Download Force of Attraction / Repulsion r QQ F = Force of Attraction

S. Achenbach: PHYS 155 (Part 1, Topic 3)
Example Soliutions
p. 1
Force of Attraction / Repulsion
Example 1 (Coulomb’s Law)
Find the force of attraction between two particles
if one has a net charge of 3⋅10-6 C and the other has a net charge of -7⋅10-6 C.
The particles are located 2 mm apart.
Solution:
F=
κ Q1 Q2
r2
= 47.3 ⋅103 N = 47.3 kN
Phys155 • 1-3: Separation of Charge – Electric Fields
S. Achenbach
1
Force of Attraction / Repulsion
Example 2 (Coulomb’s Law)
Find the force between the electron and proton in a hydrogen atom.
rH
+
Solution:
with radius of hydrogen atom rH = 5.29⋅10-11 m
and the magnitude of the charge of 1 proton or electron |Q| = [1 / (6.242⋅1018)] C
F=
κ Q1 Q2
r2
9 ⋅109
=
N m2
C2
⎛
⎞⎛
⎞
1
1
⎜⎜
C ⎟⎟ ⎜⎜
C ⎟⎟
18
18
⎝ 6.242 ⋅10
⎠ ⎝ 6.242 ⋅10
⎠
2
−11
(5.29 ⋅10 m )
= ...
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S. Achenbach: PHYS 155 (Part 1, Topic 3)
Example Soliutions
p. 2
Shell Theorem
Example 3 (Extended Shell Theorem)
Two uniformly charged spheres have net charges of Q1 = +17 C and Q2 = +23 C.
They are separated in space by r = 1 m.
What is the force of repulsion?
r1
+
+
r
r2
Solution:
F=
sphere 1, r1 = 10 cm
contains net charge Q1
κ Q1 Q2
sphere 2, r2 = 8 cm
contains net charge Q2
r2
= 3.519 ⋅1012 N
Phys155 • 1-3: Separation of Charge – Electric Fields
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Shell Theorem
Example 4 (Charge in a Plane)
Find the force on a charge QA = +7 C located at point A.
r1 = 10 m
A
+
+
Q1 = 1 C
Solution:
r2 = 6 m
Find force due to Q1 first.
Then find force due to Q2.
Then add the two force vectors.
+
Q2 = 2 C
F1 =
F2 =
κ Q1 Q A
2
1
r
κ Q2 Q A
r2
2
= 0.63 ⋅109 N
= 3.5 ⋅109 N
v
F = − F1 iˆ + F2 ˆj
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Phys155 • 1-3: Separation of Charge – Electric Fields
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S. Achenbach: PHYS 155 (Part 1, Topic 3)
Example Soliutions
p. 3
Energy Levels due to Point Charge
Example 5
-1
1
m
r2
=
∞
Consider a hydrogen atom with its electron removed (called a hydrogen ion), i.e. a single
proton.
What is V12 if point 1 is on the normal orbit path of the electron
and point 2 is a long distance away (e.g. inifinite)?
2
5.
29
10
⋅
r1
=
Solution:
• r1 = 5.29 ⋅ 10-11 m; r2 = ∞; Q = 1 C / (6.242⋅1018) (proton charge)
• V12 is positive
1 1
−
r1 r2
2
⎞
1
1
9 N m ⎛
⎜⎜
= 9 ⋅10
C ⎟⎟
−0
2
18
−11
C ⎝ 6.242 ⋅10
⎠ 5.29 ⋅10 m
1
+
V12 = + κ Q
= + 27.3
Nm
= + 27.3 V
C
Phys155 • 1-3: Separation of Charge – Electric Fields
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Energy Levels due to Point Charge
Example 6
Based on results from example 5, how much work is required
to remove an electron from a hydrogen atom (ignoring the kinetic energy of the electron)?
Solution:
• equivalent to moving an electron in an electric field set up by the nucleus (1 proton)
from r = 5.29⋅10-11 m to r = ∞.
• with V21 =
W
Q
⇒
W = V21 ⋅ Q
(Q is charge of electron = -1C / (6.242⋅1018) )
• sign: - reversed direction of movement: V21 = -V12 = -27.3 V
- movement of negative charge (electron): Q = - ...C
pos
⎞
J ⎞⎛
1
⎛
W = ⎜ − 27.3 ⎟ ⎜⎜ −
C ⎟⎟ = + 4.37 ⋅ 10−18 J
18
C ⎠ ⎝ 6.242 ⋅10
⎝
⎠
Note: About half of the electric potential energy of an electron orbiting in a hydrogen atom is its kinetic energy.
Therefore, really only about 2.2⋅10-18 J are required to ionize a hydrogen atom.
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Phys155 • 1-3: Separation of Charge – Electric Fields
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S. Achenbach: PHYS 155 (Part 1, Topic 3)
Example Soliutions
p. 4
Electric Potential in a Field set up by 2 Point Charges
b
Example 7
Find Vab for Q1 = +3 C and Q2 = -2 C.
3m
a
Q1 +
4m
Solution:
- Q2
4m
• I) calculate |Vab| due to Q1
Vab ( due to Q1) = κ Q1
1
1
−
r1b r1a
= 1.35 ⋅ 109
with r1a = 4 m; r1b = √(4m)2 + (3m)2 = 5 m
Nm
= 1.35 ⋅ 109 V
C
Phys155 • 1-3: Separation of Charge – Electric Fields
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Electric Potential in a Field set up by 2 Point Charges
b
Example 7
Find Vab for Q1 = +3 C and Q2 = -2 C.
3m
a
Q1 +
4m
Solution:
- Q2
4m
• II) calculate |Vab| due to Q2
Vab ( due to Q2 ) = κ Q 2
1
1
−
r2 b r2 a
Nm 2
1
1
= 9 ⋅ 10
−2C
−
2
C
5m 4m
9
= 0.9 ⋅ 109 V
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Phys155 • 1-3: Separation of Charge – Electric Fields
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S. Achenbach: PHYS 155 (Part 1, Topic 3)
Example Soliutions
p. 5
Electric Potential in a Field set up by 2 Point Charges
b
Example 7
5m
5m
Find Vab for Q1 = +3 C and Q2 = -2 C.
4m
Q1 +
4m
a
- Q2
Solution:
• III) calculate Vab
Vab = Vab ( due to Q1) + Vab ( due to Q 2 )
signs - moving a positive test charge
from b (5 m from a positive charge Q1) to a (4 m from a positive charge Q1)
closer to a positive charge (up against electric field) requires pos. work
⇒ sign of Vab due to Q1 is pos.
- moving a positive test charge
from b (5 m from a negative charge Q2) to a (4 m from a negative charge Q2)
closer to a negative charge (down with electric field) requires neg. work (gives off)
⇒ sign of Vab due to Q2 is neg.
• ⇒ Vab = 1.35⋅109 V – 0.9⋅109 V = 0.45⋅109 V
Phys155 • 1-3: Separation of Charge – Electric Fields
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Electric Field due to a uniformly charged Plane
+
+
+
+
+
+
Example 8
What is the electric field strength of a field
set up by a charged infinite plane?
The charge density is σ = 1 μC/m2.
field lines
uniformly charged plane
Solution:
E=
=
σ
2 ε0
; with permittivity of free space ε0 = 8.85⋅10-12 C2/(Nm2)
10−6
C
m2
⎛
C2 ⎞
2 ⋅ ⎜⎜ 8.85 ⋅ 10−12
⎟
N m 2 ⎟⎠
⎝
= ...
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Phys155 • 1-3: Separation of Charge – Electric Fields
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S. Achenbach: PHYS 155 (Part 1, Topic 3)
Example Soliutions
p. 6
Electric Field due to a uniformly charged Plane
uniformly charged plane 1
σ1 = 1⋅10-8 C/m2
Example 9
+
+
+++
+++
++
+++
+ +++ +++
++
+
++
+
+
++
+
+++ +++ +++ ++++ ++
+
+
+
+++ +++ +++ ++++ ++
++
++
++ +++
++ +++
+++
++
+
+
+++ +++ +++ +++
+
+
+++ +
+++ +
+
+
Sketch the electric field set up by
two uniformly charged parallel planes.
Both are positively charged,
one with a charge density of 10-8 C/m2
and one with 2⋅10-8 C/m2 .
The planes are separated by 1 m.
uniformly charged
plane 2
σ2 = 2⋅10-8 C/m2
1m
Solution:
Find the electric field due to plane 1
and then find the electric field due to plane 2
and then sum them.
Phys155 • 1-3: Separation of Charge – Electric Fields
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Electric Field due to a uniformly charged Plane
Example 9
E1 =
σ1
=
2 ε0
10
−8
C
m2
⎛
C ⎞
2 ⋅ ⎜⎜ 8.85 ⋅ 10−12
⎟
N m 2 ⎟⎠
⎝
2
= 565
N
C
+
+
+
+
+
+
field lines
field strength
is 565 N/C
end view of plane 1
E2 =
σ2
=
2 ε0
2 ⋅ 10
−8
C
m2
⎛
C ⎞
⎟
2 ⋅ ⎜⎜ 8.85 ⋅ 10−12
N m 2 ⎟⎠
⎝
2
= 1130
N
C
+
+
+
+
+
+
field lines
field strength
is 1130 N/C
end view of plane 2
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Phys155 • 1-3: Separation of Charge – Electric Fields
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S. Achenbach: PHYS 155 (Part 1, Topic 3)
Example Soliutions
p. 7
Electric Field due to a uniformly charged Plane
Example 9
The electric field due to both plates is the vector sum of the two electric fields:
+
+
+
+
+
+
+
+
+
+
+
+
plane 1
field strength is
E1 + E2 = 1695 N/C
plane 2
field strength is
E2 - E1 = 565 N/C
field strength is
E1 + E2 = 1695 N/C
Phys155 • 1-3: Separation of Charge – Electric Fields
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Electric Field due to a uniformly charged Plane
Example 10
Sketch the electric field set up by two uniformly charged parallel planes.
One plane is positively charged with a charge density of σ1 = 10-9 C/m2 .
One plane is negatively charged with a charge density of σ2 = -10-9 C/m2 .
The planes are separated by 2 cm.
Solution:
• Find the electric field due to each plane separately and then sum them.
• The field strength will be the same for both plates:
σ
σ
E= 1 = 2 =
2 ε0 2 ε0
10−9 C
N
m2
= 56.5
2
C
⎛
⎞
2 ⋅ ⎜ 8.85 ⋅ 10−12 C
2⎟
Nm ⎠
⎝
• The direction of the fields is - „away from positively charged plane 1“ and
- „towards negatively charged plane 2“
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Phys155 • 1-3: Separation of Charge – Electric Fields
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S. Achenbach: PHYS 155 (Part 1, Topic 3)
Example Soliutions
p. 8
Electric Field due to a uniformly charged Plane
Example 10
+
+
+
+
+
+
-
field lines
field strength
is 56.5 N/C
+
+
+
+
+
+
end view of plane 1
plane 1
field strength is
E1-E2 = 0
E1=-E2
fields cancel
-
field lines
field strength
is 56.5 N/C
end view of plane 2
plane 2
field strength is
E1+E2 = 113 N/C
field strength is
E1=-E2
E1-E2 = 0 fields
cancel
Phys155 • 1-3: Separation of Charge – Electric Fields
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Electric Potential due to a uniformly charged Field
Example 11
A vertical plane containing the y-axis is uniformly charged
and the charge density is σ = +10-10 C/m2.
What is the electric potential of point A with respect to point B
if A is located at (x,y,z) = (1m,1m,0m) and B at (11m,6m,0m)?
+
+
+
+ y (1,1,0)
+ A
x
+
B
(11,6,0)
(11,1,0)
C
Solution (Method 1):
A and B not located on same field line
⇒ add point C which is on the same - field line as A
- equipotial line as B ⇒ VAB = VAC
VAC =
σ
⋅ Δx
2 ε0
= 56.5
Nm
C
⇒ VAB = 56.5 V
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Phys155 • 1-3: Separation of Charge – Electric Fields
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S. Achenbach: PHYS 155 (Part 1, Topic 3)
Example Soliutions
p. 9
Electric Potential due to a uniformly charged Field
Example 11
Solution (Method 2):
– compare sail boat example discussed in previous topic –
• Consider a test charge Qt to be moved from B to A.
v v
• With force F applied to move Qt, required work is W = F ⋅ d
• With definition for V
• As previously derived,
VAB
v v
F ⋅d
=
Qt
W
=
Qt
v
r
F
E =
Qt
+
+
+
+ y (1,1,0)
+ A
x
+
B
(11,6,0)
.
Here, force is not exerted by the field (compare previous derivation with annulus),
but force is applied by mover against the force due to the electric field
⇒ moving force has negative sign with respect to electric field
v v
VAB = − E ⋅ d
Phys155 • 1-3: Separation of Charge – Electric Fields
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Electric Potential due to a uniformly charged Field
Example 11
Solution (Method 2):
– compare sail boat example discussed in previous topic –
• Consider a test charge Qt to be moved from B to A.
v v
• With force F applied to move Qt, required work is W = F ⋅ d
• With definition for V
• V AB
v v
= −E ⋅d
(
C
(
C
= − 5.65 N
= + 5.65 N
VAB
,0 N
C
v v
F ⋅d
=
Qt
W
=
Qt
,0 N
C
+
+
+
+ y (1,1,0)
+ A
x
+
B
(11,6,0)
)⋅ [(1 m, 1 m, 0 m ) − (11 m, 6 m, 0 m)]
)⋅ (10 m ) + (0 N C )⋅ (5 m ) + (0 N C )⋅ (0 m )
⇒ VAB = +56.5 V
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Phys155 • 1-3: Separation of Charge – Electric Fields
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S. Achenbach: PHYS 155 (Part 1, Topic 3)
Example Soliutions
p. 10
Energy in a Capacitor
V [volts]
Example 12
50
40
Solution (1):
30
20
• Q = C ⋅ V as derived above
10
with - excess charge in one plate Q
- capacitance C [C/V=F] of the capacitor
- voltage V of one plate with resp. to the other
⇒ V =
2
4
6
8
10
Q [μC]
1
⋅ Q ⇒ the slope of the line in the V-Q-graph is 1/C
C
50 V
1
10 μC
Coulombs
⇒
slope =
=
= 0.2 ⋅ 10−6
C=
10 μC
slope 50 V
Volts
⇒ C = 0.2⋅10-6 F = 0.2 µF
Phys155 • 1-3: Separation of Charge – Electric Fields
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Energy in a Capacitor
V [volts]
Example 12
50
2) How much energy is stored in the electric field
between the plates of the capacitor
when there is an excess charge of 10 µC
on one of the plates?
40
30
20
10
2
4
6
8
10
Q [μC]
Solution:
• W=½QV
as derived above
with - excess charge in one plate Q
- voltage V of one plate with respect to the other
• from the graph at Q = 10 µC: V = 50 V
⇒ energy W = ½ (10⋅10-6 C)⋅ (50V) = ...
• Note: this energy corresponds to the area under the graph (triangle)
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Phys155 • 1-3: Separation of Charge – Electric Fields
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S. Achenbach: PHYS 155 (Part 1, Topic 3)
Example Soliutions
p. 11
Energy in a Capacitor
V [volts]
Example 12
50
40
3) How much energy is stored in the electric field
between the plates of the capacitor
when the voltage across the plates is 30 V?
30
20
10
2
4
6
8
Solution:
• W = ½ Q V as derived above
• from the graph at V = 30 V: excess charge on each plate is Q = ±6 µC
⇒ energy W = ½ (6⋅10-6 C)⋅ (30V) = 90⋅10-6 J
10
Q [μC]
9
• cross check:
from part 1)
W = ½ C V2 = ½ (0.2⋅10-6 C/V)⋅ (30 V)2 = 90⋅10-6 CV = 90⋅10-6 C J/C = 90⋅10-6 J 9
Phys155 • 1-3: Separation of Charge – Electric Fields
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Effect of Dielectrics on Capacitance
Example 13
Facts: A=0.1 m2; d=10-5 m; V=10 V; εr, polystyrene=2.6
1) What is the capacitance of the capacitor?
Solution:
C = εr
ε0 A
d
C2
C
=
Nm Nm
= 0.230 ⋅ 10−6 F = 0.230 μF
S. Achenbach
Phys155 • 1-3: Separation of Charge – Electric Fields
=
C
C
C
= =F
J
V
C
22
S. Achenbach: PHYS 155 (Part 1, Topic 3)
Example Soliutions
p. 12
Effect of Dielectrics on Capacitance
Example 13
Facts: A=0.1 m2; d=10-5 m; V=10 V; εr, polystyrene=2.6
2) What is the amount of charge (of course, that means excess charge) on each plate?
Solution:
with C = 0.230 µF as result from part 1),
charge Q (for simplicity; in fact really QP) is
Q =C⋅V
= 2.30 ⋅ 10−6 C = 2.30 μC
Phys155 • 1-3: Separation of Charge – Electric Fields
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Effect of Dielectrics on Capacitance
Example 13
Facts: A=0.1 m2; d=10-5 m; V=10 V; εr, polystyrene=2.6
3) What is the electric field strength between the plates of the capacitor?
Solution:
with equations E =
Qeff
ε0 ⋅ A
and
=
QP = ε r ⋅ Qeff
⇒
E=
QP
εr ⋅ ε0 ⋅ A
2.30 ⋅ 10−6 C
(2.6) ⋅ ⎛⎜ 8.85 ⋅ 10−12 C
⎝
E = 1.01 ⋅ 106
2
⎞
2
⎟ ⋅ (0.1 m )
Nm ⎠
2
N
C
Note: dielectric reduces strength of electric field between plates by a factor of εr = κ
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Phys155 • 1-3: Separation of Charge – Electric Fields
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S. Achenbach: PHYS 155 (Part 1, Topic 3)
Example Soliutions
p. 13
Effect of Dielectrics on Capacitance
Example 13
Facts: A=0.1 m2; d=10-5 m; V=10 V; εr, polystyrene=2.6
4) What is the charge in the thin sheet of the dielectric
that is right next to a plate of the capacitor?
Solution:
with equations QD = QP − Qeff
and QP = ε r ⋅ Qeff ⇒ QD = QP −
QP
εr
and with |QP| = 2.30⋅10-6 C as result from part 2),
QD = 1.42 ⋅ 10−6 C
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Phys155 • 1-3: Separation of Charge – Electric Fields
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