Use Green’s reciprocity theorem (Prob. 3.50) to solve the following two problems. [Hint: for distribution 1, use the actual situation; for distribution 2, remove q, and set one of the conductors at potential V0.]
(a) Both plates of a parallel-plate capacitor are grounded, and a point charge q is placed between them at a distance x from plate 1. The plate separation is d. Find the induced charge on each plate. [Answer: Q1 = q(x/d − 1); Q2 = −qx/d]
(b) Two concentric spherical conducting shells (radii a and b) are grounded, and a point charge q is placed between them (at radius r ). Find the induced charge on each sphere.
Reference prob 3.50
(a) Suppose a charge distribution ρ1(r) produces a potential V1(r), and some other charge distribution ρ2(r) produces a potential V2(r). [The two situations may have nothing in common, for all I care—perhaps number 1 is a uniformly charged sphere and number 2 is a parallel-plate capacitor. Please understand that ρ1 and ρ2 are not present at the same time; we are talking about two different problems, one in which only ρ1 is present, and another in which only ρ2 is present.] Prove Green’s reciprocity theorem:25
[Hint: Evaluate ? E1 . E2 dτ two ways, first writing E1 = −∇V1 and using integration by parts to transfer the derivative to E2, then writing E2 = −∇V2 and transferring the derivative to E1.]
(b) Suppose now that you have two separated conductors (Fig. 3.41). If you charge up conductor a by amount Q (leaving b uncharged), the resulting potential of b is, say, Vab. On the other hand, if you put that same charge Q on conductor b (leaving a uncharged), the potential of a would be Vba . Use Green’s reciprocity theorem to show that Vab = Vba (an astonishing result, since we assumed nothing about the shapes or placement of the conductors).
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