What is the magnitude of the electric field produced by a charge of magnitude 5.00x10-6 C at a distance of 2.00 m?

Answer 1

The electric field produced by a point charge is E=k Q/r^2

where k = 9 x 10^9 and r is the distance from the point charge

Therefore E = (9 x 10^9 )(5 x 10^-6)/4 = 11,250 Newtons/Coulomb or Volts/Meter