Question

If an uncharged parallel-plate capacitor (capacitance C) is connected to a battery, one plate becomes negatively charged as electrons move to the plate face (area A). In Fig. 25-26, the depth dfrom which the electrons come in the plate in a particular capacitor is plotted against a range of values for the potential difference Vof the battery. The density of conduction electrons in the copper plates is $\mathbf{8}\mathbf{.}\mathbf{49}\mathbf{\times }{\mathbf{10}}^{\mathbf{28}}\mathbf{}\mathbf{electrons}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$. The vertical scale is set by ${\mathbf{d}}_{\mathbf{s}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{pm}$and the horizontal scale is set by ${\mathbf{v}}_{\mathbf{s}}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{V}$What is the ratio C/A?

Short answer

The C/A ratio is$6.79\times {10}^{-4}\mathrm{F}/{\mathrm{m}}^2$

Step-by-step solution

Given data

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

Fig.25-26 of plot d vs V.

Determining the concept

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.

Determining the ratio of  

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,

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

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

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