Transcript Lesson 4: Electric Potential

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Potential
IV . Electric ______________
Move some _______________
charge q from point A to B in
electric
the E ( ________________
) field of a charged object:
lines
E field ________
A
charged
object
B
q
F
work W
 It takes ____________
to move q closer because it
force F
distance d .
requires a ____________
applied for a ______________
Because of the work done, the energy of the
PE
___________of
charges is stored up as an electric _______
.
system
The electric _______________________
V is defined
potential difference
as the work done per charge:
V=
W/q
no direction
scalar
V is a _____________
because it has ___________________
units of V:
[V] = [ W ] / [ q ]
[V] =
J
/
C
By definition,
volt
one _______________________
= 1 _________
joule/coulomb
V
J/C = 1 _____
or
1 _______
 Because of its units, the potential difference is often
voltage
called the ________________
.
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Ex 1: 18 J of work are required to move 2.0 C of charge
between two points in an electric field. Find the electric
potential difference between the two points.
Given:
18 J = W
2.0 C =
q
V= ?
V=W/q
V = 18 J /2.0 C
= 9.0 J/C
= 9.0 V
How much work would be required to move 4.0 C
of charge between the same two points of the problem
above?
q = 4.0 C
V=W/q
V = 9.0 V
W= ?
9.0 V
= W/ 4.0 C
36 J = W
How much potential energy is stored up as a result? 36 J
DC volts in different sizes and shapes:
AC volts in different sizes and shapes:
NOTE:
V = W/q can be used with 2 sets of units!
When working with small amounts of charge
smaller
and energy, you often use _______________
units.
•For W, instead of joules, use units:
electronvolts, eV
______________________
•For q, instead of coulombs, use units:
elementary charges, e
the number of ______________________
Using these new units, the potential difference V in:
V = W/q
will have units:
which again be volts:
[V]=[W] /[ q ]
V =
the e's cancel .
You get V because _________________
eV / e
Ex: 20 electronvolts of work are required to move 8
electrons between two points in an electric field. Find
the electric potential difference between the two points
Given:
20 eV = W
8e=
q
V= ?
V=W/q
= 20 eV / 8 e
= 2.5 eV/e
= 2.5 V
Ex: How much work (in eV) is done on a proton when
it is accelerated through a potential difference of
68,000 volts?
V= W/q
68,000 V = W / 1 e
 W = 68,000 eV
How much kinetic energy will the proton have as a
result?
DKE = W =
68,000 eV
1
Conversions between units. See PhysRT, page ____:
1 electronvolt (eV) = _____________________ J
Notice that this number is the same as the one
in the conversion:
1 elementary charge (e) = _____________________ C
Ex. Convert 9.6 x 10-19 J to eV.
9.6 x 10-19 J x ___________________ = ______________ eV
Ex. Convert 4640 eV to J.
4640 eV x ___________________ = ___________________ J
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1
Conversions between units. See PhysRT, page ____:
-19
1.60
x
10
1 electronvolt (eV) = _____________________ J
Notice that this number is the same as the one
in the conversion:
1.60 x 10-19
1 elementary charge (e) = _____________________
C
Ex. Convert 9.6 x 10-19 J to eV.
1 eV
6.0
9.6 x 10-19 J x ___________________ = ______________
eV
1.60 x 10-19 J
Ex. Convert 4640 eV to J.
1.60 x 10-19 J
-16
7.42
x
10
4640 eV x ___________________ = ___________________ J
1 eV
Summary of electric potential units:
quantity
big
units
W
(or energy)
J
eV
q
C
e
V
small units
V
V
 eV's are NOT a unit of __________________________
.
potential difference
work
They are a unit for __________
energy or _____________.
e
J with _______
Also, you cannot mix:
_____
eV
C with _______
or: _____
One of them must be converted first!
Ex. Find the work done in joules in moving 5 electrons
through a potential difference of 6.0 V.
Solution 1:
Find answer in eV, then convert:
V = W /q
6.0 V = W / 5 e
30 eV = W
-19 J
1.60
x
10
30
4.8 x 10-18 J
________ eV x _______________ = ____________
1 eV
Solution 2:
Convert e to C first, then calculate.
-19 C
1.60
x
10
8.0 x 10-19 C
5
_______ e x __________________ = ____________
1e
V = W /q
6.0 V = W / 8.0 x 10-19 C
4.8 x 10-18 J = W
NOTES:
PE
1. Electric potential difference represents the ___________
charge
per ____________
stored up in the system of charges.
2. If the charges are allowed to move freely through a
KE
potential difference, they will gain the energy as _____
in the same way that a ___________
object gains it.
falling
gravity:
electricity:
PE = 100 J
KE =
100 J
KE = 100 J
PE =
100 J
3. A van der Graff is a device that produces a strong
potential
electric
_____________
field. This provides a large _____________
work
difference
__________________
to do ______________
on charges:
V = W/q 
W = qV
KE of the charges is increased:
As a result, the _______
W = DE =
DKE
speed
This increases their _____________
because:
KE = (1/2) mv2
Neutrons cannot be accelerated by van der Graff's
neutral
because they are ________________
and are unaffected
by ______________
fields.
electric