Download Chapter 17: Electric Potential

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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Electric Potential Energy U is the energy of a charged
object in an external electric field (Unit Joule J)
Electric Potential V is the property of the electric field itself,
whether or not a charged object has been placed in it (Unit
Volt V)
A charged particle q exerts an electric force (a conservative
force) in an electric field. If the charge q is not static then work
is done in an electric field by moving the charge q. The work
is proportional to the magnitude of charge (More work is
needed to move a bigger charge). This is similar to lift/drop an
object against/with gravity.
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F=q0E
F=mg
W=Fs cosθ and ΔU=-W
ΔU=Epot =mgh
ΔU = qEd
(E-field is homogeneous, E is
constant vector everywhere)
Electric Potential Energy
Gravitational Potential Energy
ÖThe electric potential energy increase/decrease in the same
way as like the gravitational energy.
ÖOnly differences in potential energy are important
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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).
•Ue = 0 at some reference point.
•For two point particles take Ue = 0 at r = ∞.
kq1q2
Ue =
r
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Example: A proton and an electron, initially separated by a
distance r, are brought closer together. How does the
potential energy of this system of charges charge?
ke 2
For these two charges U e = −
r
Bringing the charges closer together decreases r:.
ΔU e = U ef − U ei < 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
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
ΔUe > 0.
This means that work must be done by an external
agent to bring the charges closer together.
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What is the potential energy of a system (arrangement) of
point charges? To calculate:
Begin by placing the first charge at a place in space
far from any other charges. No work is required to
do this.
Next, bring in the remaining charges one at a time
until the desired configuration is finished.
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Example: What is the potential energy of three point charges
arranged as a right triangle?
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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q2
2C
r12 1m
Are these the same?
r12
r23
q3
q1
1C
r13 1m
kq1q2 kq1q3 kq2 q3
U e = 0+
+
+
r12
r13
r23
k ⋅ 1C ⋅ 2C k ⋅ 1C ⋅ 3C k ⋅ 2C ⋅ 3C
+
+
U=
1m
1m
2
C
≈ k (2 + 3 + 4)
m
C
≈ k ⋅9
m
r23
q3
q1
3C
q2
2C
1C
r13
3C
kq1q2 kq1q3 kq2 q3
+
+
Ue = 0 +
r12
r13
r23
k ⋅ 1C ⋅ 2C k ⋅ 1C ⋅ 3C k ⋅ 2C ⋅ 3C
+
+
1m
1m
2m
C
≈ k (1.4 + 3 + 6)
m
C
≈ k ⋅ 10.4
m
U=
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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 ΔV, its potential energy change is ΔUe = q ΔV.
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Example: A charge Q = +1 nC is placed somewhere in
space far from other charges. Take ra = rb = rc = rd = 1.0 m
and re = rf = rg = 2.0 m.
f
b
c
e
a
Q
d
(a) Compare the potential at points d
and g.
Since Q>0 the potential at point d is
greater than at point g, it is closer to the
charge Q.
g
(b) Compare the potential at points a and b.
The potential at point a is the same as at point b;
both are at the same distance from the charge Q.
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Example continued:
(c) Place a charge of +0.50 nC at point e. What will the
change in potential (ΔV) be if this charge is moved to point
a?
(
)
(
)
kQ 9.0 ×109 Nm 2 /C 2 (1.0 nC )
=
= +4.5 Volts
Ve =
2m
re
kQ 9.0 × 109 Nm 2 /C 2 (1.0 nC )
=
= +9.0 Volts
Va =
1m
ra
ΔV = Vf – Vi = Va-Ve = +4.5 Volts
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Example continued:
(d) What is the change in potential energy (ΔU) of the
+0.50 nC charge ?
ΔUe =qΔV = (+0.50 nC)(+4.5 Volts)= +2.3 nJ
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Example continued:
(e) How would the results of the previous questions change
if instead of a +1.0 nC charge there is a -1.0 nC charge in its
place?
(a)The potential at point d is less than the potential at
point g.
(b) Unchanged
(c) -4.5 V
(d) -2.3 nJ
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Fp
E
.P
1m
+1C
U =0
F =0
1C
EP = k
(1m) 2
U system = 0
1C
Vp = k
1m
Equipotential line
1C ⋅ 1C
Fp = q ⋅ E = k
(1m) 2
1C
EP = k
(1m) 2
1+1C
C ⋅ 1C
U system = q ⋅ V = k
1m
1C
Vp = k
1m
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The Relationship between E and V
The circles are
called equipotentials
(surfaces of equal
potential).
f
b
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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5V
4V
3V
2V
1V
Equipotentials
and field lines
for a dipole.
0V
-1V
-5V
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Uniform E-field
V1
V2
V3
V4
E
Equipotential surfaces
ΔUe − W − qEd
Where d is the distance
ΔV =
=
=
− Ed over which ΔV occurs.
q
q
q
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If the electric field inside a conductor is zero, what is the
value of the potential?
If E=0, then ΔV=0. The potential is constant!
What is the value of V inside the conductor? It will be
the value of V on the surface of the conductor.
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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 17.31): 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 )
= − ΔU = − qΔV = − q (Vs − V p )
= −(+ 2e )(200.0 − 500.0 )kV
= +9.6 × 10
−14
J
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Example (text problem 17.32): 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 = −ΔU = −qΔV
1
2
mv f = −qΔV
2
2
2
−31
6
mv f
9.11×10 kg 7.26 ×10 m/s
ΔV = −
=−
2q
2 − 1.60 ×10 −19 C
= +150 Volts
Note: the electron moves
(
(
)(
from low V to high V.
)
)
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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 ∝ ΔV
∴ Q ∝ ΔV
Written as an equality: Q = CΔV, where the proportionality
constant C is called the capacitance.
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What is the capacitance for a parallel plate capacitor?
σ
Q
ΔV = Ed = d =
d
ε0
ε0 A
ε0 A
∴Q =
d
where C =
ΔV = CΔV
ε0 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 17.42): A parallel plate capacitor
has a capacitance of 1.20 nF. There is a charge of
magnitude 0.800 μC on each plate.
(a) What is the potential difference between the plates?
Q = CΔV
Q 0.800 μC
ΔV = =
= 667 Volts
1.20 nF
C
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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
ΔV = =
C ε0 A
ΔV ∝ d
If d is doubled so is the potential difference.
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Example (text problem 17.86): A parallel plate capacitor has a
charge of 0.020 μC 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?
0.020 μC
Q
=
= 8.3 × 10 −11 F = 83 pF
C=
ΔV 240 Volts
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Example continued:
(b) What is the area of a single plate?
C=
A=
ε0 A
d
Cd
ε0
=
(83 pF)(0.40 mm)
8.85 ×10 −12 C 2 / Nm 2
= 0.0038 m 2 = 38 cm 2
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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.
Dielectric slab
C = κ C0
where C0 =
ε0 A
d
and κ is the dielectric constant. Vacuum κ=1; H2O= 80.4
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A dielectric breakdown appears when the Electric field is too
strong. That means strong enough to tear atoms apart.
Substance
Dielectric strength (V/m)
Mica
100 x 106
Teflon
60 x 106
Paper
16 x 106
Pyrex glass
14 x 106
Neoprene rubber
12 x 106
Air
3.0 x 106
Next time you walk across a carpet and get a shock from the
doorknob think about the fact that you just have produced an
electric field of roughly 3 million V/m.
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Example (text problem 17.55): 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 κ = 2.5, what is the
capacitance of this capacitor?
C0 =
ε0 A
d
8.85 ×10 −12 Nm 2 /C 2 (0.40 * 0.30 )m 2
=
0.030 ×10-3 m
= 3.54 × 10 −8 F
(
)
(
)
and C = κ C 0 = (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 = QΔV
2
1
2
= C (ΔV )
2
Q2
=
2C
}
These are found by
using Q=CΔV and
the first relationship.
37
For a parallel plate capacitor
Q = ε0 E A and V = Ed
Therefore
U = ½ Q V = ½ (ε0 E A) (Ed)= ½ ε0 E2 A d
A⋅d is the volume between the plates so we can define
a density (i.e. a quantity divided by its volume)
Electric energy density
uE = electric energy/ volume = ½ ε0 E2
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Example (text problem 17.63): 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 = CΔV =
ε0 A
d
−8
ΔV = 1.77 × 10 C
(b) What energy is stored in this capacitor?
1
U = QΔV = 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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