Answer:
[tex]n=6.25\times 10^{11}[/tex]
Explanation:
We need to find the number of electrons that would have to be removed from a coin to leave it with a charge of [tex]+10^{-7}\ C[/tex]. Then the number of electrons be n. Using quantization of electric charge as :
q = ne
e is charge on an electron
[tex]n=\dfrac{q}{e}\\\\n=\dfrac{10^{-7}}{1.6\times 10^{-19}}\\\\n=6.25\times 10^{11}[/tex]
So, the number of electrons are [tex]6.25\times 10^{11}[/tex].
