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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1. 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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