# When $1.0 \times 10^{12}$ electrons are transferred from one conductor to another, a potential difference of $10\; V$ appears between the conductors. Calculate the capacitance of the two conductor system.

$Q= (1.0 \times 10^{12} )(1.6 \times 10^{-19}) C$
$\quad= 1.6 \times 10^{-7} C$
$v= 10 volt$
$C=?$
As $C= \large\frac{Q}{V}$
$C= \large\frac{1.6 \times 10^{-7}}{10}$$=1.6 \times 10^{-8}F$