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An electric field of 234,000 N/C points due west at a certain spot. What are the magnitude and direction of the force that acts on a charge of -7.25 µC at this spot?

Respuesta :

Answer:

F = 1.69 N

Explanation:

Given that,

Electric field, E = 234,000 N/C

Charge, Q = -7.25 µC

We need to find the electric force acting on the charge. It can be given as follows :

[tex]F=qE\\\\F=7.25\times 10^{-6}\times 234000\\\\F=1.69\ N[/tex]

As the charge is negative, the force will act in the opposite direction of electric field. Hence, the electric force is 1.69 N.

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