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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? 0. 32 N/C left 0. 32 N/C right 50 N/C left 50 N/C right.

Sagot :

The magnitude of the electric field must be 50 N/C. Option D is correct.

How to calculate the magnitude of the electric field?

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.

Learn more about  the electric field:

https://brainly.com/question/8781325

Answer:

D. 50 N/C right

Explanation:

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