a point charge 50*10 raise to the power -6 C is located in the XY plane at the position vector r0= 2i^+3j^. what is the electric field at the point of position vector r=8i^+5j^?

Answer 1

E=kq (rvector)/(magnitude rvector)^3

rvector= displacement=(6i+2j)

and rvector= sqrt(6^2+2^2)=sqrt40
so magnitude rvector ^3=40sqrt40
check that