example : What is the potential difference between
points A and B. What work is done by the E-field if a +2 mC
charge is moved from A to B?
VA = -1800 V
VB = +900 V
B
VAB= VA – VB = -1800 V – 900 V
VAB = -2700 V
Q1
Note point B is at higher
potential.
WorkAB = q(VA – VB) = (2 x 10-6 C )(-2700 V)
Work = -5.40 mJ

2 cm
+
+3 nC
6 cm
A
Q2

-
2 cm
-5 nC
E-field does negative work.
Thus, an external force was required to move the charge.
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Example 6 (Cont.): Now suppose the +2 mC charge is
moved from back from B to A?
VA = -1800 V
VB = +900 V
B
Q1
VBA= VB – VA = 900 V – (-1800 V)

2 cm
+
+3 nC
6 cm
VBA = +2700 V
This path is from high
to low potential.
WorkBA = q(VB – VA) = (2 x
Work = +5.40 mJ
10-6
C )(+2700 V)
A
Q2

-
2 cm
-5 nC
E-field does positive work.
The work is done BY the E-field this time !
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Example
An electron is accelerated in a TV tube through a
potential difference of 5000 V.
a) What is the change in PE of the electron?
V = DPE/q
DPE = qV = (-1.60 X 10-19 C)(+5000 V)= -8.0 X 10-16 J
What is the final speed of the electron (m = 9.1 X 10-31 kg)
DPE + DKE = 0 (Law of conservation of energy)
DPE = -DKE
DPE = - ½ mv2
v2 = (-2)(DPE) = (-2)(-8.0 X 10-16 J)
m
9.1 X 10-31 kg
v = 4.2 X 107 m/s
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Summary
• Electric potential energy:
• Electric potential difference: work done to move
charge from one point to another
• Relationship between potential difference and field:
• Equipotential: line or surface along which potential is
the same
• Electric potential of a point charge:
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1:
2:
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3:
4:
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5:
The electrons in a particle beam each have a kinetic
energy of 1.60 x 10-17 J. What are the magnitude
and direction of the electric field that stops these
electrons in a distance of 10.0 cm?
6:
An electron and a proton are each placed at rest in
an electric field of 520 N/C. Calculate the speed of
each particle 48.0 ns after being released.
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q1
d
7:
What is the potential at point P, located at the
center of the square of point charges.
Assume that d = 1.3m and the charges are
q1 = +12 n C,
q3 = +31 n C,
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q2
q2= -24 n C
q4= +17 n C
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d
P
d
d
q3
q4
41
1- The electric field has a magnitude of 3.0 N/m at a distance of 60 cm from
a point charge. What is the charge?
(a) 1.4 nC
(b) 120 pC
(c) 36 mC
(d) 12 mC
(e) 3.0 nC
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1- A conducting sphere has a net charge of −4.8 × 10−17 C. What
is the approximate number of excess electrons on the sphere?
(a) 100
(b) 200
(c) 300
(d) 400
(e) 500
Electric charge always occurs in multiples of e
Q = Ne
e = 1.60
19
× 10
C
(N =1、2、3…)
N= (-4.8x10-17 C/-1.6x10-19 C=300 electrons)
2- Two point charges, 8x10-9 C and -2x10-9 C are separated by
4 m. The electric field magnitude (in units of V/m) midway
between them is:
A) 9x109
B) 13,500 C) 135,000 D) 36x10-9 E) 22.5
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E
E
E
E
kq1 kq2
 E1  E 2  2  2
2
2
9  109  8  109 9  109  2  109


2
2
22
98 9 2 9
 2  2  2 ( 8  2)
2
2
2
90

 22.5 N / C
4
8  109 2  109

2
x
(4  x) 2
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3 - If 10000 electrons are removed from a neutral ball, its
charge is;
(a) +1.6×10-15 C
(b) +1.6×10-23 C
(c) -1.6×10-15 C
(d) -1.6×10-23 C
Q = Ne =10000 x -1.6×10-19
Q = -1.6×10-15 C
4 - A charge of 10-6 C is in a field of 9000 N/C, directed
upwards. The magnitude and direction of the force it
experiences are;
(a) 9×10-3 N, downwards
(b) 3×10-3 N, downwards
(c) 9×10-3 N, upwards
(d) 3×10-3 N, upwards
F= q E = 9000 x 10-6
F = 9 x 10 -3 N
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