The magnitude of the electric field must be 50 N/C. Option D is correct.
The magnitude of the electric field can be calculated by dividing the force exerted by the field by the charge of the particle.
[tex]E = \dfrac FQ[/tex]
Where,
[tex]E[/tex] - electric field
[tex]F[/tex] - force = 4 N
[tex]Q[/tex] - charge = 0.08 C
Put the values in the formula,
[tex]E = \dfrac 4{0.08}\\\\E = 50 \rm \ N/C[/tex]
Therefore, the magnitude of the electric field must be 50 N/C to the right since the charge is moving towards the right.
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