91Ó°ÊÓ

The density of the conduction electrons in the copper plates is,$8.49×{10}^{28}\mathrm{electrons}/{\mathrm{m}}^3$

Fig.25-26 of plot d vs V.

Using the formula for the charge on the surface of the plate for the given and equation 25-1, find the ratio C/A .

Formulae are as follows:

$\mathbf{C}\mathbf{=}\frac{\mathbf{q}}{\mathbf{v}}$

Where Cis capacitance, V is the potential difference, and q is a charge on the particle.

For a given potential difference , the charge on the surface of the plate is,

q = Ne

q = (nAd)e

$\mathrm{A}=\frac{\mathrm{q}}{\mathrm{ned}}$

Where, d is the depth from which the electron comes into the plate, and is the density of conduction electrons. The charge collected on the plates is related to the capacitance and potential difference by (equation 25-1) as

q = CV

$\mathrm{C}=\frac{\mathrm{q}}{\mathrm{v}}$

Combining the above two expressions, we get,

$\frac{\mathrm{C}}{\mathrm{A}}=\mathrm{ne}\frac{\mathrm{d}}{\mathrm{v}}$

Withrole="math" localid="1661338776090" $\frac{\mathrm{d}}{\mathrm{V}}=\frac{{\mathrm{d}}_{\mathrm{s}}}{{\mathrm{V}}_{\mathrm{s}}}$ , this gives,

$\frac{\mathrm{d}}{\mathrm{V}}=\mathrm{ne}\frac{{\mathrm{d}}_{\mathrm{s}}}{{\mathrm{V}}_{\mathrm{s}}}$

$$\begin{gathered} \frac{\mathrm{C}}{\mathrm{A}}=\mathrm{ne}\frac{{\mathrm{d}}_{\mathrm{s}}}{{\mathrm{V}}_{\mathrm{s}}} \\ \frac{\mathrm{C}}{\mathrm{A}}=\left(8.49×{10}^{28}\mathrm{electrons}/{\mathrm{m}}^3\right)×\left(1.6×{10}^{-19}\mathrm{C}\right)×\left(\frac{1.00×{10}^{-12}\mathrm{m}}{20.0\mathrm{V}}\right) \\ =6.79×{10}^{-4}\frac{\mathrm{F}}{{\mathrm{m}}^2} \end{gathered}$$

Hence, The C/A ratio is role="math" localid="1661338918521" $6.79×{10}^{-4}\mathrm{F}/{\mathrm{m}}^2$