This preview shows pages 1–3. Sign up to view the full content.
THE
ELECTRIC
FIELD
26.1. Model:
Visualize:
both the charges are positive, their electric fields at
P
are directed away from the charges.
Solve:
The electric field from
q1
is
The electric field is that of the two charges placed on the yaxis.
Please refer to Figure
Ex26.1.
We denote the upper charge by
q1
and the lower charge by
q2.
Because
(9.0
x
lo9 N m2
/
Cz)(l
x
C)
ebelow +xaxis
(cos
&

sin
e 3)
(0.05 m)’
+
(0.05 m)’
Because tan6
=
5 cm/5 cm
=
1,
the angle
6
= 45”.
Hence,
zl
=(1800N/C)
(A;$)
’
Similarly, the electric field from
q2
is
,!?,,,,,
=
El
+
E,
=
2(18OO N
/
C)

i^
=
25502
N
/
C
(4
Thus, the strength of the electric field is 2550 N/C and its direction is horizontal.
Assess:
ycomponents of their fields will cancel when added.
Because the charges are located symmetrically on either side of the xaxis and are of equal value, the
26.2.
Model:
Visualize:
Please refer to Figure
Ex26.2.
The electric field of the positive charge
q,
at
P
is away from ql. On the
other hand, the electric field of the negative charge
q2
at P is toward
These two electric fields are then added
vertically to obtain the net electric field at
P.
Solve:
The electric field from
The electric field is that of the two charges located on the yaxis.
cos
&

sin
6)
(9.0
x
lo9 N mz
/
C’)(l.O
x
C)
El=
1
=
[
(0.05 m)’
+
(0.05
m)’

(

“I1,
@below +xaxis
4m0
52
Because tan0
=
5
cm/5 cm,
8
=
45”.
So,
(45
+LiL;)
Jz
Similarly, the electric field from
*
E,,,,,
=
+
E2
= 2(1800
N
/
C)
Thus, the strength of the electric field is 2550 N/C and its direction is vertically downward.
Assess:
cancel.
A
quick visualization of the components of the two electric fields shows that the horizontal components
261
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document 262
Chapter
26
26.3. Model:
The electric field
is
that of the two
1
nC charges located on the
yaxis.
Visualize:
Please refer to Figure Ex26.3. We denote the top 1 nC charge by
q1
and the bottom 1 nC charge by
q2
The electric fields
(
E,
and
E,)
of
both the positive charges are directed away from their respective charges. With
vector addition, they yield the net electric field
Erie,
at the point P indicated by the dot.
Solve:
The electric fields from q1 and
are
(9.0
x
lo9 N m'
/
C')(1
x
lo4 C)
(0.05 m)'
along +xaxis
i
=36001N/C
1
@above +xaxis
Because tan@
=
10 cm/5 cm,
8
=
tan'(2)
=
63.43'. So,

(9.0
x
lo9 N m'
/
C')(I
x
C)
E?
=
(cos63.43'
1 +
sin63.43'33)
=
(3221
+
6443) N
/
C
(0.10 m)'
+
(0.05 m)'
The net electric field is thus
E,,
a,P
=
E,
+
i2
=
(39222
+
6443) N
/
C
To find the angle this net vector makes with the x axis, we calculate
*
@=
9.3'
644N/C
3922 N
/
C
tan@
=
Thus, the strength of the electric field at P is
=
J(3922 N
/
C)'
+
(644
N
/
C)'
=
3975 N/C
and
makes an angle of 9.3'above the +xaxis.
Assess:
special symmetry relative to the charges, we expected the net field to be at an angle relative to the xaxis.
26.4. Model:
Visualize:
Please refer to Figure Ex26.4. We denote the positive charge by q1 and the negative charge by
q2.
The
electric field
E,
of the positive charge
q,
is directed away from
ql,
but the field
2,
is toward the negative charge
We will add
El
and
Solve:
Because of the inverse square dependence on distance,
E?
This is the end of the preview. Sign up
to
access the rest of the document.
This note 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

Click to edit the document details