physics, Electric Field, question is in the description

Question

What is the magnitude and direction of the electric field 0.25 meters away from a point charge with 5.0 C. Draw a diagram. Use proportional reasoning to find the field strength at positions 0.125 meters away and 0.75 meters away.

Answer 1

The electric field E is vector value, E=F/Q, where Q is the charge creates the field.

Pay attention to the sign of charge Q, as by convention we use positive test charge +q.

F=kQq/R^2, that means the direction of E and F will be the same for +Q and opposite for -Q.

In case of unchangeable Q=5.0 C and positions of the test charge two time closer or two times farther from Q, you will have corresponding forces and electric fields.

As force and field are proportional to 1/R^2, that means the following:

1. In case of 0.25 m→ 5.0 m, that is two time farther the E→E_in/4
2. In case of 0.25 m→ 0.125 m, that is two time closer the E→4E_in