Question

A charge Q acts on a point charge to create an electric field. Its strength, measured a distance of 40 cm away is 100 N/C. What is the magnitude of the electric field strength at a distance of 20 cm

Answer 1

Answer:

$E_2=80N/C$

Explanation:

From the question we are told that:

Initial Distance $d_1=40cm=>0.4m$

Initial Electric field strength $E_1=100N/C$

Final Distance $d_2=20cm=>0.20m$

Generally the equation for Electric field is mathematically given by.

 $E=\frac{kq}{d^2}$

 $q=\frac{100*(0.4)^2}{K}$

Substituting q for $d=20cm$

 $E_2=\frac{k}{0.2}*\frac{100*(0.4)^2}{K}$

 $E_2=80N/C$