Question

Switch S is closed to connect the uncharged capacitor of capacitance C=0.25mf to the battery of potential difference V=12V. The lower capacitor plate has thickness L= 0.5cm and face area A= 2*10–⁴, and it consist of copper, in which the density of conduction electrons is n= 8.49*10²⁸electrons/m³. From what depth d the plate must electrons move to the plate face as the capacitor becomes charged?

Answer 1

Charge of an electron $(e)=1.6×10^{-19}C$

For a charge Q, the number of electrons

$=\frac{Q}{e}$

But $Q=CV$

$C=0.25×10^{-6}F$

$V=12V$

$\therefore\>Q=0. 25×10^{-6}×12$

$=3×10^{-6}C$

Number of electrons $=\frac{3×10^{-6}}{1.6×10^{-19}}=1.875×10^{13}$ Electrons

Number of electrons $=$ Density of conduction of electrons $×$ Volume

$8.49×10^{28}×2×10^{-4}×d=1.875×10^{13}$

$d=1.104×10^{-12}M$