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Resolução - Halliday - Volume 3 - Eletricidade E Magnetismo - P23 018

Exercícios Resolvidos.

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18. We use Eq. 23-3, assuming both charges are positive. Eleft q1 R 4πε0 (R2 + ring 3/2 R2 ) = = Simplifying, we obtain q1 =2 q2 Eright ring evaluated at P q2 (2R) 3/2 4πε0 ((2R)2 + R2 )  3/2 2 ≈ 0.51 . 5