P25_017 - R of the combined drop is given by R 3 = 2 R 3 and R = 2 1 3 R The charge is twice the charge of original drop q = 2 q Thus V = 1 4 πε

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17. (a) The electric potential V at the surface of the drop, the charge q on the drop, and the radius R of the drop are related by V = q/ 4 πε 0 R .Thu s R = q 4 πε 0 V = (8 . 99 × 10 9 N · m 2 / C 2 )(30 × 10 12 C) 500 V =5 . 4 × 10 4 m .
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Unformatted text preview: R of the combined drop is given by ( R ) 3 = 2 R 3 and R = 2 1 / 3 R . The charge is twice the charge of original drop: q = 2 q . Thus, V = 1 4 πε q R = 1 4 πε 2 q 2 1 / 3 R = 2 2 / 3 V = 2 2 / 3 (500 V) ≈ 790 V ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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