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

Mar 15, 2017

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

#### Explanation:

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

Mar 15, 2017

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

#### Explanation:

Since one electron has a charge of $1.6 \cdot {10}^{-} 19 C$ 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

$\frac{6.4 \cdot {10}^{-} 4 C}{1.6 \cdot {10}^{-} 19 C}$

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