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 1012 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 and the pad with subscripts 1 and 2, respectively. Using Eq. 22-26 for Er2
, the net electric field Er at a distance z = 2.56 cm along the central axis is then
1 2 2 2 2 2
0 0 0
12 2 2
12 2 2 2 2 2 2
? ?
k 1 k k
2 2 2
(4.50 10 C/m )(2.56 10 m) k? (0.227 N/C) k.
2(8.85 10 C /N m ) (2.56 10 m) (1.30 10 m)
z z
E E E
z R z R
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:
2 .
dV rLdr The charge enclosed is, with A2.5 10 C/m 6 5,
2 4
enc 0 2 .
2 q r A r r L dr ALr
By Gauss’ law, we find | | (2E rL)qenc/0; we thus obtain
3
0
4 . E A r
(a) With r = 0.030 m, we find |Er | 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,
0.04 2 11
0
1 2 1.0 10 C/m.
q Ar r L dr
L L
And the result, for r = 0.050 m, is |E| /2 0r3.6 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
enc 2
2 2 0
0 0
( ) 1 ( )4
4 4
q r
E r r r dr
r r
to solve for (r) and obtain
( )r ( ) .
r d
dr r E r r
d
dr Kr K r
02 2 20
c h
6 6 0 34. 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 2RL, and the flux through it is 2RLE. We assume there is no flux through the ends of the cylinder, so this is the total flux. Gauss’ law yields q = 20RLE.
Thus,
12 2 2
4 92 8.85 10 C /N m (0.014 m)(0.16 m) (2.9 10 N/C) 3.6 10 C.
q
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, where ρ is the volume charge density. (b) Write an expression for E when r > R.
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 qV r2 , where V r2l 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
cylinder (2 ).
EA E r
Now, Gauss’ law leads to
2 0
0
2 .
2
r E r E r
(b) Next, we consider a cylindrical Gaussian surface of radius r > R. If the external field Eext then the flux is 2r Eext. The charge enclosed is the total charge in a section of the charged cylinder with length l . That is, qR2 . In this case, Gauss’ law yields
2 2
0 ext ext
0
2 .
2
r E R E R
r
