Electric Charge and Electric Field
21-39
The charge is negative, so the field at the origin is directed toward the small segment. The small segment is located at
angle
θ
as shown in the sketch. The electric field due to
dQ
is shown in Figure 21.97b, along with its components.
2
0
1
4
dQ
dE
a
π
=
P
2
2
0
2
Q
dE
d
a
θ
π
=
P
Figure 21.97b
(
)
2
2
0
cos
/2
cos
x
dE
dE
Q
a
d
θ
π
θ θ
=
=
P
(
)
/2
/ 2
0
2
2
2
2
2
2
0
0
0
0
cos
sin
2
2
2
x
x
Q
Q
Q
E
dE
d
a
a
a
π
π
θ θ
θ
π
π
π
=
=
=
=
∫
∫
P
P
P
(
)
2
2
0
sin
/2
sin
y
dE
dE
Q
a
d
θ
π
θ θ
=
=
P
(
)
/2
/ 2
0
2
2
2
2
2
2
0
0
0
0
sin
cos
2
2
2
y
y
Q
Q
Q
E
dE
d
a
a
a
π
π
θ θ
θ
π
π
π
=
=
=
−
=
∫
∫
P
P
P
E
VALUATE
:
Note that
,
x
y
E
E
=
as expected from symmetry.
21.98.
I
DENTIFY
:
Apply
0
x
F
=
∑
and
0
y
F
=
∑
to the sphere, with
x
horizontal and
y
vertical.
S
ET
U
P
:
The free-body diagram for the sphere is given in Figure 21.98. The electric field
E
G
of the sheet is
directed away from the sheet and has magnitude
0
2
E
σ
=
P
(Eq.21.12).
E
XECUTE
:
0
y
F
=
∑
gives
cos
T
mg
α
=
and
cos
mg
T
α
=
.
0
x
F
=
∑
gives
0
sin
2
q
T
σ
α
=
P
and
0
2
sin
q
T
σ
α
=
P
.
Combining these two equations we have
0
cos
2
sin
mg
q
σ
α
α
=
P
and
0
tan
2
q
mg
σ
α
=
P
. Therefore,
0
arctan
2
q
mg
σ
α
⎛
⎞
=
⎜
⎟
⎝
⎠
P
.
E
VALUATE
:
The electric field of the sheet, and hence the force it exerts on the sphere, is independent of the
distance of the sphere from the sheet.
Figure 21.98
21.99.
I
DENTIFY
:
Each wire produces an electric field at
P
due to a finite wire. These fields add by vector addition.
S
ET
U
P
:
Each field has magnitude
2
2
0
1
4
Q
x
x
a
π
+
P
. The field due to the negative wire points to the left, while
the field due to the positive wire points downward, making the two fields perpendicular to each other and of equal
magnitude. The net field is the vector sum of these two, which is
E
net
= 2
E
1
cos 45° =
2
2
0
1
2
cos45
4
Q
x
x
a
π
°
+
P
. In
part (b), the electrical force on an electron at
P
is
eE
.
E
XECUTE
:
(a)
The net field is
E
net
=
2
2
0
1
2
cos45
4
Q
x
x
a
π
°
+
P
.
E
net
=
(
)(
)
9
2
2
6
2
2
2 9.00
10 N m /C
2.50
10
C cos45
(0.600 m)
(0.600 m)
(0.600 m)
−
×
⋅
×
°
+
= 6.25
×
10
4
N/C.
The direction is 225° counterclockwise from an axis pointing to the right through the positive wire.

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