Transcript Electric Potential

Physics 102: Lecture 3
Electric Potential
Recall from last lecture….
• E field has magnitude and direction:
– E due to point charge Q: E=kQ/r2
– Force on charge q due to E: F=qE
– E and F are vectors
• Electric field lines
– Density gives strength (# proportional to charge.)
– Arrow gives direction (Start + end on -)
• Conductors
– E=0 inside a conductor
Overview for Today’s Lecture
• Electric Potential Energy/ Work
– Uniform fields
– Point charges
• Electric Potential (like height)
– Uniform fields
– Point charges
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Recall Work from P101
• Work done by the force given by:
W = F d cos(q)
Positive: Force is in direction moved
Negative: Force is opposite direction moved
Zero: Force is perpendicular to direction moved
• Careful ask WHAT is doing work!
Opposite sign for work done by you!
• Conservative Forces (not friction)
Change in Potential Energy = -Wconservative
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Work and D Potential Energy
Gravity
Uniform Electric Field
• Brick raised yi yf
• FG = mg (down)
• WG = -mgh
• DUG= +mgh
• +charge q moved d to left
• FE = qE
• WE = -qEd
• DUE= +qEd
(right)
yf
h
yi
Fg=mg
Fg=mg
Fg=mg
Fg=mg
Fg=mg
Fg=mg
Fg=mg
Fg=mg
q
d
q
Uniform E
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Work and D Potential Energy
W = F d cos(q)
Gravity
• Brick raised yi yf
• FG = mg (down)
• WG = -mgh
• DUG= +mgh
yf
Electric
• + Charge moved ∞  rf
• FE = kq1q2/r2
• WE = -kq1q2/rf
• DUE= +kq1q2/rf
rf
h
yi
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Preflight 3.1
ACT
C
F - A
B
Uniform E
In what direction does the force on a
negative charge at point A point?
53% 1) left
43% 2) right
4% 3) up
Force on charge is in same direction of
field if POSITIVE and opposite
direction if NEGATIVE.
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Preflight 3.2
“I would say zero because
the path is perpendicular
to the field”
motion
F
F
F
F
F
- C
- A
B
Uniform E
When a negative charge is moved from A to C
the ELECTRIC force does
10% 1) positive work.
85% 2) zero work.
05% 3) negative work.
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Preflight 3.3
ACT
“because the direction of
the displacement is 180
degrees from direction of
the force ”
C
A
B
F
F - F -F -F - motion
Uniform E
When a negative charge is moved from A to B
the ELECTRIC force does
66% 1) positive work.
7% 2) zero work.
21% 3) negative work.
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Preflight 3.5
C
ACT
A
-
-
-
-
B
-
Uniform E
When a negative charge is moved from A to B,
the electric potential energy of the charge
33% 1) Increases
14% 2) is constant
53%
3) decreases
D(EPE) = -WE field
Electric force did negative
work so electric potential
energy increased. Just like
pushing mass uphill.
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ACT: Electric Potential Energy
+
E
A-
AC: W=0
C
B
-
-
-
CB: W<0
-
When a negative charge is moved from A to B, the electric
potential energy of the charge
(A) increases
(B) is constant
(C) decreases
1) The electric force is directed to bring the electron
closer to be proton.
2) Since the electron ends up further from the proton
the electric field did negative work.
3) So the electric potential energy increased
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Work done by YOU
to assemble 3 charges
• 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
DUE = +19.8 mJ
5m
1
3
5m
mJ
5m
2
Note the units and watch signs:
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ACT: Work done by YOU
to assemble 3 negative charges
How much work would it take YOU to assemble 3 negative
charges?
Likes repel, so YOU will still do positive
work!
A.W = +19.8 mJ
B.W = 0 mJ
C.W = -19.8 mJ
5m
1
3
5m
5m
2
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Preflight 3.11
1
5m
Negative because i really hate physics.
And i dont know what is being ask in
this question.
2
+
+
5m
5m
-
3
The total work required by you to assemble
this set of charges is:
57% (1)
positive
16% (2)
zero
27% (3)
negative
Bring in (3): zero work
Bring in (2): negative work
Bring in (1): zero work
(see next page for explanation)
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Preflight 3.11
5m
2
+
+
5m
Bring in (3): zero work because the other charges are far away so
the electric field due to those charges is zero.
Bring in (2): negative work. why? Let’s figure out the work done by
the electric field, which is just the negative of the work done. The
electric field felt by charge 2 is the field due to charge 3, which
points toward charge 3. So, we are moving charge 2 in the same
direction of the field. Therefore the work done by the field is
positive, so the work done by you is negative
Bring in (1): zero work. why? We must do negative work due to
charge 3 and an equal amount of positive work due to charge 2, so
the net work is zero. Another way to think about it is that the electric
potential energy of charge 1 is zero, since it has equal but opposite
contributions from charges 2 and3
Net result is the sum of 0,negative, and 0 and is therefore negative.
5m
-
3
1
Preflight 3.11
5m
2
+
+
5m
Yet another way to work the problem:
Wyou = -WE = Electric Potential Energy (EPE) of the three charges.
EPE = kq1q2/r + kq2q3/r +kq1q3/r
where r is the separation between the charges (5 m). All three terms
have the same magnitude, since all the charges have the same
magnitude. The first term is positive but the next two are negative.
Therefore, the EPE is negative, so that Wyou is negative.
As a practice exercise, try calculating Wyou, assuming the magnitude
of each charge is 5 C. Answer = -0.045 J.
5m
-
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Work and D Electric Potential
Gravity
Uniform Electric Field
• Brick raised yi yf
• +charge q moved d to left
DUE= +qEd = change in
DUG= +mgh = change in
electric potential energy
gravitational potential
energy
DUE= +q(Ed) = charge(q) x
change in electric potential
DUG = m(gh) = mass (m) x
(Ed) VE
change in gravitational
potential (gh) VG yf m Moving opposite to E increases
VE
h
Moving from low to high
increases VG
yi m
q
d
Uniform E
q
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Electric Potential
• Units Joules/Coulomb Volts (symbol V)
– Batteries
– Outlets
– EKG
• Really Potential differences
• Equipotential lines at same “height”
• Field lines point “downhill”
from higher to lower potential
• V = k q/r (distance r from charge q)
V(∞) = 0
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Preflight 3.7
Electric field points from
greater potential to lower
potential
The electric potential at point A is _______ at point B
55%1) greater than
26%2) equal to
19%3) less than
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Preflight 3.9
89%
conductor
The electric potential at point A is _______ at point B
1) greater than
2) equal to
3) less than
“The electric field within a conductor is
zero, and therefore, the potential for
points A and B are the same
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Preflight Summary
Path
Vfinal - Vinitial
AB
Negative
AC
Zero
CB
Negative
Charge D U = q DV
Negative
+
Positive
-
WE field
Positive
Negative
+
+
-
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ACT: Electric Potential
+
E
C
A
B
The electric potential at A is ___________ the electric potential
at B.
1) greater than
1) Electric field lines point “down hill”
2) AC is equipotential path (perpendicular to E)
2) equal to
3) CB is down hill, so B is at a lower potential than
(“down hill from”) A
3) less than
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Electric Potential: Summary
• E field lines point from higher to lower potential
• For positive charges, going from higher to lower
potential is “downhill”
• For negative charges, going from lower to higher
potential is “downhill”
• For a battery, the + terminal is at a higher potential
than the – terminal
Positive charges tend to go “downhill”, from + to Negative charges go in the opposite direction, from - to +
Comparison:
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 U would be if a charge were located there:
U = Vq
• Usually we talk only about changes in potential or
potential energy when moving from one location to
another
• Neither U nor V has direction, just location. Sign matters!
not covered during lecture
Electric Potential due to a point
charge
What is the electric potential V a distance r from a point
charge, assuming V=0 at 
answer: V = k q/ r
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not covered during lecture
Two Charges
• Calculate electric potential at point A due to charges
– Calculate V from +7C charge
– Calculate V from –3.5C charge
– Add (EASY!)
• V = kq/r
V7=(9109)(710-6)/5 = 12.6103V
V3=(9109)(-3.510-6)/5 = -6.3103V
Vtotal = V7+V3 = +6.3103V
Q=+7.0C
4m
A
6m
Q=-3.5 C
How much work do you have to do to bring W=DU=DVq3
=(+6.310 V)(2C)
a 2 C charge from far away to point A?
=+12.6 mJ
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not covered during lecture
ACT: Two Charges
• In the region II (between the two charges) the
electric potential is
1) always positive
2) positive at some points, negative at others.
3) always negative
I
II
Q=+7.0C
III
Q=-3.5 C
Very close to positive charge potential is positive
Very close to negative charge potential is negative
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To Do
• Read 17.5-6
• Extra problems from textbook Ch 17:
– Concepts 1-8
– Problems 1, 9, 15, 19, 23, 25
• Bring “Problem Solver” to discussion section
• Complete preflight before Monday 6:00am.
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