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A cylinder of diameter 1.72 m is in a region where the electric field is as shown in the figure below. If E1 = 40.2 N/C and E

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D # 옵 ê GP area of the Circular Cross - Section A = 772 2 A = = 2-32 m AL 10.8672 ¢, - El A Cos (180) flun (1j through - 2:32D # 옵 ê GP area of the Circular Cross - Section A = 772 2 A = = 2-32 m AL 10.8672 ¢, - El A Cos (180) flun (1j through - 2:32

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