7. Pull the required charges to obtain the arrangements shown below (one at the time).Use the superposition principle and:a. Draw the net electric field at the position of the E-field sensor.b. Write an expression for the net electric field at the position of the E-field sensor(show all your work).c. Find the magnitude and direction of the net electric field at the position of theE-field sensor (click on "show numbers" and verify your answersNote all charges are fixed at the grid, and 1 box = 0.1 m.

Given, Charge configuration let q=-1 nCand at (-0.5,0) or (0.5,0)(1) (a) (b) E→net=E→1+E→2E→net=kq1r13r→1+kq2r23r→2wherer→1 & r→2 are position vectors (c) r→1=0,0.5--0.5,0=0.5,0.5r→2=0,0.5-0.5,0=-0.5,0.5r1=r2=0.52+0.52=0.707 mThen E→net=kq1r13r→1+kq2r23r→2E→net=9*109-1*10-90.70730.5,0.5+9*109-1*10-90.7073-0.5,0.5E→net=-25.50.5,0.5-25.5-0.5,0.5E→net=0,-25.5 N/Cmagnitude E=25.5 N/Cdirection= - y axis or down wards(2) (b) E→net=E→1+E→2E→net=kq1r13r→1+kq2r23r→2wherer→1 & r→2 are position vectors (c) r→1=0,0--0.5,0=0.5,0r→2=0,0-0.5,0=-0.5,0r1=r2=0.5 mThen E→net=kq1r13r→1+kq2r23r→2E→net=9*109-1*10-90.530.5,0+9*109-1*10-90.53-0.5,0E→net=-720.5,0-72-0.5,0E→net=0,0 N/Cmagnitude E=0 N/Cdirection= zero (3) (b) E→net=E→1+E→2E→net=kq1r13r→1+kq2r23r→2wherer→1 & r→2 are position vectors (c) r→1=0,-0.5--0.5,0=0.5,-0.5r→2=0,-0.5-0.5,0=-0.5,-0.5r1=r2=0.52+0.52=0.707 mThen E→net=kq1r13r→1+kq2r23r→2E→net=9*109-1*10-90.70730.5,-0.5+9*109-1*10-90.7073-0.5,-0.5E→net=-25.50.5,-0.5-25.5-0.5,-0.5E→net=0,25.5 N/Cmagnitude E=25.5 N/Cdirection= y axis or Upwards

Answered: 7. Pull the required charges to obtain…