HW1-phy2049-Spring-2012-solution

22 19 a consider the figure below the magnitude

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Unformatted text preview: −9 C) | | FAC | = A C 2 = = 2.70 ×10−6 N. 2 4πε 0 d (0.200 m) (c) We also obtain PHY2049 Spring 2012 HW1 r | q q | | (8.99 ×109 N ⋅ m 2 C2 ) (−4.00 ×10−9 C)(−4.00 ×10−9 C) | | FBC | = B C 2 = = 3.60 ×10−6 N. 2 4πε 0 d (0.200 m) 22- 15. By symmetry we see that the contributions from the two charges q1 = q2 = +e cancel each other, and we simply use Eq. 22- 3 to compute magnitude of the field due to q3 = +2e. (a) The magnitude of the net electric field is 1 2e 1 2e 1 4e | Enet |= = = 2 2 4πε 0 r 4πε 0 ( a / 2 ) 4πε 0 a 2 −19 C) 9 2 2 4(1.60 × 10 = (8.99 × 10 N ⋅ m C ) = 160 N/C. (6.00 × 10−6 m )2 (b) This field points at 45.0°, counterclockwise from the x axis. 22- 19. (a) Consider the figure below. The magnitude of the net electric field at point P is ⎡1 ⎤ r q d /2 1 qd Enet = 2 E1 sin θ =...
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