2_1 - | = q 2 2 × 4 ^ P d Notice parts d and e are not...

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PHY 5346 HW Set 2 Solutions – Kimel 4. 2.1 We will work in cylidrical coordinate, ( _ , z , d , with the charge q located at the point d 3 = dz ! , and the conducting plane is in the z = 0 plane. Then we know from class the potential is given by d x 3 = 1 4 ^ P 0 q x 3 ? d 3 ? q x 3 + d 3 E z = ? q 4 ^ P 0 / / z 1 z ? d 2 + _ 2 1/2 ? 1 z + d 2 + _ 2 1/2 E z = q 4 ^ P 0 z ? d z ? d 2 + _ 2 3/2 ? z + d z + d 2 + _ 2 3/2 a) a = P 0 E z z = 0 a = P 0 q 4 ^ P 0 ? d ? d 2 + _ 2 3/2 ? + d + d 2 + _ 2 3/2 = ? q 2 ^ d 2 1 1 2 + _ d 2 3 2 Plotting ? 1 1 2 + r 2 3 2 gives -1 -0.8 -0.6 -0.4 -0.2 0 1 2 3 4 5 r b) Force of charge on plane F 3 = 1 4 ^ P 0 q ? q 2 d 2 ? z ! = 1 4 4 ^ P 0 q 2 d 2 z ! c) F A = w = P 0 2 E 2 = a 2 2 P 0 = 1 2 P 0 ? q 2 ^ d 2 1 1 2 + _ d 2 3 2 2 = 1 8 P 0 q 2 ^ 2 d 4 1 + _ 2 d 2 3 F = 2 ^ q 2 8 P 0 ^ 2 d 4 X 0 K _ 1 + _ 2 d 2 3 d _ = 2 ^ q 2 8 P 0 ^ 2 d 4 1 4 d 2 = 1 4 4 ^ P 0 q 2 d 2
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d) W = X d K Fdz = q 2 4 4 ^ P 0 X d K dz z 2 = q 2 4 4 ^ P 0 d e) W = 1 2 1 4 ^ P 0 > i , j , i fi j q i q j | x 3 i ? x 3 j | =
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Unformatted text preview: | = ? q 2 2 × 4 ^ P d Notice parts d) and e) are not equal in magnitude, because in d) the image moves when q moves. f) 1 Angstrom = 10 ? 10 m, q = e = 1.6 × 10 ? 19 C. W = q 2 4 × 4 ^ P d = e e 4 × 4 ^ P d = e 1.6 × 10 ? 19 4 × 10 ? 10 9 × 10 9 V = 3.6 eV...
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