A charge q1 of -5.00 × 10‐⁹ C and a charge q2 of -2.00 × 10‐⁹ C are separated by a distance of 40.0 cm. Find the equilibrium position for a

Question

A charge q1 of -5.00 × 10‐⁹ C and a charge q2 of -2.00 × 10‐⁹ C are separated by a distance of 40.0 cm. Find the equilibrium position for a third charge of +15.0 × 10‐⁹ C.

P.S. 10-⁹ is 10^-9

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Thái Dương 1 year 2021-09-03T04:34:16+00:00 1 Answers 16 views 0

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    0
    2021-09-03T04:35:27+00:00

    Answer:

    Explanation:

    The equilibrium position will be in between q₁ and q₂ .

    Let this position of third charge be x distance from q₁

    q₁ will pull the third charge towards it with force F₁

    F₁ =9 x 10⁹x 5 x 10⁻⁹x15 x 10⁻⁹ / x²

    = 675 x 10⁻⁹ / x²

    q₂ will pull the third charge towards it with force F₂

    F₂ =9 x 10⁹x 2 x 10⁻⁹x15 x 10⁻⁹ /( .40-x )²

    = 270 x 10⁻⁹ / ( .40-x )²

    For equilibrium

    675 x 10⁻⁹ / x² = 270 x 10⁻⁹ / ( .40-x )²

    5  / x² = 2 / ( .40-x )²

    ( .40-x )² / x² = 2/5 = .4

    .4 – x / x = .632

    .4 – x = .632x

    .4 = 1.632 x

    x = .245 .

    24.5 cm

    so third charge must be placed at 24.5 cm away from q₁ charge.

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )