How many electrons would have to be removed from a copper penny to leave it with a positive charge of $10^{- 7} C$ ?

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Solution

$1.6 \times 10^{- 19} C c h a r g e = 1 e l e c t r o n$
$1 C c h a r g e = \frac{1}{1.6 \times 10^{- 19}} e l e c t r o n$
$= 6.25 \times 10^{18} e l e c t r o n$
So, $10^{- 7} C c h a r g e = 6.25 \times 10^{18} \times 10^{- 7} e l e c t r o n$
$= 6.25 \times 10^{11} e l e c t r o n s$ .

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How many electrons would have to be removed from a copper penny to leave it with a positive charge of 10−7C?

1.6×10−19C charge=1 electron1 C charge=11.6×10−19electron=6.25×1018electronSo, 10−7C charge=6.25×1018×10−7electron=6.25×1011electrons.

How many electrons would have to be removed from a copper penny to leave it with a positive charge of $10^{- 7} C$ ?

$1.6 \times 10^{- 19} C c h a r g e = 1 e l e c t r o n$
$1 C c h a r g e = \frac{1}{1.6 \times 10^{- 19}} e l e c t r o n$
$= 6.25 \times 10^{18} e l e c t r o n$
So, $10^{- 7} C c h a r g e = 6.25 \times 10^{18} \times 10^{- 7} e l e c t r o n$
$= 6.25 \times 10^{11} e l e c t r o n s$ .