HW_6_Solutions

HW_6_Solutions - HW 6 Solutions (due online 1 AM, Friday,...

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HW 6 Solutions (due online 1 AM, Friday, February 4) 1. Chapter 24 Problem 28 Figure 24-42 shows a thin plastic rod of length L = 14.3 cm and uniform positive charge Q = 52.4 fC lying on an x axis. With V = 0 at infinity, find the electric potential at point P 1 on the axis, at distance d = 3.76 cm from one end of the rod. Solution: Consider an infinitesimal segment of the rod, located between x and x + dx . It has length dx and contains charge dq = dx , where = Q / L is the linear charge density of the rod. Its distance from P 1 is d + x and the potential it creates at P 1 is 00 11 . 44 dq dx dV d x d x    To find the total potential at P 1 , we integrate over the length of the rod and obtain: 0 0 0 0 0 9 2 2 15 3 ln( ) ln 1 4 4 4 (8.99 10 N m C )(56.1 10 C) 0.12 m ln 1 7.39 10 V. 0.12 m 0.025 m L L dx Q L V d x d x L d     2. Chapter 24 Problem 34 Two large parallel metal plates are 2.2 cm apart and have charges of equal magnitude but opposite signs on their facing surfaces. Take the potential of the negative plate to be zero. If the potential halfway
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HW_6_Solutions - HW 6 Solutions (due online 1 AM, Friday,...

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