Question

Two charges are located in the x

Answer 1

Let us we are taking a point on origin from where we are going to find out the value of the electric field of these two charges.

For the q1 charge,the electric feild will have only single componenet along y-axis.

E1y=Kq₁/(1200)² =(910⁹)(-2.610^-9)/ (1200)²=.00001625N/c=-1.6210^-5 N/C

E1x=0

For the charge q₂ ist of all we have to find the distance of charge from the origin...By the help of the given coordinates

          r₂= 1²+0.45²=1.09m

Now   E2x=Ecos     E2y=Esin

           here =?

    From the coordinates given    y=perpendicular=o.45m and x=base=1m

     Tan=0.45/1

   =tan^-10.45/1

      =24.22⁰

E2=kq2/r2²=(910⁹)(4.210^-9)/ 1.09²

E2=31.81N/C

E2x=31.8cos24.22              and E2y=31.8sin24.22

       =29.01N/C                 and    E2y=13.04N/C

Ex=E1x+E2x                            Ey=E1y+E2y

Ex=0+29.01

Ex=29.01N/C                            Ey=-1.610^-5 +13.04= 13.03N/C