P23_057 - 57 We note that the contributions to the field...

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Unformatted text preview: 57. We note that the contributions to the field from the pair of −2q charges exactly cancel, and we are left with the (opposing) contributions from the 4q (at r = 2d) and −q (at r = d) charges. Therefore, using k = 1/4πε0 4q q |Enet | = k −k 2 =0 . 2 (2d) d The net field at P vanishes completely. ...
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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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