This problem investigates what fraction of the available charge must be transferred from one conductor to another to produce a typical contact potential. (a) As a rough approximation treat the conductors as$\mathbf{10}\mathbf{}\mathbf{\text{cm}}\mathit{x}$ 10 cm square plates$\mathbf{2}\mathbf{}\mathbf{\text{cm}}$ apart-a parallel-plate capacitors so that$\mathit{q}\mathbf{=}\mathit{C}\mathit{V}$ , where $\mathit{C}\mathbf{=}{\mathit{\sigma }}_{\mathbf{0}}\mathbf{(}\mathbf{0}\mathbf{.01}\mathbf{}{\mathbf{\text{m}}}^{\mathbf{2}}\mathbf{/}\mathbf{0}\mathbf{.02}\mathbf{}\mathbf{\text{m}}\mathbf{)}$. How much charge must be transferred from one plate to the other to produce a potential difference of $\mathbf{2}\mathbf{}\mathbf{\text{V}}$?(b) Approximately what fraction would this be of the total number of conduction electrons in a $\mathbf{100}\mathbf{}\mathbf{\text{g}}$piece of copper. which has one conduction electron per atom?

The equation for the total charge $\mathit{Q}$on the plates of a capacitor given a potential$\mathit{V}$:

$\mathit{Q}\mathbf{=}\mathit{C}\mathit{V}$

……. (1)

Here$\mathit{C}$is the capacitance of a $\mathbf{(}\mathbf{10}\mathbf{}\mathbf{\text{cm}}\mathbf{)}\mathbf{(}\mathbf{10}\mathbf{}\mathbf{\text{cm}}\mathbf{)}$parallel-plate capacitor (with a plate separation of $\mathbf{2}\mathbf{}\mathbf{\text{cm}}\mathbf{)}$, given by:

$\mathit{C}\mathbf{=}{\mathit{\epsilon }}_{\mathbf{0}}\mathbf{\left(}\frac{\mathbf{0}\mathbf{.01}{\mathbf{m}}^{\mathbf{2}}}{\mathbf{0}\mathbf{.02}\mathbf{m}}\mathbf{\right)}$ …… (2)

The number of electrons$N$that would give rise to the total charge$\mathit{Q}$would be written as:

$\mathit{N}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{e}}$

With e being the elementary unit charge on the electron. So then the $8.85\times {10}^{-12}C$is used for the$Q$ and $1.6\times {10}^{-19}$C for e:

$\begin{array}{c}N=\frac{Q}{e}\\ =\frac{(8.85\times {10}^{-12}\text{C})}{1.6\times {10}^{-19}\text{C}}\\ =5.53\times {10}^7\end{array}$

So that's the number of electrons needed to give rise to the$2\text{V}$

And${\diamond }_0$ is the permittivity of free space.

The atomic mass for copper of 63.546. Convert them to standard units:

$\left(63.546u\right)\left(\frac{1.66\times {10}^{-27}\text{kg}}{1\text{u}}\right)=1.055\times {10}^{-25}\text{kg}$

(a)

Potential difference across capacitor$=2\text{V}$

To find the total charge involved to produce a potential of$2\text{V}$ , equation (2) is inserted in for$C$ in equation (1) , with${\epsilon }_0$ being $8.85\times {10}^{-12}{C}^2/N.m^2$and Vbeing$2\text{V}$ , with${\epsilon }_0$ being $8.85\times {10}^{-12}{C}^2/N.m^2$and Vbeing$2\text{V}$ :

$\begin{array}{c}Q=CV\\ ={\epsilon }_0\left(\frac{0.01{\text{m}}^2}{0.02m}\right)\left(V\right)\\ =(8.85\times {10}^{-12}\frac{C^2}{N\cdot m^2})\left(\frac{0.01{\text{m}}^2}{0.02m}\right)(2\text{V})\\ =8.85\times {10}^{-12}\text{C}\end{array}$

(b)

Next, since there is one conduction electron per copper atom, the number of the electrons in the $100\text{g}$sample can be found by just giving a look how many copper atoms it takes to make up the $100\text{g}$. it can be expressed with the equation:

$n=\frac{m}{amu}$

With being the number of electrons or atoms, the mass of the sample, and being the atomic mass of the copper atom. So then$0.1\text{kg}$ is inserted in for the (since standard units are needed), and$1.055*{10}^{-25}$

$\text{kg}$for the amu:

$\begin{array}{c}n=\frac{m}{amu}\\ =\frac{(0.1\text{kg})}{(1.055\times {10}^{-25}\text{kg})}\\ =9.48\times {10}^{23}\end{array}$

So, there are$9.48\times {10}^{23}$ atoms/conduction electrons in the$100\text{g}$ sample.

So then just divide the two of them to get the ratio:

$\begin{array}{c}R=\frac{N}{n}\\ =\frac{(5.53\times {10}^7)}{(9.48\times {10}^{23})}\\ =5.83\times {10}^{-17}\end{array}$

The ratio of the two sets of electrons is$R=6\times {10}^{-17}$