Chapter 17: Electric Potential

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Chapter 17
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Chapter 17: Electric Potential
•Electric Potential Energy
•Electric Potential
•How are the E-field and Electric Potential related?
•Motion of Point Charges in an E-field
•Capacitors
•Dielectrics
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§17.1 Electric Potential Energy
Electric potential energy (Ue) is energy stored in the electric
field.
•Ue depends only on the location, not upon the path taken
to get there (conservative force). Not a vector.
•Ue = 0 at some reference point.
•For two point particles take Ue = 0 at r = .
kq1q2
•For the electric force U e 
r
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Example: A proton and an electron, initially separated by a
distance r, are brought closer together.
(a) How does the potential energy of this system of
charges charge?
ke2
For these two charges U e  
r
Bringing the charges closer together decreases r:.
Ue  Uef  Uei  0
This is like a mass falling near the surface of the Earth;
positive work is done by the field.
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Example continued:
(b) How will the electric potential energy change if both
particles have positive (or negative) charges?
When q1 and q2 have the same algebraic sign then
Ue > 0.
This means that work must be done by an external
agent to bring the charges closer together.
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Example: What is the potential energy of three point charges
arranged as a right triangle? (See text Example 17.2)
q2
q2
r12
q1
Ue  0 
r12
r23
q3
r13
kq1q2 kq1q3 kq2 q3


r12
r13
r23
q1
r23
q3
r13
kq1q2 kq1q3 kq2 q3


Ue  0 
r12
r13
r23
Are these the same?
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§17.2 Electric Potential
Electric potential is the electric potential energy per unit
charge.
Ue
V
qtest
Electric potential (or just potential) is a measurable scalar
quantity. Its unit is the volt (1 V = 1 J/C).
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For a point charge of charge Q:
U e kQ
V

qtest
r
When a charge q moves through a potential difference
of V, its potential energy change is Ue = qV.
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Example: A charge Q = +1 nC is placed somewhere in space
far from other charges. Take ra = 1.0 m and re = 2.0 m.
f
b
c
e
a
Q
d
g
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Example continued:
Place a charge of +0.50 nC at point e. What will the
change in potential (V) be if this charge is moved to point
a?




kQ 9.0 109 Nm2 /C 2 1.0 nC
Ve 

 4.5 Volts
re
2m
kQ 9.0 109 Nm2 /C 2 1.0 nC
Va 

 9.0 Volts
ra
1m
V = Vf  Vi = Va  Ve = +4.5 Volts
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Example continued:
What is the change in potential energy (U) of the +0.50
nC charge ?
Ue = qV = (+0.50 nC)(+4.5 Volts)= +2.3 nJ
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§17.3 The Relationship between E
and V
f
b
The circles are
called equipotentials
(surfaces of equal
potential).
c
e
a
Q
+9 V
+4.5 V
d
g
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The electric field will point in the direction of maximum
potential decrease and will also be perpendicular to the
equipotential surfaces.
f
b
c
e
a
Q
+9 V
+4.5 V
d
g
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Equipotentials
and field lines
for a dipole.
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Uniform E-field
V1
V2
V3
V4
E
Equipotential surfaces
U e
V 
  Ed
q
Where d is the distance
over which V occurs.
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§17.4 Moving Charges
When only electric forces act on a charge, its total
mechanical energy will be conserved.
Ei  E f
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Example (text problem 17.40): Point P is at a potential of
500.0 kV and point S is at a potential of 200.0 kV. The space
between these points is evacuated. When a charge of +2e
moves from P to S, by how much does its kinetic energy
change?
Ei  E f
Ki  U i  K f  U f
K f  K i  U i  U f  U f  U i 
 U  qV  q Vs  V p 
  2e 200.0  500.0 kV
 9.6 1014 J
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Example (text problem 17.41): An electron is accelerated from
rest through a potential difference. If the electron reaches a
speed of 7.26106 m/s, what is the potential difference?
Ei  E f
0
Ki  U i  K f  U f
K f   U   qV
1
2
m vf   qV
2
2
2
31
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m vf
9.11 10 kg 7.26 10 m/s
V  

2q
2  1.60 1019 C

 150 Volt s



Note: the electron moves
from low V to high V.

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§17.5 Capacitors
A capacitor is a device that stores electric potential energy
by storing separated positive and negative charges. Work
must be done to separate the charges.
+
+
+
+
+
+
+
Parallel plate
capacitor
–
–
–
–
–
–
–
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For a parallel plate capacitor:
EQ
E  V
 Q  V
Written as an equality: Q = CV, where the proportionality
constant C is called the capacitance.
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What is the capacitance for a parallel plate capacitor?

Q
V  Ed  d 
d
0
0 A
0 A
Q 
d
V  C V
where C 
0 A
d
.
Note: C depends only on constants and geometrical factors.
The unit of capacitance is the farad (F). 1 F = 1 C2/J = 1 C/V
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Example (text problem 17.56): A parallel plate capacitor has
a capacitance of 1.20 nF. There is a charge of magnitude
0.800 C on each plate.
(a) What is the potential difference between the plates?
Q  CV
Q 0.800 C
V  
 667 Volt s
C
1.20 nF
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Example continued:
(b) If the plate separation is doubled, while the charge is
kept constant, what will happen to the potential difference?
Q Qd
V  
C 0 A
V  d
If d is doubled so is the potential difference.
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Example (text problem 17.100): A parallel plate capacitor
has a charge of 0.020 C on each plate with a potential
difference of 240 volts. The parallel plates are separated by
0.40 mm of air.
(a) What is the capacitance of this capacitor?
Q
0.020 C
C

 8.3 10 11 F  83 pF
V 240 Volts
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Example continued:
(b) What is the area of a single plate?
C
A
0 A
d
Cd
0

83 pF0.40 mm
8.851012 C 2 /Nm2
 0.0038m 2  38 cm2
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§17.6 Dielectrics
As more and more charge is placed on capacitor plates,
there will come a point when the E-field becomes strong
enough to begin to break down the material (medium)
between the capacitor plates.
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To increase the capacitance, a dielectric can be placed
between the capacitor plates.
C   C0
where C0 
0 A
d
and  is the dielectric constant.
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Example (text problem 17.71): A capacitor can be made from
two sheets of aluminum foil separated by a sheet of waxed
paper. If the sheets of aluminum are 0.3 m by 0.4 m and the
waxed paper, of slightly larger dimensions, is of thickness
0.030 mm and has  = 2.5, what is the capacitance of this
capacitor?
C0 
0 A
d
8.851012 Nm2 /C 2 0.40* 0.30m 2

0.030103 m
 3.54108 F




and C   C0  2.5 3.54108 F  8.85108 F.
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§17.7 Energy Stored in a Capacitor
A capacitor will store energy equivalent to the amount of
work that it takes to separate the charges.
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The energy stored in the electric field between the plates is:
1
U  QV
2
1
2


 C V
2
Q2

2C
}
These are found by
using Q = CV and
the first relationship.
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Example (text problem 17.79): A parallel plate capacitor is
composed of two square plates, 10.0 cm on a side, separated
by an air gap of 0.75 mm.
(a) What is the charge on this capacitor when the potential
difference is 150 volts?
Q  CV 
0 A
d
V  1.77 10 8 C
(b) What energy is stored in this capacitor?
1
U  QV  1.33 10 6 J
2
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Summary
•Electric Potential Energy
•Electric Potential
•The Relationship Between E and V
•Motion of Point Charges (conservation of energy)
•Parallel Plate Capacitors (capacitance, dielectrics,
energy storage)
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