Download CHAPTER 23 SOLUTION FOR PROBLEM 19 (a) To calculate the

CHAPTER 23
SOLUTION FOR PROBLEM 19
(a) To calculate the electric field at a point very close to the center of a large, uniformly charged
conducting plate, we may replace the finite plate with an infinite plate with the same area charge
density and take the magnitude of the field to be E = σ/ 0 , where σ is the area charge density
for the surface just under the point. The charge is distributed uniformly over both sides of the
original plate, with half being on the side near the field point. Thus
σ=
6.0 × 10−6 C
q
2
=
= 4.69 × 10−4 C/m .
2A
2(0.080 m)2
The magnitude of the field is
E=
σ
0
2
4.69 × 10−4 C/m
=
= 5.3 × 107 N/C .
8.85 × 10−12 C2 /N · m2
The field is normal to the plate and since the charge on the plate is positive, it points away from
the plate.
(b) At a point far away from the plate, the electric field is nearly that of a point particle with
charge equal to the total charge on the plate. The magnitude of the field is E = q/4π 0 r 2 , where
r is the distance from the plate. Thus
2
E=
(8.99 × 109 N · m2 /C )(6.0 × 10−6 C)
= 60 N/C .
(30 m)2
CHAPTER 23
SOLUTION FOR PROBLEM 28
To find an expression for the electric field inside the shell in terms of A and the distance from
the center of the shell, select A so the field does not depend on the distance.
Use a Gaussian surface in the form of a sphere with radius rg , concentric with the spherical shell
and within it (a < rg < b). Gauss’ law will be used to find the magnitude of the electric field a
distance rg from the shell center.
The $charge that is both in the shell and within the Gaussian sphere is given by the integral
qs = ρ dV over the portion of the shell within the Gaussian surface. Since the charge distribution
has spherical symmetry, we may take dV to be the volume of a spherical shell with radius r and
infinitesimal thickness dr: dV = 4πr 2 dr. Thus,
8 rg
8 rg
8 rg
A 2
2
qs = 4π
r dr = 4πA
ρr dr = 4π
r dr = 2πA(rg2 − a2 ) .
r
a
a
a
The total charge inside the Gaussian surface is q + qs = q + 2πA(rg2 − a2 ).
The electric field is radial, so the flux through the Gaussian surface is Φ = 4πrg2 E, where E is
the magnitude of the field. Gauss’ law yields
4π 0 Erg2 = q + 2πA(rg2 − a2 ) .
Solve for E:
1
E=
4π
0
}
]
q
2πAa2
+ 2πA −
.
rg2
rg2
For the field to be uniform, the first and last terms in the brackets must cancel. They do if
2
q − 2πAa2 = 0 or A = q/2πa2 = (45.0 × 10−15 C)/2π(2.00 × 10−2 m)2 = 1.79 × 10−11 C/m .
CHAPTER 23
HINT FOR PROBLEM 5
Think of the proton as being at the center of a cube with edge length d. According to Gauss’s
law the net electric flux through the surface of the cube is e/ 0 . Since the proton is at the center
one-sixth of the net flux is through each face of the cube.
J
ans: 3.01 nN · m2 /C
o
CHAPTER 23
HINT FOR PROBLEM 15
(a) The charge on the drum is the product of its surface charge density and the surface area of
2
its curved surface: q = σ2πRL, where σ is the charge density (2.0 µC/m from Problem 16), R
is its radius, and L is its length.
(b) The magnitude of the electric field at the drum surface is given by E = λ/2π 0 R, where
lambda is the linear charge density. Since λ = q/L, E = q/2π 0 RL. Thus q2 /R2 L2 = q1 /R1 L1 ,
where the subscripts 1 refer to the old values and the subscripts 2 refer to the new values. Solve
for q2 .
J
ans: (a) 0.32 µC; (b) 0.14 µC
o
CHAPTER 23
HINT FOR PROBLEM 27
Use Gauss’ law with a spherical Gaussian surface in the form of a sphere with radius r. The
electric flux through the surface is 4πr2 E, where E is the magnitude of the electric field. For
part (a) only the charge on the smaller shell is enclosed by the Gaussian surface and part (b) the
charge on both shells is enclosed.
J
ans: (a) 2.50 × 104 N/C; (b) 1.35 × 104 N/C
o
CHAPTER 23
HINT FOR PROBLEM 31
Use a spherical Gaussian surface with radius r, concentric with the charge distribution. The
electric field is radially outward, normal to the surface, and has a uniform magnitude over the
surface, so the electric flux through the surface is 4πr2 E, where E is the magnitude of the
electric field a distance r from the center of the charge distribution. The charge enclosed by the
Gaussian surface is
8 r
qenc =
4π(r )2 ρ dr
0
Thus
4πr
2
0E
=
8
r
0
Differentiate with respect to r and solve for ρ.
J
ans: 6 0 Kr 3
o
4π(r )2 ρ dr .