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

$$2.19\cdot 10^{13}$$

Explanation:

The electron is a fundamental particle, with a charge of

$$e=-1.6\cdot 10^{-19}C$$

which is also known as fundamental charge.

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

$$Q=Ne$$ (1)

where e is the charge of the electron.

For the object in this problem, its charge is

$$Q=+3.5\cdot 10^{-6} C$$

This can be obtained by removing a negative charge equal to

$$Q=-3.5\cdot 10^{-6}C$$

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

$$N=\frac{Q}{e}=\frac{-3.5\cdot 10^{-6}}{-1.6\cdot 10^{-19}}=2.19\cdot 10^{13}$$