Physics 102: Lecture 3 Electric Potential Energy & Electric Potential

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Physics 102: Lecture 3
Electric Potential Energy
& Electric Potential
Physics 102: Lecture 2, Slide 1
Overview for Today’s Lecture
• Electric Potential Energy & Work
– Uniform fields
– Point charges
• Electric Potential (like height)
– Uniform
if
fields
fi ld
– Point charges
Physics 102: Lecture 2, Slide 2
Work
W = F d cos(θ)
Gravity
y
• Brick raised yi→ yf
• FG = mg (down)
• WG = –mgh
• Wyou = mgh
Electric
• Charge moved xi→ xf
• FE = qE
• WE = –qEd
• Wyou = qEd
(left)
yf →
h
W=0
yi →
FG=mg
Physics 102: Lecture 2, Slide 3
F
W=0
xi
↓
−
d
E
xf
↓
CheckPoint 1.1
11
C
F – A
B
Uniform E→
In what direction does the force on a
negative charge at point A point?
68% 1) left
30% 2) right
2% 3) up
Physics 102: Lecture 2, Slide 4
Electric field ppoints in the direction a
POSITIVE charge would feel force.
CheckPoint 1.2
“I would say zero because
the p
path is perpendicular
p p
to the field”
motion
C
F − A
B
Uniform E→
When a negative charge is moved from A to C
the ELECTRIC force does
work
20% 1) positive work.
73% 2) zero work.
7% 3) negative work.
Physics 102: Lecture 2, Slide 5
CheckPoint 1.3
“because
the direction of
the displacement is 180
degrees from direction of
the force ”
C
A
F −
Uniform E→
B
motion
When a negative charge is moved from A to B
the ELECTRIC force does
work
49% 1) positive work.
12% 2) zero work.
39% 3) negative work.
Physics 102: Lecture 2, Slide 6
−W
WE field = +W
WYou
Yo
Electric force did negative work
You did positive work
ACT: Work
WA-B = work done by
g from
FE movingg charge
A to B
C
A
F −
B
Uniform E→
The negative charge is moved from A to C to B.
Is the work done by the electric force:
A) Greater than WA-B
B) Same as WA-B
C) Less than WA-B
AB
Physics 102: Lecture 2, Slide 7
Path
P
h does
d
not matter!!
Only end points matter
Work and Δ Potential Energy
WF = F d cos(θ)=-ΔU
(θ)
Gravity
y
• Brick raised yi→ yf
• FG = mg (down)
• WG = –mgh
• ΔUG= +mgh
Electric
• Charge moved xi→ xf
• FE = qE
• WE = –qEd
• ΔUE= +qEd
(left)
yf →
xi
↓
h
F −
yi →
FG=mg
Physics 102: Lecture 2, Slide 8
d
E
xf
↓
E P E for point charges
E.P.E.
E.P.E. of two charges
g q1 and q2 separated
p
a distance r:
What is the electric potential energy of an electron a
10 m from
di t
distance
r = 0.53×10
0 53 10-10
f
a proton
t (H atom)?
t )?
UE = (9 ×109)(+1.6×10-19)(-1.6×10-19)/0.53×10-10 = −4.35×10−18J
rf = 0.5×10-10 m
+
Physics 102: Lecture 2, Slide 9
−
Work done by YOU
to assemble
bl 3 + charges
h
• W1 = 0
• W2 = k q1 q2 /r
=(9×109)(1×10-6)(2×10-6)/5
=3.6 mJ
• W3 = k q1 q3/r + k q2 q3/r
(9×109)(1×10-6)(3×10-6)/5 + (9×109)(2×10-6)(3×10-6)/5 =16.2
•
•
•
Wtotal = +19.8 mJ
WE = –19.8 mJ
ΔUE = +19.8 mJ
(watch signs!)
Physics 102: Lecture 2, Slide 10
5m
1
3
5m
5m
2
mJ
ACT: Work done by YOU
to assemble 3 negative charges
How muchh workk would
ld it
i take
k YOU to assemble
bl 3 negative
i
charges?
Likes repel, so YOU will still do positive
work!
A) W = +19.8
+19 8 mJ
B) W = 0 mJ
C) W = -19.8
-19 8 mJ
Physics 102: Lecture 2, Slide 11
5m
1
3
5m
5m
2
CheckPoint 2.1
1
5m
2
+
+
5m
5m
-
3
The total work required by you to assemble
this set of charges is:
61% (1)
positive
Bring in (1): zero work
17% (2)
zero
Bring in (2): positive work
22% (3)
Physics 102: Lecture 2, Slide 12
negative
B i in
Bring
i (3):
(3) negative
ti work
kx2
Electric Potential
Electric potential energy per charge
• Units:
J l /C l b ≡ Volts
Joules/Coulomb
V lt
• Examples:
– Batteries
i
– EKG
• Only
O l potential
i l
differences matter
Physics 102: Lecture 2, Slide 13
1.5 J/C
more
more energy!
+
+
Electric Potential: like height
Devil’s Tower
Topographical map
M i to
Moving
t higher
hi h potential
t ti l Ù moving
i uphill
hill
Physics 102: Lecture 2, Slide 14
Demo: electric potential
Recall electric dipole
+
E i t ti l lilines
Equipotential
+
–
–
• + (–) charge has high (low) potential
Electric field
• Equipotential
E i t ti l lines
li
att same “height”
“h i ht”
• Electric field lines point “downhill”
Physics 102: Lecture 2, Slide 15
CheckPoint 1.7
Electric field Points from
greater potential to lower
potential
The electric potential at point A is _______ at point B
42% 1)) ggreater than
27% 2) equal to
31% 3) less than
Physics 102: Lecture 2, Slide 16
ACT
E=0
Now points A and B lie
inside a conductor…
conductor
The electric potential at point A is _______ at point B
A) greater than
B) equal to
C) less than
Physics 102: Lecture 2, Slide 17
The electric field points toward lower
potential, but the electric field is zero
inside a conductor…so
conductor so the potential
is equal everywhere!
Potential for Point charges
Electric potential a distance r from a charge q:
What is the electric potential a distance r = 0.53×10-10 m
from a proton? (V(∞)=0)
V =UE/q= k q/ r = (9×109)(1.6×10−19) /0.53×10−10 = 27.2 Volts
rf = 0.5×10-10 m
+
Physics 102: Lecture 2, Slide 18
Two Charges
• Calculate electric potential at point A due to charges
– Calculate V from +7μC charge
– Calculate V from –3.5μC charge
– Add (EASY! NO VECTORS)
A
4m
V = kq/r
V7 = (9×109)(7×10-6)/5 = 12.6×103V
V3 = (9×10
(9 109)(-3.5×10
)( 3 5 10-66)/5 = -6.3×10
6 3×103V
Vtotal = V7+V3 = +6.3×103V
Q=+7.0μC
How much work do you have to do to bring
a 2 μC charge from far away to point A?
Physics 102: Lecture 2, Slide 19
6m
Q=-3.5 μC
W=ΔU=Vq
W
ΔU Vq
= (+6.3×103V)(2μC)
= +12.6 mJ
Comparison:
p
Electric Potential Energy vs. Electric Potential
• Electric Potential Energy (U) - the energy of a
charge at some location.
• Electric Potential (V) - found for a location only –
tells what the EPE would be if a charge were
located there (usually talk about potential
differences bet
between
een two
t o locations):
U = Vq
• N
Neither
ith has
h direction,
di ti
just
j t value.
l
Sign matters!
Physics 102: Lecture 2, Slide 20
Relationship
p between F,, E,, UE, V
Vector
Number (“scalar”)
F [N]
UE [J]
Ex:
Ex:
E [N/C]=[V/m]
V
Ex:
Ex:
[J/C]=[V]
Why so many ways to describe electric force?
Physics 102: Lecture 2, Slide 21
Electron microscope
p
ΔV=10kV
Vi
−
Electron
gun
Vf
←Uniform E
motion
• What is the final velocity of the electron?
• Solve by conservation of energy:
K.E.i + P.E.i = K.E.f + P.E.f
0 + –eVi = ½mv2 + –eVf
Could solve this using F=ma &
q
((Phys
y 101))
kinematic equations
TRY AT HOME! (HARDER)
Physics 102: Lecture 2, Slide 22
See everyone Monday!
Physics 102: Lecture 2, Slide 23
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