Question

84 What is the
magnitude
of the point charge choosen so that electric field 20 cm
away from it has a
magnitude of 18 x 10^8N\C​

Answer 1

q = 8 × 10^(-3) C

Check the attached image.

Answer 2

Electric field,E= kq/r²

Electric field,E= kq/r² { r is the distance from the charge and r= 0.5m}

Electric field,E= kq/r² { r is the distance from the charge and r= 0.5m}or 2 = kq/(0.5)²

Electric field,E= kq/r² { r is the distance from the charge and r= 0.5m}or 2 = kq/(0.5)²or q = 2×(0.5)²/ 9×10^9 = 0.055 nC