The magnitude of electric intensity at a distance 'x' from a charge 'q' is E. An identical charge is placed at a distance '2x' from it. Then the magnitude of the force it experience is

Given data: E is the size of the electric field at a distance ‘X’ from a charge ‘q’. A similar charge is positioned ‘2x’ away from it.  

Concept used: electric field 

Detailed solution:

Electric field at x is E

$E=\frac{1}{4\pi {\epsilon }_0}=\frac{q}{x^2}$

Electric field at $2\mathrm{x}$ is $E_1$

 $E_1=\frac{1}{4\pi {\epsilon }_0}\frac{q}{(2x{)}^2}$ 

 $E_1=\frac{1}{4}\left(\frac{1}{4\pi {\epsilon }_0}\frac{q}{(2x{)}^2}\right)$

From equation.

$E_1=\frac{E}{4}$

hence the force experienced will be $\frac{qE}{4}$.