How many electrons must be removed from an electrically neutral silver dollar to give it a charge of +3.5×10^-6C?

Question

How many electrons must be removed from an electrically neutral silver dollar to give it a charge of +3.5×10^-6C?

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Phúc Điền 1 year 2021-09-04T12:51:48+00:00 1 Answers 13 views 0

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    2021-09-04T12:53:02+00:00

    Answer:

    [tex]2.19\cdot 10^{13}[/tex]

    Explanation:

    The electron is a fundamental particle, with a charge of

    [tex]e=-1.6\cdot 10^{-19}C[/tex]

    which is also known as fundamental charge.

    For an object having N excess electrons, the total charge on the object is

    [tex]Q=Ne[/tex] (1)

    where e is the charge of the electron.

    For the object in this problem, its charge is

    [tex]Q=+3.5\cdot 10^{-6} C[/tex]

    This can be obtained by removing a negative charge equal to

    [tex]Q=-3.5\cdot 10^{-6}C[/tex]

    Substituting into (1) and solving for N, we can find the number of electrons:

    [tex]N=\frac{Q}{e}=\frac{-3.5\cdot 10^{-6}}{-1.6\cdot 10^{-19}}=2.19\cdot 10^{13}[/tex]

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