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11 August, 15:10

A thundercloud has an electric charge of 43.2 C near the top of the cloud and - 38.7 C near the bottom of the cloud. The magnitude of the electric force between these two charges is 3.95 x 106 N. What is the average separation between these charges? (kc = 8.99 * 109 N • m2 / C2)

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Answers (2)
  1. 11 August, 15:18
    0
    1950.65m

    Explanation:

    Coulomb's law states that the force F of attraction between two charges q1 and q2 is directly proportional to the product of the charges and inversely proportional to the square of the distance d between them. Mathematically;

    |F| = k |q1||q2| / r²

    Given |q1| = |43.2| C = 43.2C

    |q2| = |-38.7|C = 38.7C

    F = 3.95 x 106 N

    d = ?

    k is the coloumbs constant = 8.99 * 109 Nm2 / C2

    Substituting this values in the formula to get distance d we have;

    3.95x10^6 = 8.99*10^9*43.2*38.7/d²

    3.95x10^6d² = 15,029,841,600,000

    d² = 15,029,841,600,000/3,950,000

    d² = 3,805,023.189

    d = √3,805,023.189

    d = 1950.65m

    Average separation between the charges is 1950.65m
  2. 11 August, 15:23
    0
    1950.65 m

    Explanation:

    Parameters given:

    First charge, Q1 = 43.2 C

    Second charge, Q2 = - 38.7 C

    Magnitude of force between them, F = 3.95 * 10^6 N

    The magnitude of the electric force between two electric charges at a distance r from one another is given as:

    |F| = |k * Q1 * Q2|/r²

    3.95 * 10^6 = |8.99 * 10^9 * 43.2 * - 38.7| / r²

    Making r² the subject of the formula and solving:

    r² = 3805025.189

    r = 1950.65 m
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