A small cork with an excess charge of +6.0 μC (1 μC = 10 -6 C) is placed 0.12 m from another cork, which carries a charge of -4.3 μC. a. Wha

Question

A small cork with an excess charge of +6.0 μC (1 μC = 10 -6 C) is placed 0.12 m from another cork, which carries a charge of -4.3 μC. a. What is the magnitude of the electric force between the corks? b. Is this force attractive or repulsive? c. How many excess electrons are on the negative cork? d. How many electrons has the positive cork lost?

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bonexptip 3 years 2021-09-02T19:43:29+00:00 1 Answers 7 views 0

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  1. Orla
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    2021-09-02T19:45:09+00:00
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    Answer:

    a.16.125 N b. The force is an attractive force. c. 2.68 × 10¹³ electrons d. 3.75 × 10¹³ electrons

    Explanation:

    a. What is the magnitude of the electric force between the corks?

    The electrostatic force of attraction between the two corks is given by

    F = kq₁q₂/r² where k = 9 × 10⁹ Nm²/C², q₁ = +6.0 μC = +6.0 × 10⁻⁶ C, q₂ = -4.3 μC = -4.3 × 10⁻⁶ C and r = distance between the corks = 0.12 m

    Substituting the values of the variables into the equation, we have

    F = kq₁q₂/r²

    F = 9 × 10⁹ Nm²/C² × +6.0 × 10⁻⁶ C × -4.3 × 10⁻⁶ C/(0.12 m)²

    = -232.2 × 10⁻³ Nm²/(0.0144 m)²

    = -16125 × 10⁻³ N

    = -16.125 N

    So, the magnitude of the force is 16.125 N

    b. Is this force attractive or repulsive?

    Since the direction of the force is negative, it is directed towards the positively charged cork, so the force is an attractive force.

    c. How many excess electrons are on the negative cork?

    Since Q = ne where Q = charge on negative cork = -4.3 μC = -4.3 × 10⁻⁶ C and n = number of excess electrons and e = electron charge = -1.602 × 10⁻¹⁹ C

    So n = Q/e = -4.3 × 10⁻⁶ C/-1.602 × 10⁻¹⁹ C = 2.68 × 10¹³ electrons

    d. How many electrons has the positive cork lost?

    We need to first find the number of excess positive charge n’

    Q’ = n’q where Q = charge on positive cork = + 6.0 μC = + 6.0 × 10⁻⁶ C and n = number of excess protons and q = proton charge = +1.602 × 10⁻¹⁹ C

    So n’ = Q’/q = +6.0 × 10⁻⁶ C/+1.602 × 10⁻¹⁹ C = 3.75 × 10¹³ protons

    To maintain a positive charge, the number of excess protons equals the number of electrons lost = 3.75 × 10¹³ electrons

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