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THE
ELECTRIC
POTENTIAL
29.1. Model:
The mechanical energy of the proton is conserved.
A
parallel plate capacitor
has
a uniform
electric field.
Visualize:
After
Before
*
.
v
=o
.
E‘
*
I
,
I
x
0
1
.o
2.0
The figure shows the beforeandafter pictorial representation. The proton has
an
initial speed
vi
=
0
m/s
and a final
speed vf after traveling a distance
d
=
2.0 mm.
Solve:
The proton loses potential energy and gains kinetic energy as it moves toward the negative plate. The
potential energy
U
is defined
as
U
=
U,
+
qEx,
where
x
is the distance from the negative plate and
is the
potential energy at the negative plate (at
x
=
0
m).
Thus, the change in the potential energy of the proton is
The change in the kinetic energy of the proton is
The law of conservation of energy is
AK
+
AUp
=
0 J. This means
2(+1.60
x
C)(50,000
N
/
C)(2.0
x
m)
(1.67
x
IO*’ kg)
=1.38x1OS
Assess: As
described in Section
29.1,
the potential energy for a charge
4
in an electric field
E
is
U
=
+
qEx,
where
x
is the distance measured from the negative plate. Having
U
=
Uo
at the negative plate (with
x
=
0 m) is
completely arbitrary. We could have taken it to be zero. Note that only
AU,
and not
has physical consequences.
29.2. Model:
electric field.
Visualize:
Before
After
The mechanical energy of the electron is conserved.
A
parallel plate capacitor has a uniform
+

I=+
*
E

+

,
I
x
()
0
0.5
1.0
mm
291
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Chapter 29
The figure shows the beforeandafter pictorial representation. The electron has an initial speed
vi
=
0
m/s
and a
final speed
v,
after traveling a distance
d
=
1
.O mm.
Solve:
The electron loses potential energy and gains kinetic energy as it moves toward the positive plate. The
potential energy
U
is defined as
U
=
U, +qEx,
where
x
is the distance from the negative plate and
U,,
is
the
potential energy at the negative plate (at
x
=
0 m). Thus, the change in the potential energy of the electron is
The change in the kinetic energy of the electron is
=
Kf

K,
=
+mv:

3mv'
1


1
2
mv:
Now, the law of conservation of mechanical energy gives AK
+
AU
=
0
J.
This means
+
qEd
=
0
J
(2)(1.60
x
C)(20,000 N
/
C)(l.O
xIO~ m)
=2.65x106
9.11~10"' kg
Assess:
Note that
AU,
=
is the change in the potential energy of the electron. It is negative because
q
=
e
for
the electron. Thus, the potential energy becomes more negative as
d
increases, that is, the potential energy of the
electron decreases with an increase in
d
(or
x).
29.3.
Model:
field.
Visualize:
The mechanical energy of the proton is conserved. A parallel plate capacitor has a uniform electric
Charge on each plate is
Q
2Q
After
Before
After
Before
4
v;=o
v,=o
*2
2
4
4
!
X
X
0
d
0
d
The figure shows the beforeandafter pictorial representation.
Solve:
The proton loses potential energy and gains kinetic energy as it moves toward the negative plate. The
potential energy is defined as
U
=
U,
+
qEx,
where
x
is the distance from the negative plate and
is the potential
energy at the negative plate (at
x
=
0 m). Thus, the change in the potential energy of the proton is
AUp
=
Vi
=
(U,
+O
J)(U,
++Ed)=
qEd
The change in the kinetic energy of the proton is
AK
=

Ki
=
+mv: +mv?
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This homework help was uploaded on 04/07/2008 for the course PHSX 211/212 taught by Professor Medvedev during the Spring '08 term at Kansas.
 Spring '08
 Medvedev

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