2_3 - PHY 5346 HW Set 3 Solutions Kimel 2 2.3 The system is...

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PHY 5346 HW Set 3 Solutions – Kimel 2. 2.3 The system is described by a) Given the potenial for a line charge in the problem, we write down the solution from the figure, d T = V 4 ^ P 0 ln R 2 Ý x 3 ? x 3 o Þ 2 ? R 2 Ý x 3 ? x 3 o 1 Þ 2 ? R 2 Ý x 3 ? x 3 o 2 Þ 2 + R 2 Ý x 3 ? x 3 o 3 Þ 2 Looking at the figure when y = 0, Ý x 3 ? x 3 o Þ 2 = Ý x 3 ? x 3 o 1 Þ 2 , Ý x 3 ? x 3 o 2 Þ 2 = Ý x 3 ? x 3 o 3 Þ 2 , so d T | y = 0 = 0 Similarly, when x = 0, Ý x 3 ? x 3 o Þ 2 = Ý x 3 ? x 3 o 2 Þ 2 , Ý x 3 ? x 3 o 1 Þ 2 = Ý x 3 ? x 3 o 3 Þ 2 , so d T | x = 0 = 0 On the surface d T = 0, so Nd T = 0, however, Nd T = / d T / x t N x t = 0 / d T / x t = 0 E t = 0 b) We remember a = ? P 0 / d T / y = ? V ^ y 0 Ý x ? x 0 Þ 2 + y 0 2 ? y 0 Ý x + x 0 Þ 2 + y 0 2 where I’ve applied the symmetries derived in a). Let a / V = ? 1 ^ y 0 Ý x ? x 0 Þ 2 + y 0 2 ? y 0 Ý x + x 0 Þ 2 + y 0 2 This is an easy function to plot for various combinations of the position of the original line charge ( x 0 , y 0 Þ . c) If we integrate over a strip of width A z , we find, where we use the integral X 0 K 1 Ý x @ x 0 Þ 2 + y 0 2 dx = 1 2 ^ 2arctan x 0 y 0 y 0 A Q = X 0 K a dx A z
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2_3 - PHY 5346 HW Set 3 Solutions Kimel 2 2.3 The system is...

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