# 2_5 - 2 r 2 qQ r On the other hand W v = 1 4 ^ P aq 2 Ý r...

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PHY 5346 HW Set 3 Solutions – Kimel 3. 2.5 a) W = X r K | F | dy = q 2 a 4 ^ P 0 X r K dy y 3 1 ? a 2 y 2 2 = q 2 a 8 ^ P 0 Ý r 2 ? a 2 Þ Let us compare this to disassemble the charges ? W v = ? 1 8 ^ P 0 > i f j q i q j | x 3 i ? x 3 j | = 1 4 ^ P 0 aq 2 r 1 r 1 ? a 2 r 2 = q 2 a 4 ^ P 0 Ý r 2 ? a 2 Þ > 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 = X r K | F | dy = q 4 ^ P 0 X r K Qdy y 2 ? qa 3 X r K Ý 2 y 2 ? a 2 Þ y Ý y 2 ? a 2 Þ 2 dy Using standard integrals, this gives W = 1 4 ^ P 0 q 2 a 2 Ý r 2 ? a 2 Þ ? q 2 a
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Unformatted text preview: 2 r 2 ? qQ r On the other hand ? W v = 1 4 ^ P aq 2 Ý r 2 ? a 2 Þ ? q Ý Q + a r q Þ r = 1 4 ^ P aq 2 Ý r 2 ? a 2 Þ ? q 2 a r 2 ? qQ r 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....
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## This note was uploaded on 01/10/2012 for the course PHYSIS 506 taught by Professor Smith during the Spring '08 term at University of Arkansas Community College at Morrilton.

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