Question

Need Help? EXAMPLE 15.3 A Charge Triangle GOAL Apply Coulomb's law in two PROBLEM Consider three point charges at the corners of a triangle, as shown in the COS 36.9 figure, where q1 6.00 x 10-9 c, q2-2.00 x 109 C, and q3- 5.00 x 10-9 C. (a) Find the components of the force F23 exerted by q2 on g3. (b) Find the components of the force F13 exerted by q1 on q3. (c) Find The foroe the resultant force on q3, in terms of components and also in terms of magnitude and direction 3.00 m 5.00 m 91 STRATEGY Coulomb's law gives the magnitude of each force, which can be split with right-triangle trigonometry into x- and y-components. Sum the vectors componentwise and then find the magnitude and directidn of the resultant vector. SOLUTION

Answer 1

F_23 bar = k Vertical-bar q_2 q_3 Vertical-bar /4^2[-i bar] = k(5.38 times 106-9)(2.04 times 10^-9)/16 (-i bar) 9 times 5.38 times 2.04/16 times 10^-9 i bar = 6.17355 times 10^-9 i bar N therefore F_23x = 6.17355 times 10^-9 F_23y = 0 F =_13 bar = K Vertical-bar q_1 q_3 Vertical-bar /5^2 [4 i bar + 3 j bar] = 9 times 10^9 times 6.37 times 10^-9 times 5.38 times 10^-9/25 (4 i bar + 3j bar) = 12.337 times 10^-9 (4 i bar + 3 j bar) = (49.249 times 10^-9)i bar + (37.011 times 10^-9)j bar N therefore F_13x = 49.349 times 10^-9, F_13y = 37.011 times 10^-9 F_net on 3 = F_23 bar + F_13 bar = (6.17355 times 10^-9 + 49.349 times 10^-9)i bar + (37.011 times 10^-9)j bar N = (55.52255 times 10^-9)i bar + (37.011 times 10^-9)j bar N F_x = 55.5255 times 10^-9 N, F_y = 37.011 times 10^-9 N