Chapter 16, Problem 57GP

To determine

To find: Number of excess electron.

Answer to Problem 57GP

  ${10}^7\text{electrons}$

Explanation of Solution

Given:

Electric field of the earth = $150\text{N/C}$

Radius of water droplet $=0.018\text{mm}=1.8\times {10}^{-5}\text{m}$

Formula used:

For the droplet to remain stationary, the magnitude of the electric force must be the same as the weight of the droplet which is given by,

  ${\mathrm{F}}_{\mathrm{E}}=\mathrm{q}\mathrm{E}=\mathrm{m}\mathrm{g}$

  $\begin{array}{l}\Rightarrow \mathrm{q}\mathrm{E}=\mathrm{m}\mathrm{g}\\ \Rightarrow (\mathrm{N}\mathrm{e})\mathrm{E}=\left[\left(\frac{4\mathrm{\pi }{\mathrm{r}}^3}{3}\right)\mathrm{e}\right]\mathrm{g}\end{array}$

  $\begin{array}{l}\text{Where,}\\ \mathrm{\rho }\text{= density of droplet}\end{array}$

r = radius of droplet

e = charge of electron

So number of electron is given by: $\mathrm{N}=\frac{4\mathrm{\pi }{\mathrm{r}}^3\mathrm{e}\mathrm{g}}{3\mathrm{e}\mathrm{E}}$

Calculation:

The radius of water $(\mathrm{r})=0.018\text{mm}$

  $\begin{array}{l}=0.018\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\hfill \\ =1.8\times {10}^{-5}\text{m}\hfill \end{array}$

Electric field E=150N/c

The density of water $\mathrm{\rho }=1.00\times {10}^3\text{g}/\text{cc}$

Acceleration due to gravity $\mathrm{g}=9.8\text{m}/{\text{s}}^2$

The magnitude of electric force on the droplet, ${\mathrm{F}}_{\mathrm{E}}=\left|\mathrm{q}\right|\mathrm{E}\dots$ (1)

Here charge on the water droplets is q and E is the electric field.

The weight of the water droplet (w)=mg

Here mass of the droplet $(\mathrm{m})=\frac{4}{3}\mathrm{\pi }{\mathrm{r}}^3\mathrm{\rho }$

Therefore weight of the droplet $(\mathrm{w})=\left(\frac{4}{3}\mathrm{\pi }{\mathrm{r}}^3\mathrm{\rho }\right)\mathrm{g}\dots .$ (2)

Let the number of excess electrons on the water droplet = n

The charge on the droplets $\mathrm{q}=\mathrm{n}\mathrm{e}$ ; here charge of the electron $\mathrm{e}=1.6\times {10}^{-19}\text{C}$

For the droplet to remain stationary, the magnitude of the electric force on the droplet must be the same as the weight of the droplets. Therefore from equation (1) and (2).

  $\begin{array}{l}\left|\mathrm{q}\right|\mathrm{E}=\mathrm{m}\mathrm{g}\\ \mathrm{n}\mathrm{e}\mathrm{E}=\frac{4}{3}\mathrm{\pi }{\mathrm{r}}^3\mathrm{\rho }\mathrm{g}\\ \mathrm{n}=\frac{4\mathrm{\pi }{\mathrm{r}}^3\mathrm{\rho }\mathrm{g}}{3\mathrm{e}\mathrm{E}}\end{array}$

Substituting the values,

  $\begin{array}{l}\mathrm{n}=\frac{4\mathrm{\pi }{\left(1.8\times {10}^{-5}\text{m}\right)}^3\left(1.00\times {10}^3\text{kg}/{\text{m}}^3\right)\left(9.8\text{m}/{\text{s}}^2\right)}{3\left(1.602\times {10}^{-19}\text{C}\right)(150\text{N}/\text{C})}\text{electrons}\\ \begin{array}{l}=9.96\times {10}^6\text{electrons}\hfill \\ \approx 1.0\times {10}^7\text{electrons}\hfill \end{array}\end{array}$

Conclusion:

The number of electrons water droplet contains is $1.0\times {10}^7\text{electrons}$ .