91Ó°ÊÓ

We are given a downward electric field of the earth E=150N/C that exerted a force on a raindrop with a diameter d=0.020mm – $0.020×{10}^{-3}$m

The drop is affected by two forces, the force due to the electric field and the force of weight, and to keep the drop motionless the force of the electric field must equal to the weight, where the weight of the raindrop could be calculated by knowing its radius

$$\begin{gathered} \mathrm{W}=\mathrm{mg} \\ \mathrm{W}=\mathrm{Òϳղ\mu } \\ \mathrm{W}=\mathrm{ÒÏ}{\left(\frac{4}{3}{\mathrm{Ï€°ù}}^3\right)}^{}\mathrm{g} \end{gathered}$$

Where $\mathrm{ÒÏ}$is the density of water and equal $1000\mathrm{kg}/{\mathrm{m}}^3$, g is the gravitational acceleration and r is the radius of the drop and equal $\mathrm{d}/2=0.010×{10}^{-3}$

As we explain to make the drop motionless, the weight must equal the force exerted by the electrical field of the earth where this force is given by

$\mathrm{F}=\mathrm{qE}$

Where q is the charge that exists on the surface of the drop and E is the electric field on the earth.

Now let equal both equations (1) and (2) to get the charge q

$$\begin{gathered} \mathrm{W}=\left(\mathrm{ÒÏ}\frac{4}{3}{\mathrm{Ï€°ù}}^3\right)\mathrm{g} \\ \mathrm{q}=\frac{4}{3}\frac{{\mathrm{ÒÏÏ€°ù}}^3\mathrm{g}}{\mathrm{E}} \end{gathered}$$

Now we can plug our value for p ,r, g, and E into equation(3) to get q

$$" width="9">q=4Ï€3ÒÏr3gE=4Ï€31000kg/m20.010×10-3m9.8m/s2150N/C=2.74×10-13C

After knowing the charge q on the drop we can obtain the number of the electron ${\mathrm{n}}_{{}^{\mathrm{e}}}$which is given by

${\mathrm{n}}_{{}^{\mathrm{e}}}=\frac{\mathrm{q}}{\mathrm{e}}$

where knowing the charge of the electron equals $1.0602×{10}^{-19}$C/electron. Now we can plug are values for e and q into equation (4) to get excess electrons

${\mathrm{n}}_{\mathrm{e}}=\frac{\mathrm{q}}{\mathrm{e}}$

${\mathrm{n}}_{\mathrm{e}}=\frac{\mathrm{q}}{\mathrm{e}}\frac{2.74×{10}^{-13}\mathrm{C}}{1.602×{10}^{-19}\mathrm{C}/\mathrm{Eelectron}}=171.0×{10}^{-4}\mathrm{electrons}$

The excess electron on the surface of the raindrop is $171.0×{10}^{-4}\mathrm{electrons}$.