Download CHAPTER 22 SOLUTION FOR PROBLEM 19 (a) The linear charge

CHAPTER 22
SOLUTION FOR PROBLEM 19
(a) The linear charge density λ is the charge per unit length of rod. Since the charge is uniformly
distributed on the rod, λ = −q/L = −(4.23 × 10−15 C)/(0.0815 m) = −5.19 × 10−14 C/m.
(b) and (c) Position the origin at the left end of
dx
P
the rod, as shown in the diagram. Let dx be an
•
infinitesimal length of rod at x. The charge in
x
this segment is dq = λ dx. The charge dq may
L
L+a
0
be taken to be a point charge. The electric field
it produces at point P has only an x component
and this component is given by
dEx =
1
4π
0
λ dx
.
(L + a − x)2
The total electric field produced at P by the whole rod is the integral
eL
8 L
e
dx
1
λ
λ
e
=
Ex =
4π 0 0 (L + a − x)2 4π 0 L + a − x e0
}
]
1
1
L
λ
λ
−
.
=
=
4π 0 a L + a
4π 0 a(L + a)
When −q/L is substituted for λ the result is
Ex = −
1
4π
2
0
(8.99 × 109 N · m2 /C )(4.23 × 10−15 C)
q
=−
= −1.57 × 10−3 N/C .
a(L + a)
(0.120 m)(0.0815 m + 0.120 m)
The negative sign indicates that the field is toward the rod and makes an angle of 180◦ with the
positive x direction.
(d) Now
Ex = −
1
4π
2
0
(8.99 × 109 N · m2 /C )(4.23 × 10−15 C)
q
=−
= −1.52 × 10−8 N/C .
a(L + a)
(50 m)(0.0815 m + 50 m)
(e) The field of a point particle at the origin is
2
(8.99 × 109 N · m2 /C )(4.23 × 10−15 C)
=
−
= −1.52 × 10−8 N/C .
Ex = −
4π 0 a2
(50 m)2
q
CHAPTER 22
SOLUTION FOR PROBLEM 21
At a point on the axis of a uniformly charged disk a distance z above the center of the disk, the
magnitude of the electric field is
}
]
σ
z
E=
1− √
,
2 0
z 2 + R2
where R is the radius of the disk and σ is the surface charge density on the disk. See Eq. 22–26.
The magnitude of the field at the center of the disk (z = 0) is Ec = σ/2 0 . You want to solve
for the value of z such that E/Ec = 1/2. This means
z
1
E
=1− √
=
Ec
z 2 + R2 2
or
√
z
z 2 + R2
=
1
.
2
2
2
2
2
2
2
2
Square both√sides, then multiply
√ them by z + R to obtain z = (z /4) + (R /4). Thus, z = R /3
and z = R/ 3 = (0.600 m)/ 3 = 0.346 m.
CHAPTER 22
HINT FOR PROBLEM 9
The particles are equidistance from P and their charges have the same magnitude. The y components of their fields sum to zero and their x components are the same, so you need calculate
only the x component of one of the fields, then double it. The x component of the field of either
particle is given by Ex = −(1/4π 0 )qx/r3 , where x is the coordinate of the particle and r is its
distance from P.
J
ans: (a) 1.38 × 10−10 N/C; (b) negative x direction
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