In free space, A charge q1=1.602*10^-12 C is located at the origin, and a second charge q2=-1.602*10^-19 C at position r =(6,8) m. What is the magnitude and direction of the electrostatic field at the position of q2?

Given the charge 9 , = 1.602x10 - @ is located at origin ( 9,0 ) . Charge 92 = - 1160zx10 c is located at 92 B ( 618) Hencs we have AB = ( 6 - 0 ) 2 + ( 8 - 0 ) y A = > AB = 62+ 8y Heney JAB = 162 +82 = 10o = 10 m Now we know according to coulomb's law the electric field vector at position of 9, due to q , charge is. E= _ Y 3 . Where r = Position vector of 92 with srespect to V, Now, here we have 1' 6 * 10- /2 AB 4 TEO IAB / 3 Where to= Permitivity of free space. = 8- 85 4 x 10 12 ( 2 Nim" Heng E = 1 . 6 * 10 " 1 2 4xX8- 854 x 10-12 X ( 6 2 + 8y ) . 10 3 = ) E = ( 8 . 6 2 8 2 3 x 10 - 5 ) x + (1. 1504 3 x 10 4 ) y Hence magnitude of field , JE ! = ) ( 8 - 6 2 8 2 3 x 10 5 ) 2 4 ( 1. 150 4 3 x 10 7 12 = 1. 43804 x 10- N/C . 1.15043 *10 - 4 Direction is = tom" 8.62 823 x 105 = 53. 13 with positive xaxis