Respuesta :

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]

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