An electric field from a charge has a magnitude of 4.5 × 104 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?

0.09 n; repel each other

−0.09 n; attract each other

−2.3 n; repel each other

2.3 n; attract each other

F = -0.09 N, attract each other

Explanation:

Electrostatic forces also known as Coulomb forces are attractive or repulsive forces between charged particles that are caused by their electric charges. Electrostatic force between opposite charges is always attractive and Electrostatic force between similar charges is always repulsive.

The strength of the electrostatic force exerted on a charge is given by

F = qE

where

F is the electrostatic force

q is the magnitude of the charge

E is the electric field magnitude

Given data:

q = +4.5 × 10⁴ N/C

E = -2.0 × 10⁻⁶ C

F = ?

F = qE

F = (+4.5 × 10⁴ N/C)(-2.0 × 10⁻⁶ C)

F = -0.09 N

The charge will be attracted towards the source of the electric field. Because electric field from a charge points outward. Its means that the source charge is positive, so the other charge which is negative is attracted towards it because opposite charges always attract each other.

he kicked it 10 feet straight up into the air? i think this works.