S. Achenbach: PHYS 155 (Part 1, Topic 2)
Example Solutions
p. 1
Electric Fields
Example 1
A particle located at point A
has an excess (net) charge of +19 C.
• What is the magnitude of the force on that particle?
• Is the direction of the force
down & to the left or up & to the right?
Solution:
• with E =
v
F
⇒ F = E ⋅Q
Q
= 1.3 N
(
C
)⋅ (19 C )
= 24.7 N
with magnitude of the electric field of E=1.3 N/C at point A and amount of charge Q=19 C.
• particle at point A is positively charged
⇒ direction of the force is in the direction of the electric field lines (down and to the left)
Phys155 • 1-2: Electric Fields
S. Achenbach
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Energy Contours (Equipotential Surfaces)
Example 2
1) What is the electric potential of point A?
Solution:
the electric potential of point A is -10 J/C
2) What is the electric potential of point B?
Solution:
the electric potential of point B is +10 J/C
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S. Achenbach: PHYS 155 (Part 1, Topic 2)
Example Solutions
p. 2
Energy Contours (Equipotential Surfaces)
Example 2
3) What is the electric potential of
point A with respect to point B?
Solution
electric potential of point A with respect to point B
(A = end point, B = start point; from B to A)
is the electric potential of point A
minus the electric potential of point B
(end condition minus start condition)
= (-10 J/C) – (+10 J/C) = ...
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Phys155 • 1-2: Electric Fields
S. Achenbach
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Energy Contours (Equipotential Surfaces)
Example 2
4) How much work is required to move a positively charged particle
with an excess charge of +23 C from point A to point B?
Solution
• by definition,
arrows on field lines show in the direction of the
force exerted by the field on a positively charged object
• particle being moved is positively charged 9
• mover has to move particle upstream, against the field: against force exerted by the field
⇒ mover must do positive work ⇒ sign will be pos.
• amount of work per unit charge required to move an object from A to B read from map:
⇒ el. pot. W/Q at B - el. pot. W/Q at A = 10J/C – (-10J/C) = 20J/C
⇒ work W = (W/Q) ⋅ Q = (20 J/C) ⋅ (23 C) = 460 J
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S. Achenbach: PHYS 155 (Part 1, Topic 2)
Example Solutions
p. 3
Energy Contours (Equipotential Surfaces)
Example 2
5) How much work is required to move a positively charged particle
with a net charge of +11 C from point D to point A?
Solution
• particle is positively charged,
so force exerted by the field on the particle
is in the direction of the field lines
• mover has to move particle downstream, with the field:
with force exerted by the field
⇒ field will do positive work (give off energy); mover „receives“ work ⇒ sign will be neg.
• amount of work per unit charge to move an object from D to A read from map:
⇒ el. pot. W/Q at A - el. pot. W/Q at D = -10J/C – (+20J/C) = -30J/C
⇒ work W = (W/Q) ⋅ Q = (-30 J/C) ⋅ (11 C) = -330 J
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Energy Contours (Equipotential Surfaces)
Example 3
1) What is the average force per unit charge
exerted by the field on a charged object
moved from A to B
along the electric field line that links them?
The straight line distance from A to B
is measured to be 0.4 m.
Solution:
• Let W be the work required to move the object from A to B,
and let Q be the net charge in the object
⇒ from map: el. pot. W/Q at B - el. pot. W/Q at A = -1J/C = -1Nm/C
• with def. W = F⋅Δx (Δx is distance moved) ⇒ average force (of the mover) Fave = W / Δx
⇒ force per unit charge
S. Achenbach
Fave ⋅
1 W 1
=
Q Q Δx
(
= − 1 Nm
Phys155 • 1-2: Electric Fields
⎛ 1 ⎞
)
⎟
⋅ ⎜⎜
C 0.4 m ⎟
⎝
⎠
= − 2. 5
N
C
6
S. Achenbach: PHYS 155 (Part 1, Topic 2)
Example Solutions
p. 4
Energy Contours (Equipotential Surfaces)
Example 3
1) What is the average force per unit charge
exerted by the field on a charged object
moved from A to B
along the electric field line that links them?
The straight line distance from A to B
is measured to be 0.4 m.
Note:
• result for assumption of positively charged object
• this negative value (-2.5 N/C) states that the mover exerts a force
in opposite direction of the movement (definition of work)
⇒ the force exerted by the mover is opposite to the force exerted by the field (upstream)
⇒ force exerted by the field is +2.5 N/C in the direction of the motion
(which is in the direction of the field lines 9)
Phys155 • 1-2: Electric Fields
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Energy Contours (Equipotential Surfaces)
Example 3
2) What is the average tangential component
of the force exerted by the field
on a charged object
if the object is moved along curve S
from point C to D, and curve S is 2 m long?
Solution:
• Now let W be the work required to move the object from C to D,
and let Q be the net charge in the object
• let Fs(x) be the force tangential to curve S at a meandered distance x from point C
⇒ average tangential force (of the mover) Fave = W / Δx
⇒ division by Q:
Fave ⋅
1 W 1
=
Q Q Δx
= −1
N
C
(-2 J/C = electr. potential at D minus el. pot. at C)
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S. Achenbach: PHYS 155 (Part 1, Topic 2)
Example Solutions
p. 5
Electric Potential (Voltage)
Example 4
1) An object with a net positive charge of 7 C is moved from point A to point B.
To do this, the mover does 21 J of work .
What is the electric potential at point B with respect to point A (i.e., what is VBA)?
Solution:
with work W required to move the object from point A to point B,
and with the excess (net) charge of the object Q,
by definition
VBA =
W
Q
= ...
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Phys155 • 1-2: Electric Fields
S. Achenbach
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Electric Potential (Voltage)
Example 4
2) What is the potential energy of an object at point B with respect to point A
if that object is positively charged with +2 C of charge?
Solution:
• potential energy of the object at point B with respect to point A
is the work W required to move the object from point A to point B
• with
VBA =
W
Q
⇒
W = VBA ⋅ Q
Pot. Energy B with respect to A = W = VBA ⋅ Q
=6J
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S. Achenbach: PHYS 155 (Part 1, Topic 2)
Example Solutions
p. 6
Electric Potential (Voltage)
Example 5
Points G and H are 2 points in space where an electric field is present.
VHG = 10 V.
How much work is required to move +19 C of charge from point G to H ?
Solution:
with
VHG =
W
Q
⇒
W = VHG ⋅ Q
= 190 J
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Electric Potential (Voltage)
Example 6
Points G and H are 2 points in space where an electric field is present.
VHG = 10 V.
How much work is required to move -19 C of charge from point G to H ?
Solution:
with
VHG =
W
Q
⇒
W = VHG ⋅ Q
= − 190 J
Note: Negative value means that the field does work on the mover
(field gives off energy; mover receives energy)
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S. Achenbach: PHYS 155 (Part 1, Topic 2)
Example Solutions
p. 7
Electric Potential (Voltage)
Example 7
1) What is the direction of the electric field?
2) What are VAB and VBA?
3) What is VCD?
Solution (1-3):
1) electric filed lines always point from higher to lower electic potentials
⇒ here, direction of field lines is upwards
2) moving from B to A: VAB = electric potential at A – el. pot. at B
= (0 J/C) – (5 J/C) = -5 J/C = -5 V
moving from A to B: VBA = electric potential at B – el. pot. at A
= (5 J/C) – (0 J/C) = 5 J/C = 5 V
3) moving from D to C: VCD = (3 J/C) – (3 J/C) = 0 J/C = 0 V
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Electric Potential (Voltage)
Example 7
4) How much work is required to move
-29 C of charge from point D to point A?
Solution:
• with VAD =
W
⇒ W = V AD ⋅ Q
Q
(work W required to move charge Q from D to A)
• what is VAD?
⇒ VAD = electric potential at A – el. pot. at D
= (0 J/C) – (3 J/C) = -3 J/C
• W = VAD ⋅ Q = (-3 J/C) ⋅ (-29 C) = 87 J
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S. Achenbach: PHYS 155 (Part 1, Topic 2)
Example Solutions
p. 8
Electric Potential (Voltage)
Example 7
5) What is the strength of the electric field
at H if dx = 1 m?
Solution:
• electric field strength is the magnitude
of force F per unit charge Q: E = F
• definition of work W = F ⋅ Δx
⇒
Q
F =W
Δx
• average magnitude of the force Fave in the middle of the 5 J/C and 6 J/C equipotential lines
⇒ E=
F Fave
≈
Q
Q
=
1 ΔW
⋅
Q Δx
F ≈ Fave =
(
= 1J
)
⎛ 1 ⎞
⎟
⋅⎜
C ⎜1 m ⎟
⎝
⎠
=1 N
Phys155 • 1-2: Electric Fields
S. Achenbach
ΔW
Δx
C
15
Electron Volt
Example 8
An electron is moved from point A to point B in an electric field where VBA = -7 V.
1) Is the work done by the mover positive or negative?
Solution:
• moving from A to B: VBA
VBA is negative ⇒ electric potential at B is lower than at A
⇒ if positive charge is moved from A to B, mover would do negative work
• however, negative charge (an electron) is moved
⇒ work done has opposite sign: work done will be positive
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S. Achenbach: PHYS 155 (Part 1, Topic 2)
Example Solutions
p. 9
Electron Volt
Example 8
An electron is moved from point A to point B in an electric field where VBA = -7 V.
2) How much energy (in units of Joules) is transfered from the mover to the field?
Solution:
⎛W ⎞
⎟⎟ ⋅ Q
⎝Q⎠
• trivially, W = ⎜⎜
(work W required to move charge Q from A to B)
• VBA = el. pot. W/Q at B - el. pot. W/Q at A
⇒ W = VBA ⋅ Q
= VBA ⋅ (-e)
= (-7 V) ⋅ (-1.602⋅10-19 C)
= 1.121⋅10-18 J
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Electron Volt
Example 8
An electron is moved from point A to point B in an electric field where VBA = -7 V.
3) How much energy (in units of eV) is transfered from the mover to the field?
Solution:
W = VBA ⋅ Q
= VBA ⋅ (-e)
= (-7 V) ⋅ (-e)
= 7 eV
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