Question

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?

Answer 1

The magnitude of the electrostatic force is 0.09 N, and the force is attractive

Explanation:

The strength of the electrostatic force acting on a charged particle is given by

$F=qE$

where

q is the charge

E is the magnitude of the electric field

For the charge in this problem, we have:

$E=4.5\cdot 10^4 N/C$ is the electric field

$q=-2.0\cdot 10^{-6}C$ is the charge

Therefore, the force experienced by this charge is

$F=(-2.0\cdot 10^{-6})(4.5\cdot 10^4)=-0.09 N$

And the negative sign means that the force is attractive: in fact, the charge is negative, while the charge that is producing the field is positive (because the electric field points outward), therefore the negative charge is attracted towards the positive charge that is generating the field.

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