Physics - EM Waves Question with Solution | TestHub

$qE =m\mathrm{g}$...(i)

$6\pi \mathrm{\eta } rv=m\mathrm{g}\mathrm{ }\mathrm{ }\mathrm{ }$

$\frac{4}{3}\pi r^3 \mathrm{\rho }\mathrm{g}=m\mathrm{g}$...(ii)

$\therefore r={\left(\frac{3m\mathrm{g}}{4\mathrm{\pi }\mathrm{\rho }\mathrm{g}}\right)}^{1/3}$...(iii)

Substituting the value of r in Eq. (ii), we get

$6\pi \mathrm{\eta }v{\left(\frac{3m\mathrm{g}}{4\mathrm{\pi }\mathrm{\rho }\mathrm{g}}\right)}^{1/3}=m\mathrm{g}$

or${\left(6\pi \mathrm{\eta }v\right)}^3 \left(\frac{3m\mathrm{g}}{4\mathrm{\pi }\mathrm{\rho }\mathrm{g}}\right)\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }={\left(m\mathrm{g}\right)}^3$

Again substituting$m\mathrm{g}=\mathrm{q}\mathrm{E},\mathrm{ }$we get

${\left(qE\right)}^2=\left(\frac{3}{4\pi \mathrm{\rho }\mathrm{g}}\right){\left(6\pi \mathrm{\eta }v\right)}^3$

Or$qE={\left(\frac{3}{4\pi \mathrm{\rho }\mathrm{g}}\right)}^{1/2} {\left(6\pi \mathrm{\eta }\mathrm{v}\right)}^{3/2}$

$\therefore q=\frac{1}{E}{\left(\frac{3}{4\pi \mathrm{\rho }\mathrm{g}}\right)}^{\frac{1}{2}}{\left(6\pi \mathrm{\eta }v\right)}^{3/2}$

Substituting the values, we get

$q=\frac{7}{81\pi \times {10}^5}\sqrt{\frac{3}{4\pi \times 900\times 9.8}\times 216{\pi }^3}$

$$\begin{gathered} \times \sqrt{{\left(1.8\times {10}^{-5}\times 2\times {10}^{-3}\right)}^3 } \\ =8.0\times {10}^{-19} \mathrm{C} \end{gathered}$$