Topic 6_2_Ext D__Electric potential energy and
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Transcript Topic 6_2_Ext D__Electric potential energy and
Topic 6.2 Extended
D – Electric potential energy and p.d.
Topic 6.2 Extended
D – Electric potential energy and p.d.
GRAVITATIONAL POTENTIAL ENERGY DIFFERENCE
In mechanics we began with vectors,
-
POINT MASS
and then found easier ways to solve
problems - namely energy
considerations.
rf
In the last chapter we looked at
electric vectors. Now we want to
look at electric energy. This will
simplify finding solutions to
electrical problems.
Consider a mass in a gravitational
field:
If we raise the mass from a height
of r0 to a height of rf and release
it, we know it will relinquish the
potential energy we gave to it converting it to kinetic energy.
r0
Topic 6.2 Extended
D – Electric potential energy and p.d.
GRAVITATIONAL POTENTIAL ENERGY DIFFERENCE
The gravitational force is given by
-
POINT MASS
GmM
r2
and the work done against gravity
is the potential energy difference
FG =
UG = FG rf - FG r0
GmM
GmM
•r0
•r
=
f
2
2
r0
rf
UG =
GmM
rf
-
GmM
r0
Gravitational Potential
Energy Difference
(Point Mass)
rf
r0
Topic 6.2 Extended
D – Electric potential energy and p.d.
ELECTRIC POTENTIAL ENERGY DIFFERENCE
- POINT CHARGE
An electric field has the same properties as a
gravitational field.
If we place a test charge q0 in an
electric field we must do work on it in
moving it from one place to another.
If we release the test charge it will
A
relinquish the potential energy we gave
to it, just as the mass did in the
q0
r
0
gravitational field.
The potential energy difference of
the test charge is given by
Ue = Fe rf - Fe r0
kq0Q
kq0Q
•r0
•r
=
f
2
2
r0
rf
Ue =
kq0Q
rf
-
kq0Q
r0
Electric Potential
Energy Difference
(Point Charge)
rf
q0
Note the missing word
Topic"Energy"
6.2 Extended
D – Electric potential energy and p.d.
ELECTRIC POTENTIAL DIFFERENCE
- POINT CHARGE
We define the electric potential difference V to be
the potential energy per unit charge. Thus
Ue
=
V =
q0
kQ - kQ
r0
rf
Electric Potential
Difference
(Point Charge)
We eliminate the need for the presence of a test
charge in the expression for electric potential energy.
We essentially have a quantity that tells us the
potential energy per unit positive charge.
We can also define the gravitational potential
difference Vg to be the potential energy per unit
mass. Thus
VG =
UG
=
m
GM - GM
r0
rf
Gravitational
Potential Difference
(Point Mass)
Topic 6.2 Extended
D – Electric potential energy and p.d.
GRAVITATIONAL POTENTIAL ENERGY DIFFERENCE
- LOCAL
Of course, at the surface of the earth we have a local
gravitational field, represented with a bunch of g's:
Consider a baseball in a gravitational field:
If we raise the ball from a height of h0 to a
height of hf and release it, we know it will
relinquish the potential energy we gave to it converting it to kinetic energy.
The potential energy of the ball in the local
h
gravitational field is given by
UG = FG hf - FG h0
= mghf - mgh0
Gravitational
Potential Energy
Difference
GRAVITATIONAL POTENTIAL DIFFERENCE
VG =
UG
=
m
W
m
= gh
hf
-
LOCAL
Gravitational
Potential
Difference
h0
FYI: The E-field will be perpendicular to the plates. Why?
Topic 6.2 Extended
FYI: The E-field will act just like the local gravitational field.
D – Electric potential energy and p.d.
ELECTRIC FIELD
IN A PARALLEL PLATE CAPACITOR
A parallel plate capacitor is essentially two metal
plates separated by a small distance.
If we connect the two plates to a battery,
electrons from the negative side of the battery will
be "loaded" onto one plate, and electrons from the
other plate will be drawn off of the other plate
into the positive side of the battery.
+
+- + + + + + + + + + + + + + + + +
----------------
Topic 6.2 Extended
D – Electric potential energy and p.d.
ELECTRIC POTENTIAL ENERGY DIFFERENCE
- PLATES
The potential energy of a test charge in the
local electric field is given by
UE = FE df - FE d0
= q0Edf - q0Ed0
ELECTRIC POTENTIAL DIFFERENCE
W
U
= Ed
V =
=
q0
q0
Electric Potential
Energy Difference
(Parallel Plates)
-
PARALLEL PLATES
Electric Potential
Difference
(Parallel Plates)
FYI: Perhaps you recall that only CONSERVATIVE forces can have
Topic The
6.2
Extended
associated potential energies.
electric
force, like the gravitational
force
fact, conservative.
This means that
the potential
energy
D is,–in Electric
potential
energy
and
p.d.
difference does NOT depend on the path of the charge.
All right.
Let’s do some practice problems.
An electron in the vicinity of a proton is moved
from a distance of 1 meter to a distance of 2
meters. What is the potential energy difference
between initial and final positions?
Note first that we seek the potential ENERGY
difference.
kq0Q
kq0Q
Ue =
(use the point charge form)
r0
rf
Ue = kq0Q 1 - 1
rf
r0
1
1
Ue = (9×109)(-1.6×10-19)(1.6×10-19)
2
1
Ue = +1.152×10-28 J
FYI: Note that the electron-proton system gained energy.
FYI: If the electron had begun at 2 m and ended at 1 m, our
change in potential energy would have been negative. Why?
Topic 6.2 Extended
D – Electric potential energy and p.d.
An electron in the vicinity of a proton is moved
from a distance of 1 meter to a distance of 2
meters. What is the potential difference between
initial and final positions?
Potential difference is just potential ENERGY
difference divided by q0 – in this case the charge of
the electron:
Ue
V =
q0
=
+1.152×10-28 J
-1.6×10-19 C
= -7.2×10-10 V
Note that the SI unit for electric potential
difference is the joule / coulomb or volt.
FYI: These are the same volts that you are familiar with.
FYI: We could have bypassed the previous problem and found V
directly by using
kQ - kQ
Ue
=
V =
r0
q0
rf
Topic 6.2 Extended
D – Electric potential energy and p.d.
An electron is located next to the negative plate of
a parallel plate pair, and released from rest. The
positive plate is located 2 cm away and the electric
field strength between the plates is 250 N/C.
(a) Find the potential energy lost by the electron
as it travels to the positive plate.
Again, choose the correct formula – that for ENERGY:
UE = q0Edf - q0Ed0
UE = q0E(df - d0)
(use the parallel plate form)
UE = (-1.6×10-19)(250)(0.02 - 0)
UE = -8×10-19 J
Topic 6.2 Extended
D – Electric potential energy and p.d.
An electron is located next to the negative plate of
a parallel plate pair, and released from rest. The
positive plate is located 2 cm away and the electric
field strength between the plates is 250 N/C.
(b) Find the speed of the electron as it strikes the
positive plate.
Use energy considerations.
K + UE = 0
K = -UE
K = 8×10-19 J
1
mv2
2
1
(9.11×10-31)v2
2
= 8×10-19
= 8×10-19
v = 1.3×106 m/s
FYI: Don’t confuse voltage V with velocity v.
Topic 6.2 Extended
D – Electric potential energy and p.d.
An electron is located next to the negative plate of
a parallel plate pair, and released from rest. The
positive plate is located 2 cm away and the electric
field strength between the plates is 250 N/C.
(c) Find potential difference between the two
plates.
Pick the right formula:
V = Ed
(use the parallel plate form)
V = (250)(0.02)
V = 5 V
FYI: The NEGATIVE sign means that these charges have a lower
potential energy thanTopic
they would6.2
if theyExtended
were farther away. The highest
potential
for these particular
charges energy
would be ZERO,
.
D – energy
Electric
potential
andat p.d.
FYI: If all the charges were POSITIVE (or NEGATIVE), the total
Find
the
total
electrostatic
potential
energy
would
be positive. The lowest potential
q3 =energy
-2Cwould
potential
energy
in ZERO.
the
occur at , and
would be
charge configuration shown.
Assume the charges have been
assembled from infinity!
q1 = +5C
q2 = +1C
0.25 m
Pick the right formula and work in pairs:
0
kqaqb
kqaqb
Ue =
(use the point charge form)
rf
kq1q2
r12
= k51
0.25
= 0.18 J
kq1q3
r13
= -k52
0.25
= -0.36 J
kq2q3
r23
= -k12
0.25
= -0.072 J
U12 =
U13 =
U23 =
U12
U13
U23
U12
U13
Utotal = 0.18 J +
-0.36
J +
-0.072
U23
J
=
-0.252
J
Question: How would you find their speeds after they have separated
Topic 6.2 Extended
to infinity? Find their speed now. Is it what you expected?
D – Electric potential energy and p.d.
Two electrons are placed 1 cm apart and released.
At what speed are they each traveling after their
separation is 10 cm?
Use energy considerations:
K + UE = 0
0
Kf + Uf = K0 + U0
kqq
kqq
1
1
mv2 + mv2 +
=
2
2
rf
r0
1
1
mv2 = kq2
r0
rf
kq2 1
1
2
v =
m r0
rf
(9×109)(-1.6×10-19)2 1
1
=
9.11×10-31
.01 .10
v = 150.87 m/s
v2