Download Home Work Solutions 3

Home Work Solutions 3
1. In Fig. 23-41, a small circular hole of radius R = 1.80 cm has been cut in the middle of an infinite, flat,
nonconducting surface that has uniform charge density σ = 4.50 pC/m2. A z axis, with its origin at the hole's
center, is perpendicular to the surface. In unit-vector notation, what is the electric field at point P at z = 2.56 cm?
(Hint: See Eq. 22-26 and use superposition.)
Figure 23-41
Sol
The charge distribution in this problem is equivalent to that of an infinite sheet of charge with
surface charge density 4.50 1012 C/m2 plus a small circular pad of radius R = 1.30 cm located at
the middle of the sheet with charge density –. We denote the electric fields produced by the sheet
r
r
and the pad with subscripts 1 and 2, respectively. Using Eq. 22-26 for E2 , the net electric field E
at a distance z = 2.56 cm along the central axis is then
   ?    
z
E  E1  E2  
1  2
k 
2 0 
z  R2
 2 0 


z
k?
k 
2
2
2 0 z  R

(4.50 1012 C/m 2 )(2.56 102 m)
2(8.85 10
12
C /N  m ) (2.56 10
2
2
2
m)  (1.30 10
2
2
m)
2
k?  (0.227 N/C) k.
2. A long, nonconducting, solid cylinder of radius 4.0 cm has a nonuniform volume charge density ρ that is a
function of radial distance r from the cylinder axis: ρ = Ar2. For A = 2.5 μC/m5, what is the magnitude of the
electric field at (a) r = 3.0 cm and (b) r = 5.0 cm?
Sol
To evaluate the field using Gauss’ law, we employ a cylindrical surface of area 2  r L where L is very
large (large enough that contributions from the ends of the cylinder become irrelevant to the
calculation). The volume within this surface is V =  r2 L, or expressed more appropriate to our needs:
dV  2 rLdr. The charge enclosed is, with A  2.5106 C/m5 ,
r

qenc   Ar 2 2 r L dr  ALr 4 .
0
2
A r3
.
By Gauss’ law, we find   |E | (2rL)  qenc /  0 ; we thus obtain E 
40
r
(a) With r = 0.030 m, we find | E |  1.9 N/C.
(b) Once outside the cylinder, Eq. 23-12 is obeyed. To find  = q/L we must find the total charge q.
Therefore,
q 1

L L

0.04
0
Ar 2 2 r L dr  1.0 1011 C/m.
And the result, for r = 0.050 m, is | E |   /2 0 r  3.6 N/C.
3. A charge distribution that is spherically symmetric but not uniform radially produces an electric field of
magnitude E = Kr4, directed radially outward from the center of the sphere. Here r is the radial distance from that
center, and K is a constant. What is the volume density ρ of the charge distribution?
Sol
We use
E (r ) 
qenc
1

2
4 0 r
4 0 r 2

r
0
 (r )4r 2 dr
to solve for (r) and obtain
 (r ) 
0 d
2
r dr
r 2 E (r ) 
0 d
r 2 dr
cKr h 6K r .
6
3
0
4. Figure 23-57 shows a Geiger counter, a device used to detect ionizing radiation, which causes ionization of atoms.
A thin, positively charged central wire is surrounded by a concentric, circular, conducting cylindrical shell with
an equal negative charge, creating a strong radial electric field. The shell contains a low-pressure inert gas. A
particle of radiation entering the device through the shell wall ionizes a few of the gas atoms. The resulting free
electrons (e) are drawn to the positive wire. However, the electric field is so intense that, between collisions with
gas atoms, the free electrons gain energy sufficient to ionize these atoms also. More free electrons are thereby
created, and the process is repeated until the electrons reach the wire. The resulting “avalanche” of electrons is
collected by the wire, generating a signal that is used to record the passage of the original particle of radiation.
Suppose that the radius of the central wire is 25 μm, the inner radius of the shell 1.4 cm, and the length of the
shell 16 cm. If the electric field at the shell's inner wall is 2.9 × 104 N/C, what is the total positive charge on the
central wire?
Figure 23-57
Sol
The electric field is radially outward from the central wire. We want to find its magnitude in the
region between the wire and the cylinder as a function of the distance r from the wire. Since the
magnitude of the field at the cylinder wall is known, we take the Gaussian surface to coincide with
the wall. Thus, the Gaussian surface is a cylinder with radius R and length L, coaxial with the wire.
Only the charge on the wire is actually enclosed by the Gaussian surface; we denote it by q. The
area of the Gaussian surface is 2RL, and the flux through it is   2 RLE. We assume there is
no flux through the ends of the cylinder, so this  is the total flux. Gauss’ law yields q = 20RLE.
Thus,


q  2 8.85 1012 C2 /N  m2 (0.014 m)(0.16 m) (2.9 104 N/C)  3.6 109 C.
5. Charge is distributed uniformly throughout the volume of an infinitely long solid cylinder of radius R. (a) Show
that, at a distance r < R from the cylinder axis,
expression for E when r > R.
where ρ is the volume charge density. (b) Write an
Sol
(a) The diagram shows a cross section (or, perhaps more appropriately, “end view”) of the charged
cylinder (solid circle).
Consider a Gaussian surface in the form of a cylinder with radius r and length , coaxial with the
charged cylinder. An “end view” of the Gaussian surface is shown as a dashed circle. The charge
enclosed by it is q  V   r 2  , where V  r 2 l is the volume of the cylinder.
If  is positive, the electric field lines are radially outward, normal to the Gaussian surface and
distributed uniformly along it. Thus, the total flux through the Gaussian cylinder is
  EAcylinder  E (2 r ). Now, Gauss’ law leads to
2 0 r E   r 2   E 
r
.
2 0
(b) Next, we consider a cylindrical Gaussian surface of radius r > R. If the external field Eext then the
flux is   2 r Eext . The charge enclosed is the total charge in a section of the charged cylinder
with length l . That is, q   R2  . In this case, Gauss’ law yields
2 0 r Eext   R 2   Eext 
R2 
.
2 0 r