Tutorial Exercise Three charged particles are located at the corners of an equilateral triangle as shown in the figure below (let q = 1.50 JC, and ( = 0.600 m). Calculate the total electric force on the 7.00-JC charge. 7.00 ALC 60 0 a -4.00 JC Part 1 of 9 - Conceptualize The 7.00-JC charge experiences a repulsive force F, due to charge q and an attractive force F2 due to the -4.00-HC charge, where F2 = (4.00/q)F1. If we sketch vectors representing F. and F2 and their sum F (see diagram), we see that the resulting vector has a magnitude somewhere between the magnitudes of the two vectors and a dire an angle some tens of degrees below the horizontal. EL 60.09 Part 2 of 9 - Categorize We find the net electric force by adding the two separate forces acting on the 7.00-JC charge. These individual forces can be found by applying Coulomb's law to each pair of charges. Part 3 of 9 - Analyze The magnitude of the electric force between two point charges q1 and 92 separated by a distance r is given by Coulomb's law Fe = Ke _ 191| | 921 where ke is called the Coulomb constant and is given by Ke = 8.9876 x 109 N . m2 /C2 Applying Coulomb's law to find the force exerted on the 7.00-JC charge by the charge q gives the following magnitude. (8.99 x 109 N . m2/C2)(7.00 x 10-6 c) 1.50 P 1.5 x 10-6 C 0.600 2 0.6 m 0.2622 0.262 N Part 4 of 9 - Analyze For the vector Fa, we have F. = (C N)( cos 600 1 [+ v sin 60 j ) i N