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22 November, 00:40

Suppose a van de Graaff generator builds a negative static charge, and a grounded conductor is placed near enough to it so that a 7.0 x 10-6 coulombs of negative charge arcs to the conductor. Calculate the number of electrons that are transferred.

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Answers (2)
  1. 22 November, 01:23
    0
    Number of electrons transferred = 4.375 x 10^ (13)

    Explanation:

    From coulombs law,

    Mathematically, we can say that a charge is the number of electrons multiplied by the charge on 1 electron. Symbolically, it is given as;

    Q = ne

    where;

    q is the symbol used to represent charge

    n is a number of electrons;

    e is a charge on 1 electron (1.6 * 10^ (-19) C).

    From the question,

    Q = 7.0 x 10^ (-6) coulombs of negative charge

    Thus, making n the subject of the formula;

    n = Q/e = 7.0 x 10^ (-6) / (1.6 * 10^ (-19)) = 4.375 x 10^ (13)
  2. 22 November, 04:26
    0
    Given Information:

    Charge = q = 7.0x10⁻⁶ C

    Required Information:

    Number of electrons = n = ?

    Answer:

    Number of electrons = 4.375x10¹³

    Explanation:

    We know that charge is given by

    q = n*e

    Where n is the number of electrons and e is the electronic charge that is 1.60x10⁻¹⁹ C

    n = q/e

    n = 7.0x10⁻⁶/1.60x10⁻¹⁹

    n = 4.375x10¹³ electrons

    Therefore, 4.375x10¹³ electrons were transferred.
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