Classical Electrodynamics - Jackson Problem 2.5.pdf

PHY 5346 HW Set 3 Solutions – Kimel 3. 2.5 a) W = ∫ r ∞ |F|dy = q2a 4π 0 ∫ r ∞ dy y3 1 − a2 y2 2 = q2a 8π 0r2 − a2  Let us compare this to disassemble the charges − W′ = − 1 8π 0 ∑ i≠j q iq j |x⃗i−x⃗j | = 1 4π 0 aq2 r 1 r 1 − a2 r2 = q2a 4π 0r2 − a2  > W The reason for this difference is that in the first expression W, the image charge is moving and changing size, whereas in the second, whereas in the second, they don’t. b) In this case W = ∫ r ∞ |F|dy = q 4π 0 ∫ r ∞ Qdy y2 − qa3 ∫ r ∞ 2y2 − a2  yy2 − a2 2 dy Using standard integrals, this gives W = 1 4π 0 q2a 2r2 − a2  − q 2a 2r2 − qQr On the other hand − W′ = 1 4π 0 aq2 r2 − a2  − qQ + a r q r = 1 4π 0 aq2 r2 − a2  − q 2a r2 − qQr The first two terms are larger than those found in W for the same reason as found in a), whereas the last term is the same, because Q is fixed on the sphere.