Kiarab1234
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A charge of +0.08 C moves to the right due to a 4 N force exerted by an electric field. What is the magnitude and direction of the electric field?

A) 0.32 N/C left

B) 0.32 N/C right

C) 50 N/C left

D) 50 N/C right

Respuesta :

Maicasso

This is late but the answer is 50 n/c right

Answer : Electric field is given by, E = 50 N/C in right.

Explanation :

It is given that,

Charge, q = 0.08 C (moves to the right)

Force exerted by the electric field, F = 4 N

We know that the electric field is defined as :

[tex]E=\dfrac{F}{q}[/tex]

[tex]E=\dfrac{4\ N}{0.08\ C}[/tex]

[tex]E=50\ N/C[/tex]

The direction of field is same as the direction of electric force i.e.

[tex]E=50\ N/C[/tex] in right direction.

So, the correct option is (D).

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