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Halliday-cap 22-vol.3- Rar - P22 035

Solução do exercícios 1 aqo 51 do Capítulo 22- Cargas Elétricas- 4 ed. VOL.3 A solução dos exercícios está em inglês

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35. (a) Eq. 22-1 gives F12 q1 q2 =k 2 = d 2   2 20.0 × 10−6 C 9 N·m 8.99 × 10 = 1.60 N . C2 (1.50 m)2 (b) A force diagram is shown as well as our choice of y axis (the dashed line). ... ....... .. ...... .... ................. .. ........... . ....... .... ....... .. .... ... ... .. 1 • q ... .. ... .. ... .. ... .. ... .. ... .. ... .. ... .. • q3 ... .. ... .. ... .. ... .. y ... .. ... .. ... .. ... .. ... .. q2 • ... .. The y axis is meant to bisect the line between q2 and q3 in order to make use of the symmetry in the problem (equilateral triangle of side length d, equal-magnitude charges q1 = q2 = q3 = q). We see that the resultant force is along this symmetry axis, and we obtain  2 q |Fy | = 2 k 2 cos 30◦ = 2.77 N . d