Three charged particles are c 7.00-MC charge. 7.00 AC 4.00 JC Part 1 or B - Conceptualize The 7.00-C charge experiences a re orce F, due to the - vectors representing F, and F, and their sum F (see diagram), we Part 2 of 8 - Categorize We find the net electric force by adding the two separate forces slamb's law to each pair of charges. Part 3 of 8 - Analyze The magnitude of the electric force between there ke Is called the Coulomb s given by ke - 8.9876 x 10' N . m' /C. applying Coulomb's law to find the force exerted on the 7.00-NC charge by the charge q gives the following magnitur (8.99 * 10' N - 12/3 )(7.00 x 10 ' =X 1. 50 | 1.5 x 10-" C) 8.600 2 0.6 my D.2622 0.262 N Part 4 of 8- Analyzs For the vector Far we have F - D.262 0.262 N) me Gumi ~ + an 50";) D.131 0.131 1 + 0.226 898656 0.227 1 ) N. Part 6 of B - Analyze Applying Coulomb's law to the force on the 7.00-uC charge by the -4.00-C charge. 8.99 x 10' N - m'/C)(7.00 = 10- cX 4.00 4 x x 10 ') D.6 0.6 m) 0. 69392 0.698 N. Part S of B - Analyze For the vector Far we have F - 0.695 0.639 N ) (cos 50 ) an 50- ; ) (0.3496 0.35 1+ 0.605 -0.606 1) N. Part 7 of 8 - Analyze Thus, the total force on the 7.00-HC chart In unit-vector notation, Is given by the following. F -F + F, - (0.481 0.481 1 + 0.379 8-D.378 1) N Part 3 of 8- Analyze From the equation above, the x and y compon F. - 0.481 0.481 N F, - -0.378 -0.378 N. and the magnitude of the total force is given by the - VD.481 - D.481 N) + (D.378 0.378 N)' - .61175567 0.612 N Part 8 of 8 . Analyze The direction of the total force F Is - tan- (0 785862 ) , we have F- 0612