Question

A typical capacitor in a memory cell have a capacitance of 3 x 10-14 F. If the voltage across the capacitor is 0.5 V, what is the number of electrons that must move to the capacitor to charge it? 

Answer 1

Given:

$C=3\times 10^{-14}\: \rm F$

$V=0.5\: \rm V$

$e=1.6\times 10^{-19}\:\rm C$

The charge on the capacitor plates

$$q=CV=3\times 10^{-14}\times 0.5=1.5\times 10^{-14}\:\rm C$$

The number of electrons

$$N=\frac{q}{e}=\frac{1.5\times 10^{-14}\:\rm C}{1.6\times 10^{-19}\:\rm C}=93750$$