Menu Close

Question-52845




Question Number 52845 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
absence of voltage  mg=6ηπrv_1 ......(i)  presence of applied voltage  qE−mg=6πηrv_2 .....(ii)  divide qn ii by i  ((qE)/(mg))−1=(v_2 /v_1 )
$${absence}\:{of}\:{voltage} \\ $$$${mg}=\mathrm{6}\eta\pi{rv}_{\mathrm{1}} ……\left({i}\right) \\ $$$${presence}\:{of}\:{applied}\:{voltage} \\ $$$${qE}−{mg}=\mathrm{6}\pi\eta{rv}_{\mathrm{2}} …..\left({ii}\right) \\ $$$${divide}\:{qn}\:{ii}\:{by}\:{i} \\ $$$$\frac{{qE}}{{mg}}−\mathrm{1}=\frac{{v}_{\mathrm{2}} }{{v}_{\mathrm{1}} } \\ $$$$ \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *