Answer: [tex]40\ N/C[/tex]
Explanation:
Given
Magnitude of charge is [tex]q=0.5\ C[/tex]
Force experienced is [tex]F=20\ N[/tex]
Electric field intensity is the electrostatic force per unit charge
[tex]\therefore E=\dfrac{F}{q}\\\\\Rightarrow E=\dfrac{20}{0.50}\\\\\Rightarrow E=40\ N/C[/tex]
Thus, the electric field intensity is [tex]40\ N/C[/tex]