A sphere has a negative charge of $6.4 \times 10^{- 7}$ $6.4 xx 10^-7$ coulombs. Approximately how many electrons must be removed to make the sphere neutral?

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A sphere has a negative charge of $6.4 \times 10^{- 7}$ $6.4 xx 10^-7$ coulombs. Approximately how many electrons must be removed to make the sphere neutral?

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Answer 1 · 2017-03-15T01:10:20

I got $4 \times 10^{12}$ $4xx10^12$ electrons

Explanation:

You know that one electron has a charge of $1.6 \times 10^{- 19} C$ $1.6xx10^-19C$ so we need to remove:
$\frac{6.4 \times 10^{- 7}}{1.6 \times 10^{- 19}} = 4 \times 10^{12}$ $(6.4xx10^(-7))/(1.6xx10^(-19))=4xx10^12$ electrons

Answer 2 · 2017-03-15T01:13:59

$4 \cdot 10^{12}$ $4*10^12$ electrons

Explanation:

Since one electron has a charge of $1.6 \cdot 10^{- 19} C$ $1.6*10^-19C$ we can simply take the overall charge of the sphere and divide it by the charge of a single electron to see how many electrons account for the overall charge. It would look like

$\frac{6.4 \cdot 10^{- 4} C}{1.6 \cdot 10^{- 19} C}$ $(6.4*10^-4C)/(1.6*10^-19C)$

Which would give us $4 \cdot 10^{12}$ $4*10^12$ electrons

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A sphere has a negative charge of $6.4 \times 10^{- 7}$ $6.4 xx 10^-7$ coulombs. Approximately how many electrons must be removed to make the sphere neutral?

2 Answers

Explanation:

Explanation:

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A sphere has a negative charge of $6.4 \times 10^{- 7}$ $6.4 xx 10^-7$ coulombs. Approximately how many electrons must be removed to make the sphere neutral?