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if a particle with a charge of +4.3 C is attracted to another particle by a force of 6.5 N, what is the strength of the electric field at this location?

Respuesta :

GeorgePP

Answer: 1.5 N/C

Explanation:

The equation is derived from Coulomb's Law, but is not exactly Coulomb's Law:

F = qE

q is the test charge

E is the electrical field produced by the source charge

F is the force in Newtons

In this problem, we were given the value of the test charge q, and the force acting on it due to the electric field from the source charge.

E = F/q

E = 6.5N/4.3C

E = 1.5 N/C

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