Home Work Solutions 31. 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 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 1012 C/m 2 )(2.56 102 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 2.5106 C/m5 , r qenc Ar 2 2 r L dr ALr 4 . 0 2 A r3 . By Gauss’ law, we find |E | (2rL) qenc / 0 ; we thus obtain E 40 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, the electric field is so intense that. for r = 0. between collisions with gas atoms. the free electrons gain energy sufficient to ionize these atoms also. Since the magnitude of the field at the cylinder wall is known. 6 3 0 4. the Gaussian surface is a cylinder with radius R and length L. The resulting free electrons (e) are drawn to the positive wire. circular. If the electric field at the shell's inner wall is 2. Thus. 3. directed radially outward from the center of the sphere. And the result. A charge distribution that is spherically symmetric but not uniform radially produces an electric field of magnitude E = Kr4. 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.050 m. More free electrons are thereby created. what is the total positive charge on the central wire? Figure 23-57 Sol The electric field is radially outward from the central wire. A particle of radiation entering the device through the shell wall ionizes a few of the gas atoms. which causes ionization of atoms. positively charged central wire is surrounded by a concentric.q 1 L L 0. The shell contains a low-pressure inert gas. a device used to detect ionizing radiation.9 × 104 N/C. However. coaxial with the wire. The resulting “avalanche” of electrons is collected by the wire. 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 )4r 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 . and the length of the shell 16 cm. Figure 23-57 shows a Geiger counter. A thin. and K is a constant.04 0 Ar 2 2 r L dr 1. generating a signal that is used to record the passage of the original particle of radiation. and the process is repeated until the electrons reach the wire. conducting cylindrical shell with an equal negative charge. is | E | /2 0 r 3. we take the Gaussian surface to coincide with the wall. the inner radius of the shell 1. Here r is the radial distance from that center.0 1011 C/m. Suppose that the radius of the central wire is 25 μm. creating a strong radial electric field.4 cm.6 N/C. . 2 0 r . 5. expression for E when r > R. Charge is distributed uniformly throughout the volume of an infinitely long solid cylinder of radius R. That is.6 109 C. “end view”) of the charged cylinder (solid circle). Gauss’ law yields 2 0 r Eext R 2 Eext R2 . at a distance r < R from the cylinder axis.85 1012 C2 /N m2 (0. In this case. so this is the total flux. (b) Write an Sol (a) The diagram shows a cross section (or.9 104 N/C) 3.16 m) (2. q 2 8. and the flux through it is 2 RLE. Gauss’ law leads to 2 0 r E r 2 E r . coaxial with the charged cylinder.Only the charge on the wire is actually enclosed by the Gaussian surface. The area of the Gaussian surface is 2RL. normal to the Gaussian surface and distributed uniformly along it. Consider a Gaussian surface in the form of a cylinder with radius r and length . we consider a cylindrical Gaussian surface of radius r > R. (a) Show that. The charge enclosed by it is q V r 2 . perhaps more appropriately. where ρ is the volume charge density. Thus. the electric field lines are radially outward. q R2 . The charge enclosed is the total charge in a section of the charged cylinder with length l . where V r 2 l is the volume of the cylinder. We assume there is no flux through the ends of the cylinder. If is positive. the total flux through the Gaussian cylinder is EAcylinder E (2 r ). Gauss’ law yields q = 20RLE. we denote it by q. An “end view” of the Gaussian surface is shown as a dashed circle.014 m)(0. 2 0 (b) Next. Now. If the external field Eext then the flux is 2 r Eext . Thus.