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Two electrons are separated by a distance of

3.00 × 10^−6 meter. What are the magnitude and

direction of the electrostatic forces each exerts

on the other?

(1) 2.56 × 10^−17 N away from each other

(2) 2.56 × 10^−17 N toward each other

(3) 7.67 × 10^−23 N away from each other

(4) 7.67 × 10^−23 N toward each other

Respuesta :

JhoanEusse

Answer:

(1) [tex]2.56\times10^{-17} N[/tex] away from each other.

Explanation:

Two charged particles interact due an electrostatic force each of them feels due the other one, and the magnitude of that electric force is:

[tex] F_{e}=k\frac{\mid q_{1}q_{2}\mid}{r^{2}}[/tex]

With q1, q2 the charges of the particles, k the Coulomb's constant and r the distance between the particles. The force is the same on each of the particles due Newton's third law. Then the force is:

[tex] F_{e}=(9.0\times10^{9})\frac{\mid e *e\mid}{(3.00\times10^{-6})^{2}}=(9.0\times10^{9})\frac{\mid (1.6\times10^{-19})^2\mid}{(3.00\times10^{-6})^{2}}[/tex]

[tex] F_{e}=2.56\times10^{-17} N[/tex]

The direction of the force is given by the sign of the charges, if the charges have the same sign the force between them is repulsive, and if the charges are opposite the force between the are attractive. Because we have two electrons and both have the same negative charge, the force between them is repulsive so the force is pointing away from each other.

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