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Problem 3.9 In Ex. 3.2 we assumed that the conducting sphere was grounded (V = 0). But with the addition of a second image charge, the same basic model will handle the case of a sphere at any potential Vo (relative, of course, to infin- ity). What charge should you use, and where should you put it? Find the force of attraction between a point charge q and a neutral conducting sphere.

Problem 3.9 In Ex. 3.2 we assumed that the conducting sphere was grounded (V = 0). But with the addition of a second image charge, the same basic model will handle the case of a sphere at any potential Vo (relative, of course, to infin- ity). What charge should you use, and where should you put it? Find the force of attraction between a point charge q and a neutral conducting sphere.

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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Example 3.2. A point charge q is situated a distance a from the center of a grounded conducting sphere of radius R (Fig. 3.12). Find the potential outside the sphere. V=0 FIGURE 312 FIGURE 3.13 Solution Examine the completely different configuration, consisting of the point charge q together with another point charge --. (3.15) placed a distance (3.16) to the right of the center of the sphere (Fig. 3.13). No conductor, now-just the two point charges. The potential of this configuration is Vr) = (3.17) where s and e' are the distances from q and q', respectively. Now, it happens (see Prob. 3.8) that this potential vanishes at all points on the sphere, and therefore fits the boundary conditions for our original problem, in the exterior region." Conclusion: Eq. 3.17 is the potential of a point charge near a grounded con- ducting sphere. (Notice that b is less than R, so the "image" charge q' is safely inside the sphere-you cannot put image charges in the region where you are cal- culating V; that would change p, and you'd be solving Poisson's equation with "Tis solution is due to William Thomson (later Loed Kelvin), who published ik in 1848, when he was just 24. It was apparently inspired by a theorem of Apollonius (200 BC) that says the locus of points with a fed ratio ef distances from two given points is a spbere. See .C. Maswell "Treatise on Electricity and Magnetism, Vel. 1," Dover, New York, p. 245. I thank Gabriel Karl for this interesting history. 3.2 The Method of Images 129 the wrong source.) In particular, the force of attraction between the charge and the sphere is q*Ra 4xeo (a - Ra 1 (3.18) 4T€0 (a - b)
Transcribed Image Text:Example 3.2. A point charge q is situated a distance a from the center of a grounded conducting sphere of radius R (Fig. 3.12). Find the potential outside the sphere. V=0 FIGURE 312 FIGURE 3.13 Solution Examine the completely different configuration, consisting of the point charge q together with another point charge --. (3.15) placed a distance (3.16) to the right of the center of the sphere (Fig. 3.13). No conductor, now-just the two point charges. The potential of this configuration is Vr) = (3.17) where s and e' are the distances from q and q', respectively. Now, it happens (see Prob. 3.8) that this potential vanishes at all points on the sphere, and therefore fits the boundary conditions for our original problem, in the exterior region." Conclusion: Eq. 3.17 is the potential of a point charge near a grounded con- ducting sphere. (Notice that b is less than R, so the "image" charge q' is safely inside the sphere-you cannot put image charges in the region where you are cal- culating V; that would change p, and you'd be solving Poisson's equation with "Tis solution is due to William Thomson (later Loed Kelvin), who published ik in 1848, when he was just 24. It was apparently inspired by a theorem of Apollonius (200 BC) that says the locus of points with a fed ratio ef distances from two given points is a spbere. See .C. Maswell "Treatise on Electricity and Magnetism, Vel. 1," Dover, New York, p. 245. I thank Gabriel Karl for this interesting history. 3.2 The Method of Images 129 the wrong source.) In particular, the force of attraction between the charge and the sphere is q*Ra 4xeo (a - Ra 1 (3.18) 4T€0 (a - b)
Problem 3.9 In Ex. 3.2 we assumed that the conducting sphere was grounded (V = 0). But with the addition of a second image charge, the same basic model will handle the case of a sphere at any potential Vo (relative, of course, to infin- ity). What charge should you use, and where should you put it? Find the force of attraction between a point charge q and a neutral conducting sphere. Chapter 3 Potentials ! Problem 3.10 A uniform line charge à is placed on an infinite straight wire, a dis- tance d above a grounded conducting plane. (Let's say the wire runs parallel to the x-axis and directly above it, and the conducting plane is the xy plane.) (a) Find the potential in the region above the plane. [Hint: Refer to Prob. 2.52.] (b) Find the charge density o induced on the conducting plane.
Transcribed Image Text:Problem 3.9 In Ex. 3.2 we assumed that the conducting sphere was grounded (V = 0). But with the addition of a second image charge, the same basic model will handle the case of a sphere at any potential Vo (relative, of course, to infin- ity). What charge should you use, and where should you put it? Find the force of attraction between a point charge q and a neutral conducting sphere. Chapter 3 Potentials ! Problem 3.10 A uniform line charge à is placed on an infinite straight wire, a dis- tance d above a grounded conducting plane. (Let's say the wire runs parallel to the x-axis and directly above it, and the conducting plane is the xy plane.) (a) Find the potential in the region above the plane. [Hint: Refer to Prob. 2.52.] (b) Find the charge density o induced on the conducting plane.
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