Download Physics 42 Chapter 24 Homework

Physics 42 HW#2 Chapter 24
Problems: 4, 15, 18, 19, 27, 31, 34, 52, 54, 57, 63, 65
4.
Consider a closed triangular box resting within a
horizontal electric field of magnitude E = 7.80 × 104 N/C as
shown in Figure P24.4. Calculate the electric flux through
(a) the vertical rectangular surface, (b) the slanted surface,
and (c) the entire surface of the box.
P24.4
(a)
A ′ = ( 10.0 cm )( 30.0 cm )
30.0 cm
=
A ′ 300
=
cm 2 0.030 0 m 2
Φ E, A ′ =
EA ′ cosθ
Φ E, A ′ =
(7.80 × 10 ) ( 0.030 0) cos180°
4
10.0 cm
60.0Þ
Φ E, A ′ =
−2.34 kN ⋅ m 2 C
(b)
=
A
4

cm )( w ) ( 30.0 cm ) 
( 30.0 =

Φ E, A =
(c)
(7.80 × 10 ) ( A ) cos60.0°
Φ E, A = EA cosθ =
10.0 cm 
cm 2 0.060 0 m 2
=
=
 600
cos60.0° 
(7.80 × 10 ) (0.060 0) cos60.0° =
4
FIG. P24.4
+2.34 kN ⋅ m 2 C
The bottom and the two triangular sides all lie parallel to E, so Φ E =
0 for each of these. Thus,
Φ E, total =−2.34 kN ⋅ m 2 C + 2.34 kN ⋅ m 2 C + 0 + 0 + 0 = 0 .
33.
Consider a long cylindrical charge distribution of radius R with a uniform charge density ρ. Find the
electric field at distance r from the axis where r < R
If ρ is positive, the field must be radially outward. Choose as the gaussian
surface a cylinder of length L and radius r, contained inside the charged rod.
Its volume is π r 2L and it encloses charge ρπ r 2L . Because the charge
distribution is long, no electric flux passes through the circular end caps;
=
E ⋅ dA EdA
=
cos90.0° 0 . The curved surface has E=
⋅ dA EdA cos0° ,
and E must be the same strength everywhere over the curved surface.
P24.33
FIG. P24.29
Gauss’s law,
q
,
∫ E ⋅ dA =
∈
0
becomes
E
∫
dA =
Curved
Surface
ρπ r L
2
∈0
Now the lateral surface area of the cylinder is 2π rL : E ( 2π r ) L =
Thus,
E=
ρr
2 ∈0
.
ρπ r 2L
∈0
.
radially aw ay from the cylinder axis .
P24.39
(a)
Inside surface: consider a cylindrical surface within the metal. Since E inside the conducting
shell is zero, the total charge inside the gaussian surface must be zero, so the inside charge/length = −λ .
cylinder is
(b)
so
Outside surface:
2λ=
 qin + qout
The total charge on the metal
q=
2λ  + λ 
out
so the outside charge/length
3λ
is
E
=
2ke ( 3λ ) 6keλ
= =
r
r
qin
=

=
0 λ  + qin
3λ
radially outw ard
2π ∈0 r
−λ
65.
An infinitely long insulating cylinder of radius R has a volume charge density that varies with the
radius as


r
b
ρ = ρ0  a − 
where ρ0, a, and b are positive constants and r is the distance from the axis of the cylinder. Use Gauss’s law to
determine the magnitude of the electric field at radial distances (a) r < R and (b) r > R.
P24.70 In this case the charge density is not uniform, and Gauss’s law is written as
1
ρdV . We use a
∫ E ⋅ dA =
∈ ∫
0
gaussian surface which is a cylinder of radius r, length  , and is coaxial with the charge distribution.
(a)
2π r  )
When r < R , this becomes E (=
ρ0 r 
∈0
r
∫  a − b dV . The element of volume is a cylindrical shell of
0
radius r, length  , and thickness dr so that dV = 2π r dr .
 2π r 2ρ0   a r 
=
E
=
E ( 2π r  ) 
  −  so inside the cylinder,
 ∈0   2 3b
(b)
ρ0 r 
2r 
 a −  .
2 ∈0 
3b
When r > R , Gauss’s law becomes
E (=
2π r  )
ρ0 R 
∈0
r
∫  a − b ( 2π r dr )
0
or outside the cylinder,
=
E
ρ0 R2 
2R 
 a −
 .
2 ∈0 r
3b 