An electric field from a charge has a magnitude of 4.5 × 10^4 N/C at a certain location that points outward. If another charge with a magnitude of −2.0 × 10^−6 C is brought near it, what is the strength of the electrostatic force that acts on this charge and how do the two charges behave?

Respuesta :

The strength of the electrostatic force acts on this charge is F =  0.09 N.

Explanation:

  • The electrostatic force is the force that occurs between two charged particles having a positive or negative charge which tends to attract or repel depends on the charged particle.

The force on the electric field can be calculated by using the formula,

                         E = F / q

                         F = -q [tex]\times[/tex] E

Given q = - 2.0 [tex]\times[/tex] 10^-6,                   E = 4.5 [tex]\times[/tex] 10^4.

                electrostatic force = (2.0 [tex]\times[/tex] 10^-6)[tex]\times[/tex](4.5 [tex]\times[/tex] 10^4)

                                           F   = 0.09 N.

  • Since the charge q is negative, the force is exerted opposite to the direction of the field. The field is a positive charge but the charge as a negative charge.