Download Chapter 24 Gauss`s Law. Solutions of Home Work Problems

Chapter 24
Gauss’s Law. Solutions of Home
Work Problems
24.1
Problem 24.6 (In the text book)
A point charge q is located at the center of a uniform ring having linear charge density
and radius a, as shown in Figure (24.6). Determine the total electric flux through a sphere
centered at the point charge and having radius R, whereR < a.
Figure 24.6:
Solution
Only the charge within the the surface contribute to the total flux. The flux through the
surface will come from the charge q at the center, and for R < a the charge on the ring will
not contribute to the net flux through the spherical surface. The flux is then:
ΦE =
q
◦
2
CHAPTER 24. GAUSS’S LAW. SOLUTIONS OF HOME WORK PROBLEMS
Note that the ring will have a flux through the surface, but all the flux that enter the surface
through one side will leave leave from the other side and the net flux from the ring through
the surface is zero.
Physics 111:Introductory Physics II, Chapter 24
Winter 2005
Ahmed H. Hussein
24.2. PROBLEM 24.13 (IN THE TEXT BOOK)
24.2
3
Problem 24.13 (In the text book)
Calculate the total electric flux through the paraboloidal surface due to a uniform electric
field of magnitude E◦ in the direction shown in Figure (24.13).
Figure 24.13:
Solution
All the electric flux that enters the object through the flat surface exits the object through
the curved surface. The flux through the flat surface is:
ΦE = E · A = EA cos θ = EA = E◦ πr2
where θ is the angle between the electric filed and the normal to the surface, and in this case
θ = 0◦
Physics 111:Introductory Physics II, Chapter 24
Winter 2005
Ahmed H. Hussein
Fe =
(b)
keq 2
r2
e8.99 × 10 N ⋅ m C jq
2 × 10 N ≈
mf
a0.25 PROBLEMS
GAUSS’S LAW. SOLUTIONS OF HOME WORK
2
9
−3
4
CHAPTER 24.
2
2
2
q ≈ 1. 2 × 10 −7 C ~ 10 −7 C or 100 nC
24.3
Problem 24.34 (In the text book)
e
je
j
8.99 × 10 9 N ⋅ m 2 C 2 1.2 × 10 −7 C
keq
= 2 ≈ charge density ρ and carries
≈ 1.7 ×
An insulating solid sphere of radius a has (c)
a uniformE volume
a
2
r
0
.
25
m
total positive charge Q. A spherical gaussian surface of radius r, which shares a common
center with the insulating sphere, is inflated starting from r = 0.
−7
q
1.2 × 10gaussian
C
(a) Find an expression for the electric flux passing
through
sphere
= 1. 4 × 10 4 N ⋅ m 2
Φ E = the≈surface of −the
(d)
12
2
2
∈
8.85 × 10
C N ⋅m
as a function of r for r < a.
0
a
f
(b) Find an expression for the electric flux for r > a.
4
*P24.34
The charge density is determined by Q = π a 3 ρ
(c) Plot the flux versus r.
3
(a)
ρ=
The flux is that created by the enclosed charge with
Solution
q
4π r 3 ρ
4π r 3 3Q
Qr 3
=
=
Φ E = in =
∈0
3 ∈0
3 ∈0 4π a 3
∈0 a 3
The charge density ρ of the insulating sphere (with volume V , radius a, and total charge Q)
is given by:
Q
Q
3Q Q
ρ=
=(b)4 3 =Φ E =3
. Note that the answers to parts (a) and (
V
4πa ∈
πa
3
0
(a) The flux ΦE is created by the charge (c)
qin within
the volume Vr of the surface with a radius r < a.
The charge qin is:
qin = ρVr =
4 3 Qr
3Q
×
πr = 3
4πa3 3
a
3
ΦE
Q
∈0
The electric flux for r < a is:
ΦE (r < a) =
qin
3Qr3
=
◦
◦ a3
0
0
(b) When r > a then the surface encloses the charge
Q and the flux is then:
ΦE (r > a) =
r
a
FIG.
P24.34(c)
Figure
24.34:
Q
◦
Note that the answers to parts (a) and (b) agree at r = a.
(c) See Figure (24.34).
Physics 111:Introductory Physics II, Chapter 24
Winter 2005
Ahmed H. Hussein
24.4. PROBLEM 24.43 (IN THE TEXT BOOK)
24.4
5
Problem 24.43 (In the text book)
A square plate of copper with 50.0-cm sides has no net charge and is placed in a region of
uniform electric field of 80.0 kN/C directed perpendicularly to the plate. Find
(a) the charge density of each face of the plate and
(b) the total charge on each face.
Solution
(a) The electric field at the surface of a conductor is given by:
E=
σ
◦
σ = E◦ = 8.00 × 104 × 8.85 × 10−12 = 7.08 × 10−7 C/m2
or
The charge density is 7.08 × 10−7 C/m2 positive on one face and negative on the other.
(b) The charge on each face would be:
σ=
Q
A
or
Q = σA = 7.08 × 10−7 × (0.500)2 = 1.77 × 10−7 C
Physics 111:Introductory Physics II, Chapter 24
Winter 2005
Ahmed H. Hussein
6
CHAPTER 24. GAUSS’S LAW. SOLUTIONS OF HOME WORK PROBLEMS
24.5
Problem 24.66 (In the text book)
A closed surface with dimensions a = b = 0.400 m and c = 0.600 m is located as in Figure
(24.66) Figure P24.66. The left edge of the closed surface is located at position x = a. The
electric field throughout the region is nonuniform and given by E = (3.0 + 2.0x2 )î N/C,
where x is in meters. Calculate the net electric flux leaving the closed surface. What net
charge is enclosed by the surface?
Figure 24.66:
Solution
The electric field throughout the region is directed along x; therefore, E will be perpendicular
to A (actually perpendicular to the normal to the surface) over the four faces of the surface
which are perpendicular to the yz plane, and E will be parallel to A over the two faces
which are parallel to the yz plane. So, there will only be flux through two faces. Therefore:
ΦE = −Ex (x = a)A + Ex (x = a + c) = − 3 + 2a2 ab + 3 + 2(a + c)2 ab = 2abc(2a + c)
Substituting the given values for a, b, and c, we find
ΦE = 0.269 N · m2 /C
Physics 111:Introductory Physics II, Chapter 24
Winter 2005
Ahmed H. Hussein
24.5. PROBLEM 24.66 (IN THE TEXT BOOK)
7
and the charge within the surface is:
Q = ◦ ΦE = 2.38 × 10−12 C = 2.38 pC
Physics 111:Introductory Physics II, Chapter 24
Winter 2005
Ahmed H. Hussein