Two point charges, A and B, are separated by a distance of 19.0 cm . The magnitude of the charge on A is twice that of the charge on B. If e

Question

Two point charges, A and B, are separated by a distance of 19.0 cm . The magnitude of the charge on A is twice that of the charge on B. If each charge exerts a force of magnitude 46.0 N on the other, find the magnitudes of the charges

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Kim Cúc 1 week 2021-07-22T00:04:14+00:00 1 Answers 0 views 0

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    2021-07-22T00:05:40+00:00

    Answer:

    The charge on point A is q_a = 2.4 × 10^{-5} C

    The charge on point B is q_b = 1.2 × 10^{-5} C

    Explanation:

    Given data

    Distance (r) = 0.19 m

    Magnitude of the charge on A is twice that of the charge on B  i.e.

    q_A = 2 q_B

    F = 46 N

    We know that force between the charges is given by

    F = \frac{k Q_A Q_B}{r^{2} }

    65 = \frac{(9) (10^{9}) (2q_b^{2} ) )}{0.2^{2} }

    q_b = 1.2 × 10^{-5} C

    q_a = 2 × q_b

    q_a = 2 × 1.2 × 10^{-5}

    q_a = 2.4 × 10^{-5} C

    Therefore the charge on point A is q_a = 2.4 × 10^{-5} C

    The charge on point B is q_b = 1.2 × 10^{-5} C

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