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Question Number 52846 by peter frank last updated on 13/Jan/19

Answered by peter frank last updated on 14/Jan/19

Commented by peter frank last updated on 14/Jan/19

a) r=(√((9ηv_1 )/(2(ρ−σ)g)))  b) drop to be stationary  qE=(4/3)πr^3 (ρ−σ)g  E=(v/d)  excess no of electron  n=(q/e)  .....

$$\left.{a}\right)\:{r}=\sqrt{\frac{\mathrm{9}\eta{v}_{\mathrm{1}} }{\mathrm{2}\left(\rho−\sigma\right){g}}} \\ $$$$\left.{b}\right)\:{drop}\:{to}\:{be}\:{stationary} \\ $$$${qE}=\frac{\mathrm{4}}{\mathrm{3}}\pi{r}^{\mathrm{3}} \left(\rho−\sigma\right){g} \\ $$$${E}=\frac{{v}}{{d}} \\ $$$${excess}\:{no}\:{of}\:{electron} \\ $$$${n}=\frac{{q}}{{e}} \\ $$$$..... \\ $$

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