A particle with charge −5 µC is located on the x-axis at the point 10 cm , and a second particle with charge 6 µC is placed on t

Question

A particle with charge −5 µC is located on
the x-axis at the point 10 cm , and a second
particle with charge 6 µC is placed on the
x-axis at −8 cm . What is the magnitude of the total electrostatic force on a third particle with charge
2 µC placed on the x-axis at −2 cm ? The
Coulomb constant is 8.9875 × 109 N · m2
Answer in units of N.

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Thành Công 5 years 2021-07-31T12:49:19+00:00 1 Answers 28 views 0

Answers ( )

  1. diemkieu
    0
    2021-07-31T12:50:30+00:00
    Reply

    Answer:

    36.25 N

    Explanation:

    The magnitude of the electrostatic force between two charges is given by Coulomb’s law:

    F=k\frac{q_1 q_2}{r^2}

    where:

    k=9\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb’s constant

    q_1, q_2 are the magnitude of the two charges

    r is the separation between the two charges

    Moreover:

    – The force is repulsive if the two  charges have same sign

    – The force is attractive if the two charges have opposite sign

    In this problem, we have 3 charges:

    q_1=-5\mu C = -5\cdot 10^{-6}C is the charge located at x=+10 cm = +0.10 m

    q_2=+6\mu C=+6\cdot 10^{-6}C is the charge located at x=-8 cm =-0.08 m

    q_3=+2\mu C=+2\cdot 10^{-6}C is the charge located at x=-2 cm=-0.02 m

    The force between charge 1 and charge 3 is:

    F_{13}=\frac{kq_1 q_3}{(x_1-x_3)^2}=\frac{(9\cdot 10^9)(5\cdot 10^{-6})(2\cdot 10^{-6})}{(0.10-(-0.02))^2}=6.25 N

    And since the two charges have opposite sign, the force is attractive, so the force on charge 3 is to the right (towards charge 1).

    The force between charge 2 and charge 3 is:

    F_{23}=\frac{kq_2 q_3}{(x_2-x_3)^2}=\frac{(9\cdot 10^9)(6\cdot 10^{-6})(2\cdot 10^{-6})}{(-0.08-(-0.02))^2}=30.0 N

    And since the two charges have same sign, the force is repulsive, so the force on charge 3 is to the right (away from charge 2).

    So the two forces on charge 3 have same direction (to the right), so the net force is the sum of the two forces:

    F=F_{13}+F_{23}=6.25+30.0=36.25 N

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