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Today 3/12
Plates if charge
E-Field
Potential
HW:
“Plate Potential”
Due Friday, 3/14
Enet = 0
How big is E?

E 
2 0
(+)

E 
2 0
E net
Enet = 0
(-)


0
Charged conducting
plate
0 = Q/A
A = Area
of one side
What’s wrong with
this picture?
Charged conducting
plate
L = 0/2
R = 0/2
Free charge
always goes to
surface of
conductor.
Charged conducting
plate
L = 0/2
EL = L/20
= 0/40
What is the
electric field inside
the conductor?
R = 0/2
Charged conducting
plate
L = 0/2
R = 0/2
ER = R/20
= 0/40
What is the
electric field inside
the conductor?
Charged conducting
plate
L = 0/2
What is the
electric field inside
the conductor?
R = 0/2
The electric field is
zero everywhere
inside the
conductor.
Always, any
conductor, no
exceptions.
What
happens if I
let it go?
q=+1C
Assume the
particle
gains 100
joules of
kinetic
energy as it
moves from
A to B.
KE = 0
A
KE = 100 J
q=+1C
B
Now I stop it
at B.
How much
work must I do
to move it back
to A? +100 J
How does the
potential
energy change
in moving from
B to A? +100 J
A
q=+1C
B
DPEBA = +100 J
DPEAB = -100 J
What if
q=+2C?
How much
work must I do
to move it back
to A? +200 J
How does the
potential
energy change
in moving from
B to A? +200 J
q=+2C
A
B
DPEBA = +200 J
DPEAB = -200 J
Now we are
back to our
original
definition.
DVBA tells us how
much PE changes
when +1C is moved
from B to A.
A
q=+1C
B
DVBA = +100 J/C
DPEBA = (+100 J/C)x(q)
DPEBA = DVBA q
What if q = -1C?
First I must
turn my hand
around.
A
q= -1C
B
What if q= -1C?
How much
work must I do
to move it back
to A? -100 J
How does the
potential
energy change
in moving from
B to A? -100 J
DVBA tells us how
much PE changes
when +1C is moved
from B to A.
A
q= -1C
B
DPEBA = DVBA q
DPEBA = (+100 J/C)x(q)
DPEBA = (+100 J/C)x(-1)
What if q= -1C?
How much
work must I do
to move it back
to A? -100 J
How does the
potential
energy change
in moving from
B to A? -100 J
DVBA does not
depend on the sign of
the point charge but
DPEBA does!!!!!
A
q= -1C
B
DPEBA = DVBA q
DPEBA = (+100 J/C)x(q)
DPEBA = (+100 J/C)x(-1)
A proton is
released
from rest at
A. What is
its speed
when it
reaches B?
m= 1.7 x 10-27 kg
q= 1.6 x 10-19 C
What happens
to the kinetic
energy?
DVAB = -100 J/C
DVAB = -100 volts


A
B
DPEAB = q DVAB
DPEAB = q (-100 J/C)
DPEAB = -1.6 x 10-17 J
A proton is
released
from rest at
A. What is
its speed
when it
reaches B?
DKEAB = +1.6 x 10-17 J
m= 1.7 x 10-27 kg
1/2 mv2 = +1.6 x 10-17 J
q= 1.6 x 10-19 C
What happens
to the kinetic
energy?


A
B
v = 1.4 x 105 m/s
DPEAB = -1.6 x 10-17 J
What direction
is the force on
an electron?
F
A
E

B
An electron
is released
from rest at
B. What is
its speed
when it
reaches A?
m= 9.1 x 10-31 kg
q= -1.6 x 10-19 C
DVAB = -100 J/C
DVBA = +100 J/C


A
B
DPEBA = q DVBA
DPEBA = q (+100 J/C)
What happens
to the kinetic
energy?
DPEBA = -1.6 x 10-17 J
An electron
is released
from rest at
B. What is
its speed
when it
reaches A?
DKEBA = +1.6 x 10-17 J
m= 9.1 x 10-31 kg
1/2 mv2 = +1.6 x 10-17 J
q= -1.6 x 10-19 C
What happens
to the kinetic
energy?


A
B
v = 5.9 x 106 m/s
DPEBA = -1.6 x 10-17 J
DVAB = ?


B
A
How does
doubling the
E-field affect
DVAB = ?
DVAB ?
DVAB
doubles


B
A
How does
moving point
B affect
DVAB = ?
DVAB ?
DVAB is
halved
Anything
else?
How does
DVAB
depend on
E and D?
D



Bold
Bnew
A
How much
work must I
do to move
the charge
from A to B?
KEA=1/2 mv02
KEB=1/2 mv02
DKEAB = 0
D
v0
v0
A
B
constant
speed
How much
work must I
do to move
the charge
from A to B?
FHq
B
WAB = FHq x D
WAB = FE x D
WAB = qED
FE = qE
A
WAB = qED
What is the
change in
potential
energy in
going from
A to B?
PEAB = qED
PEAB = qDVAB
DVAB = EDAB
WAB = qED
FHq
B
FE = qE
A
Only applies when the
field is uniform over the
distance. DVAB‘s sign
depends on the direction
of E. In this case it’s
positive.