302
Chapter 30
v=
+loooV

x
(m)
30.3. Visualize:
1Ooov
Ex
1.0
0
+
1.0
f~i
Solve:
30.4:
The electric potential difference
AV
between two points in a uniform electric field is given by Equation
V(xJ

V(XJ
=
Ex
(xf

x,)
Choosing
x,
=
1 .O m and
=
+1
.O
m,
+lo00 V (1000 V)=E,[l.O m(1.0 m)]a E,=lOOO Vlrn
Alternatively,
x,
=
1
.O m and
=
1
m. For this choice,
1000 V

(+lo00 V)
=
Ex
[1.0 m

(1.0 m)]
*
Ex=
1000 Vlrn
Assess:
The choice of initial and final positions does not change the physical nature of the electric field or the
potential difference.
30.4.
Solve:
potential,
Model:
The electric field is the negative of the derivative of the potential function.
From Equation 30.8, the component of the electric field in the sdirection is E,
=
dV/ds.
For the given
dV
du

d
=
(loox'
v)
=
200x
V
m

E,
=
200x
Atx= Om,
=
0 Vlm ,and at
x=
1 m, E, =200 (1) V/m=200 Vlrn.
Assess:
The potential increases with
x,
so the electric field must point in the
x
direction.
30.5.
Solve:
Model:
The electric field is the negative of the slope of the graph of the potential function.
There are three regions of different slope. For 0 cm
<
x
<
10 cm and 20 cm
<
x
<
30 cm,
=OVlm*
E,=OVIm
Ax
For lOcm
<x<
?Ocm,
AV
100 V

(100 V)
=
2000 V
I
m
3
=
+2000
Vlrn
_

Ax
0.20m0.lOm
I
10
20
30
Assess:
Because E,
=
dV/ds
,
the electric field is zero where the potential is not changing.
30.6.
Solve:
Model:
The electric field is the negative of the slope of the graph of the potential function.
There are three regions of different slope. For 0 cm
<
x
<
1 cm,
=
50v0v =5000~/m
~E~=~OOOV/~
Ax
0.01m

Om
For1 cm<x<?cm,,
= 10,000
V
/
m
*
=
10,000 Vlm
AV
50 V

(50 V)
dx
0.02 m0.01 m
For
2
cm
<x
<
3 cm,
=
5000
VI m
a
E,= 5000 Vlm
AV
OV (5OV)
0.03 m

0.02
m