Question

How many electrons would have to be removed from a coin to leave it with a charge of +2.2 x 10^- 8 C?

Answer 1

`N=1.375*10^(11)` electrons

Step-by-step explanation:

The total charge Q+ at the coin is equal, but with opposite sign, to the charge negative Q- removed of it. Q- is the sum of the charge of the N electrons removed from the coin:

`Q_(-)=N*q_(e)`

`Q_(-)=-Q_(+)`

q_{e}=-1.6*10^{-19}C charge of a electron

We solve to find N:

`N=-Q_(+)/q_(e)=-2.2*10^(-8)/(-1.6*10^(-19))=1.375*10^(11)`