2_9 - a = 3 P E cos S + Q 4 ^ R 2 = 3 P E x + Q 4 ^ R 2 = 3...

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PHY 5346 HW Set 4 Solutions – Kimel 2. 2.9 The system is pictured below a) We have treated this probem in class. We found the charge density induced was a = 3 P 0 E 0 cos S We also note the radial force/unit area outward from the surface is a 2 /2 P 0 . Thus the force on the right hand hemisphere is, using x = cos S F z = 1 2 P 0 X a 2 z ! 6 da 3 = 1 2 P 0 Ý 3 P 0 E 0 Þ 2 2 ^ R 2 X 0 1 x 3 dx = 1 2 P 0 Ý 3 P 0 E 0 Þ 2 2 ^ R 2 /4 = 9 4 ^ P 0 E 0 2 R 2 An equal force acting in the opposite direction would be required to keep the hemispheres from sparating. b) Now the charge density is
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Unformatted text preview: a = 3 P E cos S + Q 4 ^ R 2 = 3 P E x + Q 4 ^ R 2 = 3 P E x + Q 12 ^ P E R 2 F z = 1 2 P X a 2 z ! 6 da 3 = 1 2 P 3 P E 2 2 ^ R 2 X 1 x x + Q 12 ^ P E R 2 2 dx Thus F z = 9 4 ^ P E 2 R 2 + 1 2 QE + 1 32 P ^ R 2 Q 2 An equal force acting in the opposite direction would be required to keep the himispheres from separating....
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