Question

A 2 microcoulomb charge is placed at a distance of 0.25 m away from a 3.6 microcoulomb charge. Describe the type of electrostatic force between the charges and calculate the magnitude of the electrostatic force between them.

Answers

  1. Answer: 1.04N

    Explanation:

    Given

    q1 = 2*10^-6C

    q2 = 3.6*10^-6C

    r = 0.25m

    k = 9*10^9

    Magnitude of electrostatic force can be calculated by using coulomb’s law. Coulomb’s law states that, “the magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.”

    F =(kq1q2) / r²

    F = (9*10^9 * 2*10^-6 * 3.6*10^-6) / 0.25²

    F = 0.0648/0.0625

    F = 1.04N

    The type of electrostatic force between the charges is the repulsive force

  2. Answer:

    The force between the charge is a repulsive force

    The magnitude of the electrostatic force between them = 1.0368 N

    Explanation:

    Since both charges are positive charge, and they have the same sign, from the law of electrostatics, The type of force between the charges is Repulsive force.

    Using

    F = kqq’/r²…………………….. Equation 1

    Where F = Force between the charges, q = first charge, q’ = second charge, r = distance between the charges, k = coulombs constant.

    Given: q = 2 μC = 2×10⁻⁶ C, q’ = 3.6 μC = 3.6×10⁻⁶ C, r = 0.25 m, k = 9×10⁹ Nm²/C²

    Substitute into equation 1

    F = 9×10⁹(2×10⁻⁶)(3.6×10⁻⁶)/0.25²

    F = 64.8×10⁻³/0.0625

    F = 1036.8×10⁻³

    F = 1.0368 N

    Hence the magnitude of the electrostatic force between them = 1.0368 N

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