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PHY 5346 HW Set 3 Solutions – Kimel 4. 2.6 We are considering two conducting spheres of radii ra and rb respectively. The charges on the spheres are Qa and Qb. a) The process is that you start with qa1 and qb1 at the centers of the spheres, and sphere a then is an equipotential from charge qa1 but not from qb1 and vice versa. To correct this we use the method of images for spheres as discussed in class. This gives the iterative equations given in the text. b) qa1 and qb1 are determined from the two requirements ∑ j=1 ∞ qaj = Qa and ∑ j=1 ∞ qbj = Qb As a program equation, we use a do-loop of the form ∑ j=2 n qaj = −raqbj − 1 dbj − 1 and similar equations for qbj, xaj, xbj, daj,dbj. The potential outside the spheres is given by φx⃗ = 1 4π 0 ∑ j=1 n qaj x⃗ − xajk̂ +∑ j=1 n qbj x⃗ − dbjk̂ This potential is constant on the surface of the spheres by construction. And the force between the spheres is F = 1 4π 0 ∑ j,k qajqbk d − xaj − xbk2 c) Now we take the special case Qa = Qb, ra = rb = R, d = 2R. Then we find, using the iteration equations xaj = xbj = xj x1 = 0, x2 = R/2, x3 = 2R/3, or xj = j − 1 j R qaj = qbj = qj qj = q, q2 = −q/2, q3 = q/3, or qj = −1 j+1 j q So, as n → ∞ ∑ j=1 ∞ qj = q∑ j=1 ∞ −1 j+1 j = q ln2 = Q → q = Q ln2 The force between the spheres is F = 1 4π 0 q2 R2 ∑ j,k −1 j+k jk 2 − j−1 j − k−1 k 2 = 1 4π 0 q2 R2 ∑ j,k −1 j+kjk j + k2 Evaluating the sum numerically F = 1 4π 0 q2 R2 0.0739 = 1 4π 0 Q2 R2 1 ln22 0.0739 Comparing this to the force between the charges located at the centers of the spheres Fp = 1 4π 0 Q2 R24 Comparing the two results, we see F = 4 1 ln22 0.0739Fp = 0.615Fp On the surface of the sphere φ = 1 4π 0 ∑ j=1 ∞ qj R − xj = q 4π 0R ∑ j=1 ∞ −1 j+1 Notice 1 1 + 1 = ∑ j=1 ∞ −1 j+1 So φ = 1 4π 0 q 2R = 1 4π 0 Q 2 ln2R = Q C → C 4π 0R = 2 ln2 = 1. 386
