In Example 3.2, 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 infinity). 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. 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 r 'r' R Ꮎ a q b q' a FIGURE 3.12 r FIGURE 3.13 Solution Examine the completely different configuration, consisting of the point charge q together with another point charge placed a distance R q' = --q, a R² b = a (3.15) (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 1 V (r) ΑΠΕ0 r (2 + %). (3.17) where and ' are the distances from 9 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 7 This solution is due to William Thomson (later Lord Kelvin), who published it 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 fixed ratio of distances from two given points is a sphere. See J. C. Maxwell, "Treatise on Electricity and Magnetism, Vol. I," 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 1 qq' 1 q² Ra F = (3.18) 4лЄo (a - b)² 4л€ (a² - R²)².

In Example 3.2, 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 infinity). 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. 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 r 'r' R Ꮎ a q b q' a FIGURE 3.12 r FIGURE 3.13 Solution Examine the completely different configuration, consisting of the point charge q together with another point charge placed a distance R q' = --q, a R² b = a (3.15) (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 1 V (r) ΑΠΕ0 r (2 + %). (3.17) where and ' are the distances from 9 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 7 This solution is due to William Thomson (later Lord Kelvin), who published it 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 fixed ratio of distances from two given points is a sphere. See J. C. Maxwell, "Treatise on Electricity and Magnetism, Vol. I," 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 1 qq' 1 q² Ra F = (3.18) 4лЄo (a - b)² 4л€ (a² - R²)².