Question

How many electrons would have to be removed from a coin to leave it with a charge of \[ + 1.6 \times {10^{ - 7}}C\] ?

Answer 1

Hint: To solve such question we should know about:
Charge of an electron is $ - 1.6 \times {10^{ - 19}}C$ . So, to make anything positive charge or negative charge we add or remove electrons from it. In this question we first try to understand we have made our coin positive charged. So, we will remove electrons from it.

Complete answer:
As given in the question:
We have to make a coin that is positively charged which has \[ + 1.6 \times {10^{ - 7}}C\] charge.
So, that means we have to remove that quantity of electrons so that it will become positively charged.
As we know, the charge of an electron is $ - 1.6 \times {10^{ - 19}}C$ .
We have to remove \[ - 1.6 \times {10^{ - 7}}C\] amount of electrons.
So, number of electron to be removed $ = ne$
$Q = ne$
Keeping value in it. We get,
$ \Rightarrow - 1.6 \times {10^{ - 7}}C = n \times - 1.6 \times {10^{ - 19}}C$
n is the number of electrons to be removed.
Calculate value of n,
$ \Rightarrow \dfrac{{1.6 \times {{10}^{ - 7}}C}}{{1.6 \times {{10}^{ - 19}}C}} = n$
$ \Rightarrow n = {10^{12}}$
So, we have to remove the $n = {10^{12}}$ number of electrons. To make it positive charge \[ + 1.6 \times {10^{ - 7}}C\] .

Note:
Because every electron in an atom possesses a negative charge, it can behave as our charge carrier. We can create electricity by releasing an electron from an atom and forcing it to move.
Consider the atomic model of a copper atom, which is one of the most commonly used elements for charge flow. Copper contains 29 protons in its nucleus and an equal amount of electrons orbiting around it in its balanced form. Electrons orbit the nucleus of the atom at different distances. Electrons in near orbits feel a higher attraction to the nucleus than those in far orbits. The valence electrons are the atom's exterior electrons, and they require the least amount of force to be released.