Answer to Question #211037 in Physics for Festus

Question

The electric field intensity at a point situated 4 meters from a point charge is 200N/C. If the distance is reduced to 2 meters, the field intensity will be?

Answer 1

$$E=k\frac{q}{r^2}\\E'=200\frac{4^2}{2^2}=800\frac{N}{C}$$