Two identical 9.60-g metal spheres (small enough to be treated as particles) are hung from separate 500-mm strings attached to the same nail

Question

Two identical 9.60-g metal spheres (small enough to be treated as particles) are hung from separate 500-mm strings attached to the same nail in a ceiling. Surplus electrons are added to each sphere, and then the spheres are brought in contact with each other and released. Their equilibrium position is such that each string makes a 17.0 ∘ angle with the vertical.How many surplus electrons are on each sphere?

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Ladonna 4 years 2021-07-16T19:32:23+00:00 1 Answers 5 views 0

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  1. ngocdiep
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    2021-07-16T19:33:51+00:00
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    Answer:

    1.349 × 10¹² electrons

    Explanation:

    At equilibrium, the tension T in the string is resolved into horizontal and vertical components. Tcos17 being the vertical component and Tsin17 the horizontal component. The electrostatic force of repulsion at equilibrium, F = kq²/r² where q = excess charge on sphere and r = distance apart at equilibrium,F acts horizontally to the left and its weight, mg acts vertically downwards.

    For equilibrium, sum of horizontal components = 0 and  sum of horizontal components = 0. So, Tsin17 – F = 0

    Tsin17 = F ⇒ Tsin17 = kq²/r² (1)

    Also Tcos17 – mg = 0  ⇒ Tcos17 = mg (2)

    Dividing (1) by (2), we have

    Tsin17/Tcos17 = kq²/r² ÷ mg

    tan17 = kq²/mgr²

    q = r√(mgtan17/k)

    We find r using cosine rule r = √(500² + 500² -2 × 500²cos 2 × 17) since the string and the masses form an isosceles triangle at equilibrium.

    r = √(2 × 500² -2 × 500²cos34) = 500√2(1 – cos 34) = 500√2 × 0.1710 =120.89 mm = 0.1209 m

    substituting m = 9.60 g = 9.6 × 10⁻³ g, k = 9 × 10⁹ Nm²/C² and r into q, we have,

    q = r√(mgtan17/k)

      = 0.1209 m√(9.6 × 10⁻³ g × 9.8 m/s² × tan17/9 × 10⁹ Nm²/C²)

      = 0.1209 m√(28.76 × 10⁻³ N/9 × 10⁹ Nm²/C²)

      =  0.1209 m√(3.195 × 10⁻¹² C²/m²)

      =  0.1209 m × 1.7877 × 10⁻⁶ C/m

      = 0.2161 × 10⁻⁶ C

      = 2.161 × 10⁻⁷ C

    To find the number of surplus electrons, n on each sphere, we divide q by e the electron charge.

    So, n = q/e = 2.161 × 10⁻⁷ C ÷ 1.602 × 10⁻¹⁹ C = 0.6825 × 10¹² = 1.349 × 10¹² electrons

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